Word break problem

Word break problem

This problem is commonly asked in the Google and Amazon interview. We all know that if you typed string in Google search box does not make sense, Google breaks that into meaningful words and asks us back if we meant those words instead of a single word. This post discusses how can we find if the given string can be broken into meaningful dictionary words. For example, if I typed algorithmsandme and given dictionary is [“algorithms”, “and”, “me”], this string is breakable in meaningful words. but if the string is algorithmsorme this is not breakable into meaningful words. You can find this problem for practice at leetcode.

Word break problem : thoughts

We start with the first character of the string, check if the character itself is a word in the dictionary? If yes, then our problem reduces to the smaller problem, that is to check if substring from index 1 to s.length is breakable or not.
If not, then we check two characters and then three characters and so on till we can check the whole string. As with every character inclusion, the problem reduces in size but remains the same, so ideal case for recursive implementation.

package AlgorithmsAndMe;

import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

public class WordBreak {

    public boolean wordBreak(String s, List<String> wordDict) {
        return wordBreakUtil(s, wordDict, 0, table);
    }

    private boolean wordBreakUtil(String s, 
                                   List<String> wordDict, 
                                   int index) {

        if (index == s.length()) return true;

        boolean isBreakable = false;
        for(int i=index; i<s.length(); i++) {
            isBreakable = isBreakable 
                   || wordDict.contains(s.substring(index, i+1))
                    && wordBreakUtil(s, wordDict, i + 1);
        }

        return isBreakable;
    }
}

If you notice we are solving the same problems again and again in recursive function wordBreakUtil, how can we save that repeated calculations? Best way to save the already solve problems in a cache, that way we can refer to the cache if the problem is already solved or not. If yes, do not solve it again and use the cached value. This approach is called a Top Down approach and uses memoization to avoid repeated subproblems.

package AlgorithmsAndMe;

import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

public class WordBreak {

    public boolean wordBreak(String s, List<String> wordDict) {
        int [] table =  new int[s.length()];
        for(int i=0; i<s.length(); i++){
            table[i] = -1;
        }
        return wordBreakUtilTopDown(s, wordDict, 0, table);
    }

    private boolean wordBreakUtilTopDown(String s, 
                            List<String> wordDict,
                            int index,
                            int[] table) {

        if (index == s.length()) return true;

        if(table[index] < 0) {
            boolean isBreakable = false;
            for (int i = index; i < s.length(); i++) {
                isBreakable = isBreakable 
                        || wordDict.contains(s.substring(index, i + 1))
                        && wordBreakUtilTopDown(s, wordDict, i + 1);
            }
            table[index] = isBreakable ? 1 : 0;
        }
        return table[index] == 1 ? true : false;
    }
  }

If you run the first solution, it will exceed the time limit on leetcode, however, the second implementation should be accepted with 4ms as the time to run. Now you can appreciate the efficiency by memoization.

Word break problem using dynamic programming

In the last two implementations, two things are evident: first, the optimal solution of a subproblem leads to the optimal solution of the original problem. Second, there are overlapping subproblems. These are two must have conditions for applying dynamic programming. We already saw the memoization and top-down approach of DP to avoid repeated solving of subproblems. How can we do it bottom up?

What if store an information if the string till index i is breakable? What will be the base case? The string before index 0 is alway breakable as empty string. So table[0] can be always true. To check if string till index i is breakable or not, we check from index 0 to index i-1 if there is any index j till which string is breakable. If yes, then we just check if substring from index j to i, that will make table[i] as true.

package AlgorithmsAndMe;

import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

public class WordBreak {

    public boolean wordBreak(String s, List<String> wordDict) {
        return wordBreakBottomUp(s, wordDict, 0, table);
    }

    private boolean wordBreakUtilBottomUp(String s, List<String> wordDict){

        if(s == null || s.length() == 0) return false;

        boolean[] table  = new boolean[s.length()+1];

        table[0] = true;
        for (int i = 1; i <= s.length(); i++) {
            for (int j = i - 1; j >= 0; j--) {
                if (table[j] && wordDict.contains(s.substring(j, i))) {
                        table[i] = true;
                    }
                }
            }
        }
        return table[s.length()];
    }
}

The time complexity of the above implementation of the word break problem is O(n2)

If you want to store all the strings which can be generated by breaking a particular word, below is the code.

package AlgorithmsAndMe;

import java.util.*;

public class WordBreak2 {

    public List<String> wordBreak(String s, List<String> wordDict) {
        Map<String, List<String>> map = new HashMap<>();
        return wordBreakUtil2(s, wordDict, map);
    }

    private List<String> wordBreakUtil2(String s,
                                        List<String> wordDict,
                                        Map<String, List<String>> map) {

        if(map.containsKey(s)){
            return map.get(s);
        }

        List<String> result = new ArrayList<String>();
        if (wordDict.contains(s)){
            result.add(s);
        }

        for(int i=1; i<=s.length(); i++) {
            String prefix = s.substring(0, i);
            if(wordDict.contains(prefix)){
                List<String> returnStringsList = wordBreakUtil2(s.substring(i), wordDict, map);

                for(String returnString :returnStringsList ){
                    result.add(prefix + " " + returnString);
                }
            }
        }
        map.put(s,result);

        return result;
    }
}

Please share if there is something is wrong or missing. If you are preparing for an interview and need any help with preparation, please reach out to us or book a free session.

LRU cache (Least Recently Used cache)

LRU cache implementation in Java

This is commonly asked question in interview, especially Microsoft and Amazon interviews. Problem statement is very simple

Implement LRU cache or Least Recently Used cache

Before going further into solution, first let’s understand what is cache?  In computer architectural terms, a cache is small buffer of pages OS maintains in-order to avoid more expensive main memory accesses.

Usually cache accesses are faster than main memory access, improving overall performance. Whenever process needs to access content from a specific memory location, it tries to search that page in cache first. If it finds page in cache, it uses it and does not access memory. It’s called cache hit.

However, caches are very small in size as compared to main memory, there is probability that page requested by process is not present in cache. In that case, new page is moved into cache and one of existing page is swapped out. When this happens, it is called cache miss.

Caching also happens at application layer too, for example, caching visited pages on browser, caching frequently accessed data from DB to in memory caches like Redis.

Based on how cache miss is handled in cache, there different type of caches like first in first out cache, least recently use cache, least frequently use cached.

In first in first out cache, in the event of cache miss, entity which came first into cache is evicted first; where as in least recently used cache, page which was used least recently gets evicted. In the same vain, least frequently used cache is where page which is least frequently used among all the pages in cache.

LRU cache implementation

Consider that you have a cache with space for additional page. If cache miss happens, we bring page from memory and put it in cache for future access.

Now, if cache is full, and cache miss happens, we have to bring in a new page and evict a page from cache. In LRU cache, this page will  be the page which accessed longest time ago.

What if a page was accessed at the start, and then accessed just before the cache miss? Is it the least recently used page? On the contrary, it is the most recently used page and hence should be last to evict. Question now is which page should we evict? In this case, page which came after the first page goes out.

If page is brought into cache first, it is first candidate for eviction, if it is not accessed again before cache miss happens.

Why queue?

In principle, LRU cache is first in first out cache with special case, that if a page is accessed again, it goes to end of eviction order. Which data structure is best to implement FIFO pattern? Of course it’s queue. So our LRU cache will be a queue where each node will store a page. This queue will have specific size as cache as limited size. Whenever a new page is brought in, it is added at the rear of queue. When eviction happens, it happens from the front of cache.

Why hash?

There is one requirement of LRU cache which does not map directly to queue data structure, which is to move a node corresponding recently accessed page to end of the queue. This poses two problems : First, how to find the node in the queue corresponding to page id being accessed? Second, how to move it to end as efficiently possible? Both problems are very similar to what we solved in first non repeated character in stream.

We will use hash which will store the node address in queue corresponding to page id. This will give us immediate access to the node to be reshuffled.

Why doubly linked list based queue?

Still problem remains to move nodes around with moving all elements of the queue. Which data structure removes an element in O(1), given the element? Doubly linked list it is. If we implement queue as doubly linked list, removing and adding pages from queue will be O(1) operation.

LRU cache algorithm

  • Cache miss happens :
    • If cache has free entry, enqueue page to queue.
    • If cache is full,  remove the page from from of queue and add new page at the end of queue.
  • Cache hit happens :
    • delete node from current location in queue.
    • Enqueue page at the end of queue.
  • If page is present in hash, it’s a cache hit, if page is not present in hash map, it’s a cache miss.

LRU cache implementation

Queue interface

package com.company;

/**
 * Created by sangar on 8.10.18.
 */
public interface Queue<E> {
    public ListNode<E> peek();
    public ListNode<E> remove();
    public ListNode<E> enqueue(E data);
    public ListNode<E> deleteNode(ListNode<E> node);
    public boolean isEmpty();
    public int size();
}

Queue implementation

package com.company;

/**
 * Created by sangar on 8.10.18.
 */
public class QueueImplementation<E> implements Queue<E>{
    ListNode<E> head;
    ListNode<E> tail;
    int size;

    public QueueImplementation(){
        head = null;
        tail = null;
        this.size = 0;
    }

    @Override
    public ListNode<E> deleteNode(ListNode<E> node){
        if(this.isEmpty()) {
            return null;
        }

        if(this.head == node){
            if(this.head.getNext() != null) this.head.getNext().setPrev(null);
            this.head = this.head.getNext();
            this.size--;
            return node;
        }

        if(this.tail == node){
            if(this.tail.getPrev() != null) this.tail.getPrev().setNext(null);
            this.tail = this.tail.getPrev();
            this.size--;
            return node;
        }
        /*
            We are deleting node in between. So following things happen
            1. If node has prev, set node.prev.next = node.next.
            2. If node has next, set node.next.prev = node.prev
        */
        if(node.getPrev() != null) node.getPrev().setNext(node.getNext());
        if(node.getNext() != null) node.getNext().setPrev(node.getPrev());

        this.size--;
        return node;
    }


    @Override
    public ListNode peek() {
        if(this.isEmpty()) {
            return null;
        }
        return this.head;
    }

    @Override
    public ListNode remove() {
        if(this.isEmpty()) {
            return null;
        }
        /*
            We are deleting node at head. So following things happen
            1. Set temporary node point to head.
            2. Move head to next of node.
            3. Set prev of new head to NULL.
            4. Free the temp node.
          */
        ListNode<E> tempNode = this.head;
        this.head.setPrev(null);
        this.head = this.head.getNext();

        this.size--;
        return tempNode;
    }

    @Override
    public ListNode enqueue(E data) {
        if(this.isEmpty()) {
            this.head = new ListNode<E>(data);
            this.tail = this.head;
            this.size++;
            return this.head;
        }
        ListNode<E> newNode = new ListNode<E>(data,null, this.tail);
        this.tail.setNext(newNode);
        this.tail = newNode;

        this.size++;
        return newNode;
    }

    @Override
    public boolean isEmpty() {
        return this.head == null;
    }

    @Override
    public int size() {
        return this.size;
    }
}

LRU cache implementation in java

package com.company;

import java.util.ArrayList;
import java.util.HashMap;

/**
 * Created by sangar on 9.10.18.
 */
public class LRUCache {
    private Queue<Integer> queue;
    private HashMap<Integer, ListNode> pages;
    private int cacheSize;

    public LRUCache(int cacheSize){
        this.cacheSize = cacheSize;
        queue = new QueueImplementation<>();
        pages = new HashMap<>();
    }

    /* This function implements the LRU cache */
    public void readPage(int pageId) {

        /*Cache Miss and we can add the page in the cache */
        if (!pages.containsKey(pageId) && queue.size() < cacheSize) {
            pages.put(pageId, queue.enqueue(pageId));
            return;
        }

        /* Cache Miss and we cannot add new page to cache,
        remove the LRU page */
        if (!pages.containsKey(pageId) && queue.size() >= cacheSize) {
            //First remove it from the queue.
            ListNode evictedPage = queue.remove();
            //Remove node from hash table
            pages.remove(evictedPage.getData());
            //Enqueue new node and add it at tail
            queue.enqueue(pageId);
            return;
        }

        /* Cache is Hit */
        if (pages.containsKey(pageId)) {
            updateCache(pageId);
        }

        return;
    }

    /* This function modifies queue when there is cache hit */
    public void updateCache(int pageId){

        /* Case where queue may be empty - defensive programing*/
        if(queue.isEmpty() && queue.size() < cacheSize){
            pages.put(pageId,queue.enqueue(pageId));
        }
        /* Update the pointers of neighbor nodes and tail in dummy node */
        else{
            ListNode node = queue.deleteNode(pages.get(pageId));
            if(node != null){
                queue.enqueue((Integer)node.getData());
            }
        }
    }

    public ArrayList<Integer> cacheState(){
        ListNode current = queue.peek();
        ArrayList<Integer> cacheState = new ArrayList<>();
        while(current != null){
            cacheState.add((Integer)current.getData());
            current = current.getNext();
        }
        return cacheState;
    }
}

Test case for LRU cache implementation

package test;

import com.company.LRUCache;
import org.junit.jupiter.api.Test;

import java.util.ArrayList;
import java.util.Arrays;

import static org.junit.jupiter.api.Assertions.assertEquals;

/**
 * Created by sangar on 23.9.18.
 */
public class LRUCacheTest {

    LRUCache tester = new LRUCache(5);

    @Test
    public void testCacheInsertion() {

        ArrayList<Integer> cacheState = new ArrayList<>(Arrays.asList(1,2,3,4,5));

        tester.readPage(1);
        tester.readPage(2);
        tester.readPage(3);
        tester.readPage(4);
        tester.readPage(5);

        assertEquals(cacheState, tester.cacheState());
    }

    @Test
    public void testCacheHit() {

        ArrayList<Integer> cacheState = new ArrayList<>(Arrays.asList(1,4,5,2,3));

        tester.readPage(1);
        tester.readPage(2);
        tester.readPage(3);
        tester.readPage(4);
        tester.readPage(5);
        tester.readPage(2);
        tester.readPage(3);

        assertEquals(cacheState, tester.cacheState());
    }

    @Test
    public void testCacheMiss() {

        ArrayList<Integer> cacheState = new ArrayList<>(Arrays.asList(2,3,4,5,6));

        tester.readPage(1);
        tester.readPage(2);
        tester.readPage(3);
        tester.readPage(4);
        tester.readPage(5);
        tester.readPage(6);

        assertEquals(cacheState, tester.cacheState());
    }

    @Test
    public void testEvictionAndInsertion() {

        ArrayList<Integer> cacheState = new ArrayList<>(Arrays.asList(3,4,5,6,1));

        tester.readPage(1);
        tester.readPage(2);
        tester.readPage(3);
        tester.readPage(4);
        tester.readPage(5);
        tester.readPage(6);
        tester.readPage(1);

        assertEquals(cacheState, tester.cacheState());
    }


    @Test
    public void testEmptyCache() {
        ArrayList<Integer> cacheState = new ArrayList<>();

        assertEquals(cacheState, tester.cacheState());
    }
}

Let’s take an example and see how whole thing works. Let’s say we have cache of size 5, which is empty to start with. Application accesses page id 1,2,3,4 and 5. As they are all cache miss to start with and there is space in cache, all these pages are brought into cache.

implement lru cache in java
Cache state after reading first 5 pages

Now, application accesses page 6. We have a cache miss. What happens? As cache is full, we evict the page which is at the front and add page 6 to it.

Page 1 is evicted, and page 6 added to queue

Application accesses page 2 again, which is already present in cache, it’s a cache hit. As page 2 is most recently used page, it has to go to end of the queue.

implement lru cache
Page 2 moves at the end

Let’s say next page accessed is page 7, it is cache miss. Evict a page from cache, which is the first node in queue (3).

lru cache implementation
New page 7 added to end after evicting page 3 from front

Insertion and removal from queue is O(1) complexity.

Please share if there is something wrong or missing. If you are interested in taking personal coaching from our expert teachers, please contact us at communications@algorithmsandme.com

Matching parenthesis problem

Matching parenthesis problem

We understood the concept of stack data structure in last post. Let’s discuss matching parenthesis problemwhich applies those concept. Problem statement is :

Given a string of parenthesis ‘(‘ and ‘)’, write a function which returns true if there are matching pairs and false if there are not. A matching pair means, there should be a closing parenthesis for an opening one, in correct order.

For example : '((()))', function should return TRUE, but ')(())' will return FALSE. Special case would be when there are equal number of parenthesis, closing and opening, but they are not in perfect order, hence function should return false in that case.

Parenthesis matching problem : Line of thoughts

For each closing parenthesis, we need a corresponding opening parenthesis. There are two possibilities : Either we find one or we do not find one. If we do not find one, we can say that string does not contain matching parenthesis.
If there is corresponding opening parenthesis, then that opening parenthesis cannot be matched with any other closing parenthesis.

Also, note that every closing parenthesis will match with the most recent opening parenthesis if there is one. That means we are looking at a order where the parenthesis which came last needs to be fetched first, typical last in first out pattern, which is best implemented by stack.

For asserting that the current closing parenthesis is in sync with what we have already seen, we just need to check if current parenthesis completes a pair with opening parenthesis we last seen. Next closing parenthesis should complete pair with the one prior to last and so on.

Parenthesis matching problem : Algorithm

  1. For each character of input string
  2. If character is opening parenthesis '(', put it on stack.
  3. If character is closing parenthesis ')'
    1. Check top of stack, if it is '(' , pop and move to next character.
    2. If it is not '(', return false
  4. After scanning the entire string, check if stack is empty. If stack is empty, return true else return false.

Matching parenthesis problem : Implementation

package com.company;

import java.util.Stack;

/**
 * Created by sangar on 21.9.18.
 */
public class ParenthesisMatch {

    public static boolean isMatchingParenthesis(String s) {
        Stack<Character> stack = new Stack<>();

        if (s == null) return true;

        int len = s.length();
        for(int i=0; i<len; i++){
            switch(s.charAt(i)){
                case '(' :
                    stack.push(s.charAt(i));
                    break;
                case ')':
                    //If stack is empty, then there is an extra closing parenthesis
                    if(stack.isEmpty()) return false;
                    stack.pop();
                    break;
                default:
                    return false;
            }
        }
        //If stack is empty that means it's a matching parenthesis
        return stack.empty();
    }

    public static void main(String[] args) {
        String s = "()))";
        System.out.println(isMatchingParenthesis(s));
    }
}

Complexity of parenthesis matching algorithm is O(N) time to scan N length string and O(N) extra space for stack.
This problem is quite simple to solve, so if you are asked this question in interview, most probably interviewer wants to understand can you think of good test cases and put your code to test against them. Below are the few test cases you can try your code on.
Test cases

1. ((())) : True
2. ())) : False
3. )((( : False
4. ( : False
5 Empty string : True
6. NULL pointer : False

Can you try solving this problem now on HackerEarth

Please share if there is something wrong or missing. If you are interested to take personal coaching from our experience teachers, please reach out to us at communications@algorithmsandme.com

Longest Common Subsequence

Longest common subseuence

A subsequence of a string is set of all the characters which are left to right order and not necessarily contiguous. For example, string ABCDEG has ABC, ADG, EG, BCDEG subsequences; whereas BDA is not a subsequence of the given string, even though all the characters are present in the string, they do not appear in the same order.

longest common subsequence lcs

Given two strings X and Y, find longest common subsequence (LCS) Z. For example, X = ABCDSEFGD Y = ACFEFXVGAB; LCS Z would be ACEFG.

Longest common subsequence: line of thoughts

First of all, notice that it is an optimization problem, it is a hint that it may be a dynamic programming problem but we are not sure yet.

Let’s say that the length of the string 1 and the string of 2 are N and M. Can I know the longest common subsequence in length N and M if I already know the LCS in N-1 and M-1? The direct question is can I divide the original problem into subproblems and solve those subproblems to get the answer for original problem? In this case, the answer is yes. (This is the second hint towards dynamic programming application, optimal subproblem substructure).

How can we divide the problem into subproblems? The length of X is N and length of Y as M. Start from the end of both strings. Check if X[N] == Y[M]. If yes, the problem reduces to find the longest common subsequence in X[1..N-1] and Y[1..M-1].

What if they are not equal? Then one by one we have to exclude character from string X and Y. Why?
First, we exclude the character from the X and find LCS in remaining characters of X and all the characters of Y. The problem reduces to X[1..N-1] and Y[1..M]. Next, we exclude a character from Y, the problem reduces to X[1..N] and Y[1..M-1]. Any of the two choices can give us the longest common subsequence, so we would take maximum from both the cases.

LCS(i,j)  =  1 + LCS[i-1, j-1] if X[i] == Y[j]
  =   MAX (LCS[i-1,j], LCS[i, j-1]) if X[i] != Y[j]
=   0 if i or j is 0

Interesting to see why LCS(i,j) is 0 when either i or j is 0? Because the longest common subsequence in two strings, when one string is empty is 0.

Can we implement the recursive function?

    public int longestCommonSubsequence(String s1, String s2, int i, int j){

        //If any of the string is nul return 0
        if(s1 == null || s2 == null) return 0;

        //We reached at the end of one of the string
        if(i == s1.length() ||  j == s2.length())
            return 0;

        if(s1.charAt(i) ==  s2.charAt(j)){
            return  1 + longestCommonSubsequence(s1, s2, i+1, j+1);
        }

        return Integer.max(longestCommonSubsequence(s1, s2, i+1, j),
                longestCommonSubsequence(s1, s2, i, j+1));

If we follow the execution cycle of the above code, we will see something like below

longest common subsequence lcs

It is evident from the partial tree that there are some problems which are solved again and again. This is the third hint (overlapping subproblems) that we can apply dynamic programming.

It will be more evident if you implement the recursive function with reverse traversal of the strings. In that implementation, the base case will be when one of the string is empty, and at that point, LCS of two strings will be 0. If we take a two dimensional table such that T[i][j] represents longest common subsequence till ith and jth characters of string S1 and S2 then T[0][i] = 0 and T[i][0] = 0.

T[i][j] = T[i-1][j-1] + 1 if X[i] = Y[j]
T[i][j] = max(T[i-1][j], T[i][j-1]) if X[i] != Y[j]

Dynamic programming implementation of LCS

package com.company;

/**
 * Created by sangar on 4.2.19.
 */
public class LongestCommonSubsequence {

    public int longestCommonSubsequenceDP(String s1, String s2){

        //If any of the string is nul return 0
        if(s1 == null || s2 == null) return 0;

        int len1 = s1.length();
        int len2 = s2.length();

        int[][] table = new int[len1+1][len2+1];

        for (int i=0; i<=len1; i++){
            for (int j=0; j<=len2; j++) {
                if (j == 0 || i == 0) {
                    table[i][j] =  0;
                }

                else if (s1.charAt(i-1) == s2.charAt(j-1)) {
                    table[i][j] = 1 + table[i - 1][j - 1];
                } else {
                    table[i][j] = Integer.max(table[i - 1][j], table[i][j - 1]);
                }
            }
        }

        return table[len1][len2];
    }
}

Above implementation has time and space complexity of O(n2). Please share if there is anything wrong or missing.

What is dynamic programming?

What is Dynamic Programming or DP

Dynamic programming is an approach to solve a larger problem with the help of the results of smaller subproblems. It is a technique used to avoid computing multiple time the same subproblem in a recursive algorithm. I find a lot of students asking me question around, how do I know this problem is a dynamic programming problem? There is a definite way to arrive at the conclusion if a problem is a dynamic programming problem or not?

The first thing I would recommend you to read before going down is this beautiful explanation of dynamic programming to 4 years old.

The first thing you will notice about dynamic programming problems (not all problems) is they are optimization problem. Either it will be finding minimum or maximum of some entity. For example, find minimum edit between two strings or find longest common subsequence etc. However, problems like Fibonacci series are not exactly like an optimization problem, these are more like Combinatorial problems. Still, this can be a good hint that a problem can be a DP problem.

Second, you will notice that the problem can be divided into a pattern like an fn(n) = C + fn(n-k) where k can be anything between 1 and n.
This property is called optimum subproblem structure, where an optimum solution to the subproblem leads to the optimum solution to the larger problem.
Once you get the equation, it is very easy to come with a recursive solution to the problem. I would advise you to write the recursive solution and try to calculate the complexity of the solution. It will exponential in big-O notation.

Then why did recursion work so well with a divide and conquer approach? The key point is that in divide and conquer, a problem is expressed in terms of subproblems that are substantially smaller, say half the size. For instance, mergesort sorts an array of size n by recursively sorting two subarrays of size n/2. Because of this sharp drop in problem size, the full recursion tree has only logarithmic depth and a polynomial number of nodes. In contrast, in a typical dynamic programming formulation, a problem is reduced to subproblems that are only slightly smaller than the original. For instance, fn(j) relies on fn(j − 1). Thus the full recursion tree generally has polynomial depth and an exponential number of nodes.
However, it turns out that most of these nodes are repeats, that there are not too many distinct subproblems among them. Efficiency is therefore obtained by explicitly enumerating the distinct subproblems and solving them in the right order.
Reference

This will lead us to the third property, which is overlapping subproblems. Once, you draw the execution tree of the recursive solution of the problem, it will appear that a lot of problems are being solved again and again at different levels of recursion.

The intuition behind dynamic programming is that we trade space for time, i.e. to say that instead of calculating all the subproblems taking a lot of time but no space, we take up space to store the results of all the subproblems to save time later. The typical characteristics of a dynamic programming problem are optimization problems, optimal substructure property, overlapping subproblems, trade space for time, implementation via bottom-up/memoization.

Dynamic programming in action

Enough of theory, let’s take an example and see how dynamic programming works on real problems. I will take a very commonly used but most effective problem to explain DP in action. Problem is known as the Fibonacci series. Fibonacci series is a series of integers where each integer is the sum of previous two integers. For example, 1,1,2,3,5,8,13,17 is a Fibonacci series of eight integers. Now, the question is given a number n, output the integer which will be at the nth integer in Fibonacci series. For example for n = 4, the output should be 3 and for n=6, it should 8.

First hint: It is a combinatorial problem, so maybe a DP problem. Second, it is already given in the problem that current integer depends on the sum of previous two integers, that means f(n) = f(n-1) + f(n-2). This implies that the solution to subproblems will lead to a solution to the bigger problem which is optimal substructure property.

Next step is to implement the recursive function.

 public int fibonacci (int n) {
    if (n < 2) //base case
        return 1;

    return fibonacci(n-1) + fibonacci(n-2);
 }

Great, next step is to draw the execution tree of this function. It looks like below for n = 6. It is apparent how many times the same problem is solved at different levels.

what is dynamic programming
Recursive tree of Fibonacci series function

So, now we know three things about the Fibonacci problem: It is combinatorial problem, there is optimal substructure and there are overlapping subproblems. As in dynamic programming, we side with more space than time, we will try to use extra space to avoid recalculating subproblems.

The first way is to use a case, which stores the value of fab(n) if it is already calculated. This is called memoization or top-down approach.

Map<Integer, Integer> cache = new HashMap<>();

public int fibonacci(int n){
    if (n == 0)
       return 0;
    if (n == 1)
        return 1;

    if(cache.containsKey(n))
        return cache.get(n);

    cache.put(n, fibonacci(n - 1) + fibonacci(n - 2));

    return cache.get(n);
}

Another approach is bottom up, where the smaller problems are solved in an order which helps us with solving bigger problems. Here also, we use memoization but in a different way. We store the solution of smaller subproblems and directly use this to build the solution.

int[] fib = new int[n];
fib[0] = fib[1] = 1;
public int fibonacci(int n){
   for(int i=2; i<=n; i++){
       fib[n] = fib[n-1] + fib[n-2];
   }
   return fib[n];
}

Above solution requires extra O(n) space, however, the time complexity is also reduced to O(n) with each subproblem solved only once.

Follow longest increasing subsequence problem, how we have applied the same pattern while we solved the problem.

Final thoughts
Where to apply dynamic programming : If you solution is based on optimal substructure and overlapping sub problem then in that case using the earlier calculated value will be useful so you do not have to recompute it. It is bottom up approach. Suppose you need to calculate fib(n) in that case all you need to do is add the previous calculated value of fib(n-1) and fib(n-2)

Recursion : Basically subdividing you problem into smaller part to solve it with ease but keep it in mind it does not avoid re computation if we have same value calculated previously in other recursion call.

Memoization : Basically storing the old calculated recursion value in table is known as memoization which will avoid re-computation if its already been calculated by some previous call so any value will be calculated once. So before calculating we check whether this value has already been calculated or not if already calculated then we return the same from table instead of recomputing. It is also top down approach.
Reference: Answer by Endeavour

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0/1 knapsack problem using dynamic programming

0/1 Knapsack problem

0/1 Knapsack is typical problem which is used to demonstrate application of greedy algorithm as well as dynamic programming. There are cases when applying greedy algorithm does not give optimal solution. There are many flavors in which Knapsack problem can be asked.  

1. A thief enters a museum and want to steal artifacts from there. Every artifact has a weight and value associated with it. Thief carries a knapsack (bag) which can take only a specific weight. Problem is to find the combination of artifacts thief steals so that he gets maximum value and weight of all taken artifacts is less capacity of  knapsack he has. Thief cannot take any artifact partially. Either he takes it or leaves it. Hence the problem is 0/1 knapsack.

2. We have N files each having a size say Si. We have a total storage capacity of W bytes. For each file to be stored re-computation cost is Vi. Problem is to store as many files on storage that combined size of all files is less than W and their re-computation value is maximum. We can either store or leave a file, we cannot store partial file. Hence this is a case of 0/1 knapsack problem.

0/1 knapsack problem : Line of thoughts

Brute force method would try all subsets of set of items, whose weight adds up to maximum capacity of knapsack and see which one gives maximum value. Complexity of brute force algorithm would be of exponential order as there will be 2n possible subsets for n items.

Can we do better?  If we consider each item, there are two possibilities associated with it.
First, current item is included in optimal subset. Then we need to find out all the items in remaining N-1 items which can optimize the subproblem for weight W-wk. Value of this item is added to candidate maximum value.

Second, current item is not included in optimal subset. In that case, we need to find out items in remaining N-1 items which can optimize the the original problem. Value of current item is not added into candidate maximum value.

Inclusion depends on two conditions :

  1. Weight of the item is less than the total capacity of knapsack.
  2. Inclusion of item increases current max value with K-1 items with W-Wk weight.  

Since every steps reduces the problem to a smaller problem in terms of items of weight, recursive solution would be our first refuge. To implement this problem, what are the base cases? First, we cannot add any items to knapsack capacity is zero i.e. W == 0. Second, no item can be taken if there are no items remaining, i.e. n == 0.

Recursive implementation of 0/1 knapsack problem

package com.company;
/**
	* Created by sangar on 19.8.18.
*/
public class KnapsackProblem {
	static int knapSack(int W, int[] weights, int[] val, int n) {
		/*
			If there is no item or weight that can be carried is
			zero, return zero
		*/
		if (n &lt; 0 || W == 0)
			return 0;

		/* 
			If weight of the nth item is more than Knapsack 
			capacity W,then this item cannot be included
			in the optimal solution
		*/
		if (weights[n] &gt; W)
			return knapSack(W, weights, val, n - 1);

		/* Consider two cases, including item and excluding item.*/
		else return Integer.max(
			(val[n]
				+ knapSack(W - weights[n], weights, val, n - 1)),
			(knapSack(W, weights, val, n - 1))
		);
	}

	public static void main(String args[]) {
		int[] val = {60, 100, 120};
		int[] wt = {10, 20, 30};
		int W = 50;
	
		int n = val.length;
		System.out.println(knapSack(W, wt, val, n - 1));
	}
}

If we look at the execution trace of the function, it looks like this.

0/1 knapsack problem

There are seven problems to be solved at the leaf level. For N = 3, there are 7 problems to be solved before we start optimizing for the max value. For N, in general, it will take 2N subproblems to be solved. Hence, complexity of recursive implementation is O(2N).

If we take another example, it will become evident that there are some subproblems which are solved again and again. Overlapping subproblems is one of the criteria, we should be thinking about dynamic programming. Also, notice that optimal solution to a smaller problem leads to optimal solution to bigger problem, which is second condition for DP.  This problem satisfy both these conditions, hence let’s design DP solution for it.

0/1 knapsack problem : Dynamic programming approach

We define two dimensional array V[N,W] where N is number of items and W is capacity. For 1<= i <= n and  0<=w<=W, V[i,w] represents the optimal solution for items I1, I2, ..In with maximum weight of w.  If we can compute all the entries of this array, then the array entry V[N,W] is the solution to our problem

For i =0 and w=0, all values will be zero. So, first column and first row will be filled with all zero values.
Recursively, we can fill the table bottom up as follows.

V[i, w ] = max (V[i-1, w], V[i-1, w-w[i]) + V[i] )
V[0, w ] = 0; there are no items
V[i, 0 ] = 0; no items can be picked.
package com.company;

/**
	* Created by sangar on 19.8.18.
*/
public class KnapsackProblem {
	public static int knapsackDP(int W, int[] weights, int[] val, int n) {
		int[][] V = new int[n+1][W + 1];
		for(int i = 1 ; i &lt; V[0].length; i++){
			/*
				If weight of item is less than current value
				we can achieve minimum value V[i] with 1..i items
			*/
			if(weights[0] &lte; i){
				V[0][i] = val[0];
			}else{
				V[0][i] = 0;
			}
		}

		//Loop for all items
		for (int i = 1; i &lt; V.length; i++) {
			for (int j = 1; j &lt; V[i].length; j++) {
				/*if a weight is more than the allowed weight, 
				that weight cannot be picked. */
				if(weights[i] &gt; j){
					V[i][j] = V[i-1][j];
				}else{
					V[i][j] = Math.max(V[i-1][j], 
							val[i] + V[i-1][j-weights[i]]);
				}
			}
		}
		return V[V.length-1][W];
	}

	public static void main(String args[]) {
		int[] val = {60, 100, 120};
		int[] wt = {10, 20, 30};
		int W = 50;

		int n = val.length;
		System.out.println(knapsackDP(W, wt, val, n - 1));
	}
}

One similar problem which can be solved with same approach is minimum number of coins to be used to get change of a particular amount. I am skipping the whole analysis and directly pasting the code here.  

Complexity of the dynamic programming implementation of knapsack problem is O(N *W). Space complexity is again O(N*W). It is thumb rule that we trade space for time in dynamic programming.

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Longest increasing subsequence : Dynamic Programming

Longest increasing subsequence

Given an array of integers, find the longest increasing subsequence i.e a subsequence such that every element of subsequence satisfy this condition i < j  and a[i] < a[j].For example, in array {2,4,6,3,5,7,9} longest increasing subsequence is of length 5  = {2,4,6,7,9} longest increasing subsequence
More examples :

Input  : arr[] = {3, 10, 2, 1, 20}
Output : Length of LIS = 3
The longest increasing subsequence is 3, 10, 20

Input  : arr[] = {3, 2}
Output : Length of LIS = 1
The longest increasing subsequences are {3} and {2}

Input : arr[] = {50, 3, 10, 7, 40, 80}
Output : Length of LIS = 4
The longest increasing subsequence is {3, 7, 40, 80}

Longest common subsequence : Line of Thoughts

We have to find longest increasing subsequence till last element of array, i.e Nth element, question is does depend on LIS till N-1 element? Idea is to see if any increasing subsequence already present till current index i,  can include  A[i] and still remain increasing? To confirm that, check every element at index j such that j>=0 and j < i and A[j] < A[i].
If element A[j] is less than A[i], then A[i] can be part of increasing subsequence ending with element j. Length of such increasing subsequence can be (length of increasing subsequence ending at j )+1. Check for each such element and take maximum length.Let’s see an example and see how it works.

Before we start solution,, let’s think why are we applying dynamic programming here. First because, solution to bigger problem depends on optimal solution of subproblems and second, because if we do not store solutions to subproblems we may end up calculating them again and again. Both Optimal subproblem property and overlapping subproblem property are satisfied for this problem and hence we will use dynamic programming to solve it.

Longest increasing subsequence : Implementation

Define an array LIS of size N, LIS[i] will represent longest increasing subsequence length till element i.

LIS[i] = 1 + max(LIS[j]) for all  0<=j<i and A[j]<A[i] 
       = 1 if no such element exists where j< i and A[j]<A[i]

Below is C code to implement it.

#include <stdio.h>
#include <stdlib.h>

int maximumLIS(int a[], int end, int *lis){
    for (int i=0; i<end; i++){
        if( a[i] < a[end] && lis[i] > lis[end] )
            lis[end] = lis[i];
    }
    return lis[end];
}

int lis(int a[], int size){

    int *lis = (int *)malloc(sizeof(int)*size);
    lis[0] = 1;

    for(int i=1; i<size; i++){
    	lis[i] = 1 + maximumLIS(a,i,lis);
    }
    
    int result = lis[size-1];
    free(lis);
    
    return result;
}

int main(void) {
	int a[] = { 2,4,6,3,5,7,9 };
	int size = sizeof(a)/sizeof(a[0]);

	printf("Length of Longest increasing subsquence : %d" , lis(a, size));
	
	return 0;
}

Java implementation for the same algorithm

package com.company;

/**
 * Created by sangar on 7.1.18.
 */
public class LIS {

    private static int maximumLIS(int a[], int end, int [] LIS){
        for (int i=0; i<end; i++){
            if( a[i] < a[end] && LIS[i] > LIS[end] )
            LIS[end] = LIS[i];
        }
        return LIS[end];
    }

    private static int lis(int [] A){

        int [] LIS = new int[A.length];
        LIS[0] = 1;

        for(int i=1; i<A.length; i++){
            LIS[i] = 1 + maximumLIS(A,i,LIS);
        }

        return LIS[A.length - 1];
    }
    public static void main(String[] args) {
        int[] A = {2,4,6,3,5,7,9};
        System.out.println("Longest increasing sybsequence : " + lis(A));
    }
}

Let’s take an example and see how this code works? For example, given array A = {2,4,6,3,5,7,9}

Initialize LIS[0] =1, that means there is an increasing subsequence of length 1 at index 0.
For i = 1 i.e 4, check for j=0 to 1, excluding index 1. A[0] < A[1], hence LIS length is 2 (LIS[0] + 1 ).

For i = 2, i.e. 6 , check j = 0 to 2.  and check that LIS[0] = 1 and LIS[1] =2. Max LIS[j] for j=0 to  2 is 2. LIS[2] = 3 (LIS[1] +1).
For i =3 i.e 3, check from 0 to 3, Max LIS till index 3 will be LIS[3] = 2 because only A[0] is less than A[3]. Hence longest subsequence ending with i = 3 will have length only of 2.  LIS[3] = 2
For i = 4, i.e.5 Max LIS is again 3 which includes {2,4,5} or {2,3,5}
For i = 5, i.e 7, Max LIS is 4 which includes {2,4,5,7} or {2,3,5,7} or {2,4,6,7}
For i = 6, i.e 9, Max LIS is 5 which includes {2,4,5,7,9} or {2,3,5,7,9} or {2,4,6,7,9}

Therefore, longest increasing subsequence is 6 for given array.

Other problems which are variance of longest increasing subsequence and can be solved by finding longest increasing subsequence are :

1. Given two river banks (visualization : two parallel lines), one bank has numbers written (1….n) in sorted order. On the other bank the numbers (1…n) are arranged randomly. A bridge can be formed from the ith point from bank 1 to ith point in bank 2. Find the max number of non-intersecting bridges you can form?
Just find longest increasing subsequence in non ordered number and that will be the solution.

2. Given a set of n types of rectangular 3-D boxes, where the ith box has height h(i), width w(i) and depth d(i) (all real numbers). You want to create a stack of boxes which is as tall as possible, but you can only stack a box on top of another box if the dimensions of the 2-D base of the lower box are each strictly larger than those of the 2-D base of the higher box.

3. Another problem which borrow heavily from approach is find longest zigzag subsequence in an array

Algorithm to find longest increasing subsequence works in O(N^2) in time complexity with O(N) space complexity.

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Longest palindrome substring

Longest palindrome substring

Given a string, find the longest palindrome substring in that string. For example, in the string given below, the longest palindrome substring is DCBABCD

longest palindrome substring dp

longest palindrome substring dynamic programming

Brute force solution to the problem is pretty easy. Idea is to start from each character in the string and go on checking on the left and right side of that character until they are same. Once the left and right side characters differ, we check if the number of characters in substring centered character is greater than earlier such substring for any other character processed earlier. If it is greater, then we update the length and look for subsequent characters till the end of the string.

Implementation

    private String longestPalindrome(String s) {

        int longest = Integer.MIN_VALUE;
        int start = 0; int end = -1;
        
        for(int i=0; i<s.length(); i++){
            int currentLength = 1;
            for(int j= i-1, k= i+1; j>=0 && k < s.length() 
                              && s.charAt(j) == s.charAt(k); j--, k++){
                    currentLength+= 2;
            }
            if(currentLength >= longest) {
                longest = currentLength;
                start = i - currentLength / 2;
                end = i + currentLength / 2;
            }
        }

        for(int i=0; i<s.length(); i++){
            int currentLength = 0;
            for(int j= i, k= i+1; j>=0 && k < s.length()
                          && s.charAt(j) == s.charAt(k); j--, k++){
                    currentLength+=  2;
            }

            if(currentLength >= longest) {
                start = (i - currentLength/2) + 1;
                end = (i + currentLength /2 );
                longest = currentLength;
            }
        }

        return s.substring(start,end+1);
    }

The complexity of the implementation is O(n2).

Longest palindrome string: a dynamic programming approach

To apply dynamic programming to any problem, there are two conditions which must satisfy: First, the optimal subproblem property meaning that solution to smaller subproblems leads to the solution to the larger problem. Second, overlapping subproblem substructure meaning there are subproblems which are solved again and again in a recursive way.

The first property gives us the recurrence relationship, whereas the second property points us towards memoization.

Let’s take an example and see if we can come up with an algorithm. We have to find if a string of length 4 is a palindrome or not. To string to be a palindrome, it’s first and the last character should be the same. There are two paths here, either the first and last characters are same or they are not.

If the first and last characters are not the same, we can safely say that string is not a palindrome. If first and last characters are same, can we say the string is a palindrome? No, not yet. We have to check if the substring from the second character to the second last character is a palindrome or not.

longest palindrome substring

This is optimum subproblems property here. If substring s[i+1..j-1] is a palindrome, then string s[i..j] is palindrome if and only if s[i] = s[j].

Now, substring s[i+1..j-1] will be calculated for all the indices k where 0<=k<=i and for the indices k where j <=k<=length. This hints at overlapping substructure property. Can we precalculate this stuff before and save it in memory?
We can store this information in a matrix M, where M[i][j] stores if a substring starting at i and length j is palindrome or not. Notice that meaning of j is length and not an index in original string.

M[i][1] will be always true as a character is a palindrome in itself. M[i][2] is true only if s[i]=S[i+1]

For M[i][3], the first character of the substring will be s[i], the last character will be s[i+3-1], if s[i] = s[i+2] and if M[i+1][2] is true (substring starting at i+1 and length 2 is palindrome or not), then only M[i][3] can be true.

We fill the table bottom up, starting with substrings with length 1 and 2 from each character in the string.

M[i][j] = (s[i] == [j+i-1] && M[i+1][j-2]) for i = 3 to n-j+1 and j = 1 to n

Longest palindrome substring: DP implementation

class Solution {
    public String longestPalindromeSubstring(String s) {
        if(s.length() == 0 ) return "";

        int longest = 0;
        int start = 0;
        int end = 0;
        boolean[][] table = new boolean[s.length()][s.length()];
        table[0][0] = true;

        for (int i = 1; i < s.length(); i++) {
            //All single characters are palindrome in itself
            table[i][i] = true;
   
            //All two characters are palindrome if two characters are equal
            table[i - 1][i] = s.charAt(i - 1) == s.charAt(i);
            if(table[i-1][i] && longest <= 2){
                longest = 2;
                start = i-1;
                end = i;
            }

        }

        for (int L = 3; L <= s.length(); L++) {
            for (int i = 0; i <= s.length()-L; i++) {
                int j = i + L - 1;
                table[i][j] = table[i + 1][j - 1] 
                               && (s.charAt(i) == s.charAt(j));

                if(table[i][j] && longest <= L){
                    longest = L;
                    start = i;
                    end = j;
                }
            }
        }

        return s.substring(start, end+1);

    }
}

Test cases

package test;

import com.company.LongestPalindromeSubstring;
import org.junit.Test;
import static org.junit.jupiter.api.Assertions.assertEquals;
/**
 * Created by sangar on 2.1.19.
 */
public class LongestPalindromeSubstringTest {

    LongestPalindromeSubstring tester = new LongestPalindromeSubstring();

    @Test
    public void longestPalindromeSubstringTest() {
        assertEquals(4, tester.longestPalindromeSubstring("ABBA"));
    }

    @Test
    public void longestPalindromeSubstringOneCharTest() {
        assertEquals(1, tester.longestPalindromeSubstring("ACDB"));
    }

    @Test
    public void longestPalindromeSubstringInMidTest() {
        assertEquals(4, tester.longestPalindromeSubstring("CABBAD"));
    }

    @Test
    public void longestPalindromeSubstringOddLenTest() {
        assertEquals(5, tester.longestPalindromeSubstring("ABCBA"));
    }
}

The complexity of the dynamic programming approach is much better than brute force algorithm and is O(n2), It used extra memory though in order O(n2).

You can test this solution on the leetcode problem

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Merge k sorted arrays

Merge k sorted arrays

Given k sorted arrays each having n elements, merge k sorted arrays into one n*k element array in sorted order. For example, given 3 arrays are as below

merge k sorted arrays
merge k sorted arrays

Result array should be like

Merge k sorted arrays

Merge k sorted arrays

Since all the input arrays are sorted, the first element in result array will be among the first elements of input arrays. How can we find the minimum among all the elements plucked from the first index of each array ? Easy, take those k elements (there are k arrays, so k first elements) and build a min heap. The root of the min heap the least element among the first elements of all arrays, so it will be the first element in the result array.

Once, we add the first element into the result array, we have to find the second element. Second element can be from the set of first elements of all input arrays except one array from which the first element of result array was added. So, we will take second element of that array.

In order to know which array gave the minimum element at a particular time, we will store additional information of about array and index at which minimum element was.

If i represents the array number, and j represents the index of the minimum number in heap in ith array, then we add (j+1)th element to the min heap next and re-heapify. If j goes out of bound for ith array, we take min heap with k-1 size and go on, till we have no elements left in heap.

Follow the procedure for (n-1)*k times. When all array elements are processed, result array will be in the sorted array.

Merge k sorted arrays: algorithm

  • Build min heap with the first element of all k arrays.
  • Pick the root of min element and put it in the result array.
  • If there are remaining elements in the array,  put next element at the root of min heap and heapify again
  • If all elements are already of an array are processed, reduce the size of min heap by 1.
  • Repeat step 2, 3 and 4 till min heap is empty.

Merge k sorted arrays: implementation

package com.company;

import java.util.PriorityQueue;

/**
 * Created by sangar on 2.12.18.
 */
public class MergeKSortedArrays {
    private class HeapNode{
        public int arrayNum;
        public int index;
        public int value;

        public HeapNode(int arrayNum, int index, int value){
            this.arrayNum = arrayNum;
            this.index = index;
            this.value = value;
        }
    }

    public int [] mergeKSortedArrays(int[][] arrays){

        if(arrays == null) return null;

        PriorityQueue<HeapNode> minHeap =
			new PriorityQueue<>(arrays.length,
                (HeapNode a,HeapNode b)-> a.value - b.value);

        int size = 0;
        for(int i =0; i<arrays.length; i++){
            size += arrays[i].length;
        }
        int[] result = new int[size]; // k * n

        //add first elements in the array to this heap
        for(int i=0; i<arrays.length; i++){
            minHeap.add(new HeapNode(i, 0, arrays[i][0]));
        }

        //Complexity O(n * k * log k)
        for(int i=0; i< size; i++){
            //Take the minimum value and put into result
            HeapNode node = minHeap.poll();

            if(node != null){
                result[i] = node.value;
                if(node.index + 1 < arrays[node.arrayNum].length) {
                    //Complexity of O(log k)
                    minHeap.add(new HeapNode(node.arrayNum,
                           node.index + 1,
                           arrays[node.arrayNum][node.index + 1]));
                }
            }
        }
        return result;
    }
}

Test cases

package test;

import com.company.MergeKSortedArrays;
import org.junit.jupiter.api.Test;

import java.util.Arrays;

import static org.junit.jupiter.api.Assertions.assertEquals;

/**
 * Created by sangar on 23.9.18.
 */
public class MergeKSortedArraysTest {

    MergeKSortedArrays tester = new MergeKSortedArrays();

    @Test
    public void mergeKSortedArraysTest() {

        int[][] input  ={
            { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }
        };

        int[] expectedOutput = {1,2,3,4,5,6,7,8,9,10,11,12};

        int [] output = tester.mergeKSortedArrays(input);

        System.out.println(Arrays.toString(output));
        assertEquals(Arrays.toString(expectedOutput), 
					Arrays.toString(output));
    }

    @Test
    public void mergeKSortedArraysWithUnequalSizeTest() {

        int[][] input  ={
                { 1, 2 }, { 5, 6, 7}, { 9, 10, 11, 12 }
        };

        int[] expectedOutput = {1,2,5,6,7,9,10,11,12};

        int [] output = tester.mergeKSortedArrays(input);

        System.out.println(Arrays.toString(output));
        assertEquals(Arrays.toString(expectedOutput),
			Arrays.toString(output));
    }

    @Test
    public void mergeKSortedArraysWithNullTest() {

        int [] output = tester.mergeKSortedArrays(null);

        assertEquals(null, output);
    }
}

Complexity of code to merge k sorted arrays is O(n * k * log k) along with space complexity of O(k).

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Merge two sorted linked lists

Merge two sorted linked lists

Problem statement is simple : Merge two sorted linked lists, without using extra space. To refer to the basics of linked list, please follow the post : Linked list data structure. This problem is commonly asked in telephonic round of Amazon and Microsoft. Let’s take an example and understand what is required as solution. Given two linked lists as following,

merge two sorted linked lists
Two sorted linked lists

Result should be like this:

merge two sorted linked list
Resultant linked list.

Merge two sorted linked lists : Thoughts

Consider following two steps to merge sorted linked lists. First, figure out which node should be head of result list. Compare head nodes of two give lists, which ever is smaller, that should be the head of result list.

Second, compare two nodes, one from each list and decide which should go next in result linked list.  Advance the pointer to next node of the node which is added to result list.

As no new node is allocated during this merge, we have to make sure that all the references are maintained when nodes are added to merged linked list.

merge two sorted linked list
Two sorted list to be merged

We can start with one list as merge list and add nodes from second list at appropriate place in that list. Let’s say L1 is our merged list and we always compare node on L2 to see if it should be placed in L1 at current position. L1 grows as more nodes are sorted in merge list.

We compare first two nodes L1 and L2, and decide that node(2) has to go in merged list as head. If it was head of L2, we would have swapped L1 and L2 heads and still L1 will be head of merged list. Why? Because we want that L1 always points to last node in merged list and L1 to represent sorted merged list till this point and L2 switches between two input lists.

As L1 always points to the last node of merged linked list, next node to compare should be L1.next i.e node(4) and L2 i.e node(3).

As L1 follows the merged linked list, we will move L1.next to point node(3), however doing it directly will lead to lose of entire linked list following it. So we do it in four steps : store L1 next as temp; link L2 to L1.next; L2 points to temp and then move L1 to L1.next

Node temp = L1.next;
L1.next = L2;
L2 = temp;
L1 = L1.next
merge two sorted linked lists

Next nodes to be compared are node(5), which is L1.next and node(5) which is L2.

Comparing node 4 and 5 to add in sorted merge list

Of course node(4) has to be added to merged linked list, what should we do? First save L1.next in temp, temp now points to node(5). Second, point L1.next to L2, point L2 to temp, and at last, L1 moves to L1.next. State of two sorted linked lists looks as follows.

merge two sorted linked lists

By this time you must have noticed that L1 and L2 do not point to the one list all the time, L1 always points to the last node of merged list and L2 points to first node of separated list which needs to be merged.

Now, L1.next which is node(7) and L2 which is node(5) will be compared.

node(5) is to be added in merged sorted list. Again same set of steps. L1.next stored as temp, L1.next points to L2 i.e. node(5) and then L2 points to temp i.e. node(7)

merge two sorted linked lists

Again, node(9) which is L1.next will be compared to L2 i.e node(7). L1.next should point to L2. Final state will be as follows

At this point, L1.next i.e node(8) is less than L2, this is simple case, where we just move L1 to L1.next and L2 remains as is.

merge two sorted linked lists

Next two nodes follow the same pattern and added to merged sorted linked list.

At this point, special condition occurs which is L1.next is null. In this case, point L1.next to L2 and two linked lists are merged.

Two sorted linked lists are merged into a sorted list

Merge two sorted linked lists : Implementation

#include <stdio.h>
#include <stdlib.h>

typedef struct node{
    int data;
    struct node *next;
} Node;
 
Node * createNode(int val){
  Node * temp = (Node *)malloc(sizeof(Node));
  if(temp){
    temp->data = val;
    temp->next = NULL;
  }
  return temp;
}
/* This function inserts node at the head of linked list */
void push(Node **headRef, int data){
	Node * newNode  = createNode(data);
	newNode->next = *headRef;
	*headRef  = newNode;
}

void printList(Node * head){
    while(head){
        printf("%d->" , head->data );
        head = head->next;
    }

    printf("NULL");
    printf("\n");
}
Node * MergeLists(Node *list1, Node *list2) {
  if (!list1) return list2;
  if (!list2) return list1;

  Node *head;
	//Chosing head of merged list
  if (list1->data < list2->data) {
    head = list1;
  } else {
    head = list2;
    list2 = list1;
    list1 = head;
  }
	
  while(list1->next && list2) {
    if (list1->next->data > list2->data) {
	//Step 1. Save the next pointer
      Node *tmp = list1->next;
	//Step 2. Change next pointer to point L2
      list1->next = list2;
	//Step 3. Move L2 to temp
      list2 = tmp;
    }
	//Step 4. Move L1 ahead
    list1 = list1->next;
  } 
  if (!list1->next) list1->next = list2;
  return head;
}
int main(){
        Node * L1 = NULL;
        Node * L2 = NULL;
        Node * result = NULL;
        int carry = 0 ;
        /* creating list 1 */
        push(&L1,7);
        push(&L1,6);
        push(&L1,4);
        push(&L1,3);
	
        /* creating list 2 */
        push(&L2,10);
        push(&L2,8);
        push(&L2,1);
 
        L1 = MergeLists(L1,L2); 
        printList(L1);
 
        return 0;
}

Complexity of this method to merge two sorted lists into one is O(n+m) where n and m are number of nodes in two sorted linked lists.

Recursive implementation to merge two sorted linked lists

#include<stdlib.h>
#include<stdio.h>
 
typedef struct node{
    int data;
    struct node *next;
} Node;
 
Node * mergeSort(Node *a, Node *b){
    Node *result = NULL;
    if(a ==  NULL)
        return b;
    else if(b == NULL)
        return a;

    /* For the first node, we would set the result to either a or b */
      if(a->data <= b->data){
         result = a;
        /* Result's next will point to smaller one in lists 
           starting at a->next  and b */
         result->next = mergeSort(a->next,b);
      }
      else {
        result = b;
       /*Result's next will point to smaller one in lists 
         starting at a and b->next */
        result->next = mergeSort(a,b->next);
      }
      return result;
}

Node * createNode(int val){
  Node * temp = (Node *)malloc(sizeof(Node));
  if(temp){
    temp->data = val;
    temp->next = NULL;
  }
  return temp;
}
/* This function inserts node at the head of linked list */
void push(Node **headRef, int data){
	Node * newNode  = createNode(data);
	newNode->next = *headRef;
	*headRef  = newNode;
}

void printList(Node * head){
    while(head){
        printf("%d->" , head->data );
        head = head->next;
    }

    printf("NULL");
    printf("\n");
}

/* Driver program to run above code */
int main(){
        Node * L1 = NULL;
        Node * L2 = NULL;
        Node * result = NULL;
        int carry = 0 ;
        /* creating list 1 */
        push(&L1,7);
        push(&L1,6);
        push(&L1,4);
        push(&L1,3);
        /* creating list 2 */
        push(&L2,10);
        push(&L2,8);
        push(&L2,1);
      
        L1 = mergeSort(L1,L2); 
        printList(L1);
        
        return 0;
}

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