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Find Kth smallest element in binary search tree

Kth smallest element in a binary a search tree

Given a binary search tree, find kth smallest element in the binary search tree. For example, 5th smallest element in below binary search tree would be 14, if store the tree in sorted order 5,7,9,10,14,15,19; 14 is the fifth smallest element in that order.

kth smallest element in binary search tree
Kth smallest element in binary search tree

Kth smallest element in binary search tree: thoughts

As mentioned earlier in a lot of posts like delete a binary tree or mirror a binary tree, first try to find the traversal order required to solve this problem. One hint we already got is that we want all the nodes on BST traversed in sorted order. What kind of traversal gives us a sorted order of nodes? Of course, it is inorder traversal.

So idea is to do an inorder traversal of the binary search tree and store all the nodes in an array. Once traversal is finished, find the kth smallest element in the sorted array.

This approach, however, scans the entire tree and also has space complexity of O(n) because we store all the nodes of tree in an array. Can we avoid scanning the whole tree and storing nodes?

If we keep count of how many nodes are traversed during inorder traversal, we can actually stop traversal as soon as we see k nodes are visited. In this case, we do not store nodes, just a counter. 

Kth smallest element in binary tree: implementation

package com.company.BST;

import java.util.Stack;

/**
 * Created by sangar on 9.11.18.
 */
public class FindKthSmallestInBST {
    private int counter;

    public FindKthSmallestInBST(){
        counter = 0;
    }
    public TreeNode findKthSmallest(TreeNode root, int k){
        if(root == null) return root;

        //Traverse left subtree first
        TreeNode left = findKthSmallest(root.getLeft(),k);
        //If we found kth node on left subtree
        if(left != null) return left;
        //If k becomes zero, that means we have traversed k nodes.
        if(++counter == k) return root;

        return findKthSmallest(root.getRight(),k);
    }
}

Test cases

package test;

import com.company.BST.BinarySearchTree;
import com.company.BST.FindKthSmallestInBST;
import com.company.BST.TreeNode;
import org.junit.jupiter.api.Test;

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

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

    FindKthSmallestInBST tester = new FindKthSmallestInBST();

    @Test
    public void kthSmallestElementTest() {

        BinarySearchTree<Integer> binarySearchTree =
			new BinarySearchTree<>();
        binarySearchTree.insert(10);
        binarySearchTree.insert(8);
        binarySearchTree.insert(15);
        binarySearchTree.insert(12);
        binarySearchTree.insert(6);
        binarySearchTree.insert(9);

        TreeNode kthNode =
			tester.findKthSmallest(binarySearchTree.getRoot(),1);
        assertEquals(6, kthNode.getValue());
    }

    @Test
    public void kthSmallestElementOnRightSubtreeTest() {

        BinarySearchTree<Integer> binarySearchTree =
				new BinarySearchTree<>();
        binarySearchTree.insert(10);
        binarySearchTree.insert(8);
        binarySearchTree.insert(15);
        binarySearchTree.insert(12);
        binarySearchTree.insert(6);
        binarySearchTree.insert(9);

        TreeNode kthNode =
			tester.findKthSmallest(binarySearchTree.getRoot(),5);
        assertEquals(12, kthNode.getValue());
    }

    @Test
    public void kthSmallestElementAbsentSubtreeTest() {

        BinarySearchTree<Integer> binarySearchTree =
				new BinarySearchTree<>();
        binarySearchTree.insert(10);
        binarySearchTree.insert(8);
        binarySearchTree.insert(15);
        binarySearchTree.insert(12);
        binarySearchTree.insert(6);
        binarySearchTree.insert(9);

        TreeNode kthNode =
			tester.findKthSmallest(binarySearchTree.getRoot(),10);
        assertEquals(null, kthNode);
    }

    @Test
    public void kthSmallestElementNulltreeTest() {

        BinarySearchTree<Integer> binarySearchTree =
				new BinarySearchTree<>();
        binarySearchTree.insert(10);
        binarySearchTree.insert(8);
        binarySearchTree.insert(15);
        binarySearchTree.insert(12);
        binarySearchTree.insert(6);
        binarySearchTree.insert(9);

        TreeNode kthNode = tester.findKthSmallest(null,10);
        assertEquals(null, kthNode);
    }
}

The complexity of this algorithm to find kth smallest element is O(k) as we traverse only k nodes on binary search tree.

There is hidden space complexity here. Recursive function requires call stack memory, which is limited to Operation System default. More deep you go in recursion, more space we use on stack. If tree is completely skewed, there are more chances of stack overflow. Also recursive function is very difficult to debug in production environments. Below is the non-recursive solution for the same problem.

Non-recursive way to find kth smallest element in BST

public int kthSmallest(TreeNode root, int k) {
        Stack<TreeNode> s = new Stack<TreeNode>();

        TreeNode current = root;
        int result = 0;

        while(!s.isEmpty() || current!=null){
            if(current!=null){
                s.push(current);
                current = current.getLeft();
            }else{
                TreeNode  temp = s.pop();
                k--;
                if(k==0)
                    result = (int)temp.getValue();
                current = temp.getRight();
            }
        }

        return result;
    }

Please share if there is something wrong or missing. If you are preparing for an interview and need help, please reach out to us at communications@algorithmsandme.com

Find Kth smallest element in array

Kth smallest element in array

Given an array of integers which is non sorted, find kth smallest element in that array. For example: if input array is A = [3,5,1,2,6,9,7], 4th smallest element in array A is 5, because if you sort the array A, it looks like A = [1,2,3,5,6,7,9] and now you can easily see that 4th element is 5.

This problem is commonly asked in Microsoft and Amazon interviews as it has multiple layers and there is some many things that can be measured with this one problem.

Kth smallest element : Line of thought

First of all, in any interview, try to come up with brute force solution. Brute force solution to find Kth smallest element in array of integers would be to sort array and return A[k-1] element (K-1 as array is zero base indexed).

What is the complexity of brute force solution? It’s O(n2)? Well, we have sort algorithms like merge sort and heap sort which work in O(n log n) complexity. Problem with both searches is that they use additional space. Quick sort is another sort algorithm. It has problem that it’s worst case complexity will be O(n2), which happens when input is completely sorted.
In our case, input is given as unsorted already, so we can expect that quick sort will function with O(n log n) complexity which is it’s average case complexity. Advantage of using quick sort is that there is no additional space complexity.

Optimising quick sort

Let’s see how quick sort works and see if we can optimize solution further?
Idea behind quick sort is to find correct place for the selected pivot. Once pivot is at correct position, all the elements on left side of pivot are smaller and on right side of pivot are greater than pivot. This step is partitioning.

If after partitioning, pivot is at position j, can we say that pivot is actually jth smallest element of the array? What if j is equal to k? Well problem solved, we found the kth smallest element.

If j is less than k, left subarray is less than k, we need to include more elements from right subarray, therefore kth smallest element is in right subarray somewhere. We have already found j smallest elements, all we need to find is k-j elements from right subarray.

What if j is greater than k? In this case, we have to drop some elements from left subarray, so our search space would be left subarray after partition.

Theoretically, this algorithm still has complexity of O(n log n), but practically, you do not need to sort the entire array before you find k smallest elements.

Algorithm to find K smallest elements in array

  1. Select a pivot and partition the array with pivot at correct position j
  2. If position of pivot, j, is equal to k, return A[j].
  3. If j is less than k, discard array from start to j, and look for (k-j)th smallest element in right sub array, go to step 1.
  4. If j is greater than k, discard array from j to end and look for kth element in left subarray, go to step 1

Let’s take an example and see if this algorithm works? A =  [4, 2, 1, 7, 5, 3, 8, 10, 9, 6 ], and we have to find fifth smallest element in array A.

Kth smallest element in array

Start with pivot as first index of array, so pivot = 0, partition the array into two parts around pivot such that all elements on left side of pivot element, i.e. A[pivot] are smaller and all elements on right side are greater than A[pivot].

Start with pivot as first index of array, so pivot = 0, partition the array into two parts around pivot such that all elements on left side of pivot element, i.e. A[pivot] are smaller and all elements on right side are greater than A[pivot].

In our example, array A will look like below after pivot has found it’s correct position.

k smallest element
After partition, correct position of pivot is index 3

If pivot == k-1 (array is represented as zero base index), then A[pivot] is kth smallest element. Since pivot (3) is less than k-1 (4), look for kth smallest element on right side of the pivot.

k remains as it is as opposed to k-j mentioned in algorithm as pivot is given w.r.t entire array and not w.r.t subarray.

In second iteration, pivot = 4 (index and not element). After second execution of quick sort array A will be like

After partition of right subarray, correct position of pivot is index 4

pivot(4) which is equal to k-1(5-1). 5th smallest element in array A is 5.

Implementation

package com.company;

/**
	* Created by sangar on 30.9.18.
*/
public class KthSmallest {
	private void swap(int[] a, int i, int j){
		int temp = a[i];
		a[i] = a[j];
		a[j] = temp;
	}
	private int partition(int[] a, int start, int end){
		int pivot = a[start];
		int i  = start+1;
		int j  = end;

		while(i < j){
			while(a[i] < pivot) i++;
			while(a[j] > pivot) j--;

			if(i < j) {
				swap(a, i, j);
			}
		}
		swap(a, start, j);
		return j;
	}

	public int findKthSmallestElement(int a[], int start, 
				int end, int k){
		if(start < end){
		int p = partition(a, start, end);
		if(p == k-1){
			return a[p];
		}
		if(p > k-1)
			return findKthSmallestElement(a, start, p, k);
		if(p < k-1)
			return findKthSmallestElement(a, p+1, end, k);
		}
		return -1;
	}
}
package test;

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

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

/**
 * Created by sangar on 28.8.18.
 */
public class KthSmallestTest {

	KthSmallest tester = new KthSmallest();
	private int[] a = {4, 2, 1, 7, 5, 3, 8, 10, 9};
	@Test
	public void kthSmallest() {
		assertEquals(7, tester.findKthSmallestElement(a,0,8,6));
	}

	@Test
	public void firstSmallest() {
		assertEquals(1, tester.findKthSmallestElement(a,0,8,1));
	}

	@Test
	public void lastSmallest() {
		assertEquals(10, tester.findKthSmallestElement(a,0,8,9));
	}

	@Test
	public void kGreaterThanSize() {
		assertEquals(-1, tester.findKthSmallestElement(a,0,8,15));
	}
	@Test
	public void emptyArray() {
		int[] a = {};
		assertEquals(-1, tester.findKthSmallestElement(a,0,0,1));
	}

	@Test
	public void nullArray() {
		assertEquals(-1, tester.findKthSmallestElement(null,0,0,1));
	}
}

Complexity of using quick sort algorithm to find kth smallest element in array of integers in still O(n log n).

Kth smallest element using heaps

Imagine a case where there are a billion integers in array and you have to find 5 smallest elements from that array. Complexity of O(n log n) is too costly for that use case. Above algorithm using quick sort does not take into consideration disparity between k and n.

We want top k elements, how about we chose those k elements randomly, call it set A and then go through all other n-k elements, call it set B, check if element from set B (n-k elements) can displace element in set A (k elements)?

What will be condition for an element from set B to replace an element in set A? Well, if the new element is less than maximum in set A, than maximum in set A cannot be in set of k smallest elements right?  Maximum element in set A would be replaced by the new element from set B.

Now, problem is how to quickly find maximum out of set A. Heap is the best data structure there. What kind of heap: min heap or max heap? Max heap as it store the maximum of set at the root of it.

Let’s defined concrete steps to find k smallest elements using max heap. 

  1. Create a max heap of size k from first k elements of array.
  2. Scan all elements in array one by one.
    1.  If current element is less than max on heap, add current element to heap and heapify.
    2. If not, then go to next element.
  3. At the end, max heap will contain k smallest elements of array and root will be kth smallest element.

Let’s take an example and see if this algorithm works? Input array is shown below and we have to find 6th smallest element in this array.

kth smallest element using heaps
input array

Step 1 : Create a max heap with first 6 elements of array.

Create a max heap with set A

Step 2 : Take next element from set B and check if it is less than root of max heap. In this case, yes it is. Remove the root and insert the new element into max heap.

Element from set B removes root from max heap and added to max heap

Step 2 : It continues to 10, nothing happens as new element is greater than root of max heap. Same for 9.  At 6, again root of max heap is greater than 6. So remove the root and add 6 to max heap.

Again, new element from set B is less than root of max heap. Root is removed and new element is added.

Array scan is finished, so just return root of max heap, 6 which is sixth smallest element in given array.

	public int findKthSmallestElementUsingHeap(int a[], int k){
	//https://stackoverflow.com/questions/11003155/change-priorityqueue-to-max-priorityqueue
	PriorityQueue<Integer>  maxHeap =
			new PriorityQueue<>(k, Collections.reverseOrder());

		if(a == null || k > a.length) return -1;
		//Create max with first k elements
		for(int i=0; i<k; i++){
			maxHeap.add(a[i]);
		}
		/*Keep updating max heap based on new element
		If new element is less than root, 
		remove root and add new element
		*/
		for(int i=k; i<a.length; i++){
			if(maxHeap.peek() > a[i]){
				maxHeap.remove();
				maxHeap.add(a[i]);
			}
		}
		return maxHeap.peek();
	}

Can you calculate the complexity of above algorithm? heapify() has complexity of log(k) with k elements on heap. In worst case, we have to do heapify() for all elements in array, which is n, so overall complexity of algorithm becomes O(n log k). Also, there is additional space complexity of O(k) to store heap.
When is very small as compared to n, this algorithm again depends on the size of array.

We want k smallest elements, if we pick first k elements from a min heap, will it solve the problem? I think so. Create a min heap of n elements in place from the given array, and then pick first k elements.
Creation of heap has complexity of O(n), do more reading on it. All we need to do is delete k times from this heap, each time there will be heapify(). It will have complexity of O(log n) for n element heap. So, overall complexity would be O(n + k log n).

Depending on what you want to optimize, select correct method to find kth smallest element in array.

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

Quick sort algorithm

Quick sort Algorithm

Quick sort like merge sort is a sorting algorithm under divide and conquer paradigm of algorithms like merge sort. Basic idea of algorithm is to divide inputs around a pivot and then sort two smaller parts recursively and finally get original input sorted.

Selection of pivot

Entire idea of quick sort revolves around pivot. Pivot is an element in input around which input is arranged in such a way that all elements on left side are smaller and all elements on right side are greater than pivot. Question is how to find or select pivot and put it into correct position.

To make things simpler to start with, let’s assume first element of input is pivot element.

To put this pivot at correct position in input, start with next element of pivot in input space and find first element which is greater than pivot. Let that be ith position.

At the same time, start from end of array and find first element which is smaller than pivot. Let it be jth position.

If i and j have not crossed each other i.e i < j, then swap element at ith and jth positions, and continue moving right on input to find element greater than pivot and moving left to find element smaller than pivot.
Once i and j cross each other, swap pivot with element at jth position.  After this step, pivot will be at its correct position and array will be divided into two parts. All elements on left side will be less than pivot and all elements on right side will be greater than pivot.

Quick sort partition example

This is too much to process, I know! Let’s take an example and see how it does it work? We have an array as follows

quick sort

Let’s select first element as pivot, pivot = 3.

quick sort pivot selection

Start from next element of pivot, move towards right of array, till we see first element which is greater than pivot i.e. 3.

From end of array, move towards left till you find an element which is less than pivot.

Now, there are two indices, i and j, where A[i] > pivot and A[j] < pivot. See that i and j not yet crossed each other. Hence, we swap A[i] with A[j]. Array at the bottom of pic, shows resultant array after swap.

quick sort partition

Again, start with i+1 and follow the same rule : Stop when you find element greater than pivot. In this case, 10 is greater than 3, hence we stop.

Similarly, move left from end again, till we find an element which is less than pivot. In this case, we end up at index = 2 which is element 1.

Since, i > j, than means paths have been crossed. At this time, instead of swapping element at i and j index, swap element at j index with pivot.

After swapping pivot with jth index, we have array divided into two parts, pivot as boundary. All elements on left side of pivot are smaller (they may not be sorted) and all elements on right side of pivot are greater than pivot (again may not be sorted).

quick sort partitions

We, apply this same partition process to left and right arrays again, till base condition is hit. In this case, base condition would be if there is only one element in array to be partitioned.

Quick sort algorithm

quickSort([], start, end)
 1. If array has more than one elements i.e (start < end):
    1.1 Find correct place for pivot.
    pivot = partition(arr, low, high)
    1.2. Apply same function recursively to left of pivot index
         quickSort(arr, start, pivot -1 )
         and to the right of pivot index
         quickSort(arr, pivot + 1, end)

Quick sort implementation

#include <stdio.h>

void swap(int a[], int i, int j){
	int temp = a[i];
	a[i] = a[j];
	a[j] = temp;
}

int partition(int a[], int start, int end){
	
	int pivot = a[start];
	int i  = start+1;
	int j  = end;
	
	while(i < j){
	    while(a[i] < pivot) i++;
            while(a[j] > pivot) j--;
	
            if(i < j) {
                swap(a, i, j);
	    }
	}
	swap(a, start, j);
	return j;
}

void quickSort(int a[], int start, int end){
    if(start < end){
        int p = partition(a, start, end);
	quickSort(a,start, p-1);
	quickSort(a, p+1, end);
    }
}

int main(void) {
	int a[]= {4,3,2,5,6,8,1};
	int size = sizeof(a)/sizeof(a[0]);
	
	quickSort(a, 0, size-1);
	
	for(int i=0; i < size; i++){
		printf(" %d", a[i]);
	}
	return 0;
}

There is another implementation which is based on Lomuto partition scheme, in this scheme, we make last element as pivot. The implementation is compact but complexity is bit higher than the original partition methods in terms of number of swaps.

#include<stdlib.h>
#include<stdio.h>
 
void swap(int *a, int *b){
    int temp = *a;
    *a = *b;
    *b = temp;
}
 
int partition(int a[], int low, int high)
{
    // set pivot as highest element
    int x  = a[high];
 
    //Current low points to previous of low of this part of array. 
    int i = low - 1;
 
    for (int j = low; j <= high-1; j++)
    {
    	/*Move in the array till current node data is 
        less than the pivot */
        if (a[j] <= x){
            //set the current low appropriately
            i++;
            swap(&a[i], &a[j]);
        }
    }
    //Now swap the next node of current low with pivot
 
    swap(&a[i+1], &a[high]);
 
    printf("\n Pivot : %d\n", a[i+1]);
    for(int j=0; j<=high; j++){
 
    	printf("%d ", a[j]);
    }
    //return current low as partitioning point.
    return i+1;
}
 
/* A recursive implementation of quicksort for linked list */
void quickSortUtil(int a[], int low, int high)
{
    if (low < high)
    {
        int p = partition(a,low, high);
        quickSortUtil(a,low, p-1);
        quickSortUtil(a, p+1, high);
    }
}
 
/* Driver program to run above code */
int main(){
 
    int a[] = {5,4,2,7,9,1,6,10,8};
 
    int size = sizeof(a)/sizeof(a[0]);
    quickSortUtil(a, 0, size-1);
 
    for(int i=0; i<size; i++){
    	printf("%d ", a[i]);
    }
    return 0;
}

Complexity analysis of quick sort algorithm

If pivot splits original array into two equal parts (which is the intention), complexity of quick sort is O(n log n). However, worst case complexity of quick sort happens when input array is already sorted in increasing or decreasing order. In this case, array is partitioned into two subarrays, one with size 1 and other with size n-1. Similarly, subarray with n-1 elements, it again is divided into two subarrays of size 1 and n-2. In order to completely sort array it will split for n-1 times and each time it requires to traverse n element to find correct position of pivot. Hence overall complexity of quick sort comes out as O(n2).

There is a very interesting question, which tests your understanding of system basics. Question is what is space complexity of this algorithm? There is no apparent memory is used. However, recursive implementation internally puts stack frames on stack for partitioned indices and function call return address and so on. In worst case, there can be n stack frames, hence worst case complexity of quick sort will be O(n).

How can we reduce that? If the partition with fewest elements is (recursively) sorted first, it requires at most O(log n) space. Then the other partition is sorted using tail recursion or iteration, which doesn’t add to the call stack. This idea, was described by R. Sedgewick, and keeps the stack depth bounded by O(log n) and hence space complexity will be O(log n).

Quick sort with tail recursion

Quicksort(A, p, r)
{
 while (p < r)
 {
  q = Partition(A, p, r)
  Quicksort(A, p, q)
  p = q+1
 }
}

Selection of Pivot
If array is completely sorted, then worst case behavior of quick sort is O(n2), so there comes another problem. How can we select pivot so that two subarrays are almost equal size. There are many solutions proposed.
1. Taking median of array as pivot. So how to select median of an unsorted array. We look into this problem separately, but yes it guarantees two halves of same size.
2. Selecting pivot randomly. This requires heuristics algorithms to select pivot.

Please leave your comment in case you find something wrong or you have some improved version.

Arranging people according to country of origin : Count sort in play

Introduction

Let’s consider a problem where we have nationals of different countries in a room and our task is to arrange these nationals in such a way that all the nationals of same country are clubbed together.
Simple application of count sort isn’t? Count nationals of each country and place them in order as discussed in previous post.

Problem:

Now let’s add some complexity. What if we need all nationals of country we discovered early to be before the nationals of country we encountered later?
So input be
A = [I,I,I, U,U,S, S, E,E,E,M,M,M,I, U,M,E]
Output should be
A = [I,I,I,I, U,U,U,S,S,E,E,E,E, M,M,M]

Solution:

So there is additional information we need to track that what was the order of the visiting country while traversing the country and rest all follows as per the count sort.
How do we keep track of order?
We can keep an array of countries and put countries in such a way that index of first encounter country is lower than the country later. We can do this while counting the nationals of each country by checking that if this was the first national of this country we encountered then we can go and add the country in country array.

Complete code

void arrange(int a[], int n, int K){
        /*
            n is number of people
            K is no of distinct countries
        */
        char country[K];
        int  count[K];
        int i,j;
        int index =0;
        int next_index = 0;

        for(i=0;i<=K; i++){
                count[i] =0;
        }

        for(i=0;i<n;i++){
                if(count[a[i]] == 0){
                 /*
                   If this is the first person for that country, add that country to our country array
                 */
                        country[index++] = a[i];
                }
                count[a[i]]++;
        }

        for(i=0; i<index;i++){
                j = next_index;
                next_index += count[country[i]];
                for(; j<next_index; j++){
                        a[j] = country[i];
                }

        }


}

Conclusion

It is fairly easy to apply count sort on any arranging problem where we have a certain attribute we need to arrange input on.

Count sort : Sorting in linear time

Are there any sorting algorithm which has worst case complexity of O(n)? There are a few like count sort and decision tree algorithms. In this post we would discuss about count sort and couple of problems where this counting sort algorithm can be applied.
Counting sort was invented by Harold H. Seward

To apply counting sort, we need to keep in mind following assumptions:
  1. There should be duplicate values in the input
  2. There should be at most K different type of values in input.
  3. The input data ranges in 0 to K
Count sort goes for O(n^2) if K is very close to n i.e. a very few elements are duplicate and rest all are unique. So above three conditions are necessary to have this algorithm work in linear time.

Count Sort -Approach

Let’s see how does it work?
There are three steps involved in algorithm:
  1. Sampling the data, counting the frequencies of each element in input set.
  2. Accumulating frequencies to find out relative positions of each element.
  3. Third step distributes each element to its appropriate place.
Now let’s take one example and go through above three steps:
Input is  an array 
A = [1,2,3,2,2,3,3,1,1,1,1,3,2,2]

Here we can see that K=3.
Now let’s perform the first step of the method, i.e. count the frequency of each elementWe keep a hash and keep track of each element’s count. Since values are upper bound by K, we need at most K sized hash.
We initialize this hash as zero.
A is input array and n is size of the array.


int char count [K];

for(i=0; i<K;i++){
count[i]=0;
}

for(i=0;i<n;i++){
count[a[i]]++;
}
Count looks likes
count  =[5,5,4]
Complexity of this step is O(n).

Second step involves accumulation of frequencies where we add frequencies of previous elements to the current element.
F(j) = summation of f[i] for all i=0


for(i=1;i<K;i++){
Count[i] +=count[i-1];
}

Second step runs for K times hence the complexity of O(K)
Third step involves distribution.
Let’s create an output array called as B of size n
For every element in A we check the corresponding count in count array, and place the element at that position. So, first 1 will be at position 4 (there are total 5 and array is index from 0), first two will be placed at 9 and first three will be at 13. Subsequently, lower positions will be filled as we encounter them in input array.

for(i=0; i<n;i++){
B[--count[a[i]]] = a[i];
}

This step has complexity of O(n)

Complexity Analysis

Total complexity of the algorithm being O(K)+ O(n) + O(K) +O(n) = O(K+n) . Please refer to master theorem, how this is derived.
Now if we see, K+n remains in order of n till the time K<n, if K goes on to n^2, the complexity of algorithm goes for polynomial time instead of linear time.

Applications

There is immediate application of this algorithm in following problem:
Let’s there is an array which contains Black, White and Red balls, we need to arrange these balls in such a way that all black balls come first, white second and Red at last.
Hint: assign black 0, white 1 and Red 2 and see. 🙂

In next post we would discuss how extra space can be saved, how initial positions of elements can be maintained. We would go through an interesting problem to discuss above optimization.

Complete Code

Output of the above code after every iteration is :

Iter 0 :0  0  0  0  1  0  0  0  0  0  0  0  0  0  
Iter 1 :0 0 0 0 1 0 0 0 0 2 0 0 0 0
Iter 2 :0 0 0 0 1 0 0 0 0 2 0 0 0 3
Iter 3 :0 0 0 0 1 0 0 0 2 2 0 0 0 3
Iter 4 :0 0 0 0 1 0 0 2 2 2 0 0 0 3
Iter 5 :0 0 0 0 1 0 0 2 2 2 0 0 3 3
Iter 6 :0 0 0 0 1 0 0 2 2 2 0 3 3 3
Iter 7 :0 0 0 1 1 0 0 2 2 2 0 3 3 3
Iter 8 :0 0 1 1 1 0 0 2 2 2 0 3 3 3
Iter 9 :0 1 1 1 1 0 0 2 2 2 0 3 3 3
Iter 10 :1 1 1 1 1 0 0 2 2 2 0 3 3 3
Iter 11 :1 1 1 1 1 0 0 2 2 2 3 3 3 3
Iter 12 :1 1 1 1 1 0 2 2 2 2 3 3 3 3
Iter 13 :1 1 1 1 1 2 2 2 2 2 3 3 3 3


Final O/P :1 1 1 1 1 2 2 2 2 2 3 3 3 3


Why this filling from the end of the span? Because this step makes count sort stable sort.

Conclusion

Here we have seen a sorting algorithm which can give us o/p linear time given some particular conditions met on the i/p data.