Balanced partition problem

Balance partition problem

Given a set of integers, partition those integers into two parts where the difference between the two parts is minimum. This problem is known as balanced partition problem. For example, array A = {1,7,4,11}, two subsets can be: {1,11} and {7,4}, two have a difference of 1, which is the minimum difference we can get by splitting this array.

Mathematically, you have a set of n integers each in the range 0, . . . , K. Partition these integers into two subsets such that you minimize |S1 − S2|, where S1 and S2 denote the sums of the elements in each of the two subsets.

Balance partition problem can be asked in many other ways, for instance,
given a list of 22 players and their strengths, divide those 22 players into two teams so that both teams are balanced.
Another version can be that you have “n” candy, each candy has a value associated with it. You want to distribute those candies between two kids as equally as possible.

No matter what version is asked, the approach remains the same.

Balance partition problem: thoughts

The brute force method will be to list down all the subsets of the given set and find the sum of each one of them. Then scan through the sum of all the subsets and find the two closest ones. For a set of n elements, there can be 2n subset. Therefore, the complexity of this brute force solution is already exponential.

Let me tweak balance partition problem a bit. We find if there are two subsets of the set of integers such that the difference between “sum” of these two subsets is zero.
Essentially, this is a special case of the original problem. If the difference between the sum of two subsets is zero that means the sum of both subsets should be exactly equal to half of the sum of all elements in the set.

So problem reduces to a smaller problem that is to find if there is a subset of integers which add up to half the sum of all integers in the set? This is the subset sum problem which we have already solved. 

How can we use information provided by subset set problem above?
Let’s say S is the sum of all the integers in the set. S/2 will be half of that sum. We have to find a subset with sum i such that S/2 -i is minimum.

Whether or not, there is a subset with sum i in the set is given by solving subset sum problem. For the sums, i, which are possible with subsets of the set, find the one which is the least distance from S/2. That will give us other subsets which is least greater than half of the sum of all elements of the set and that will be minimal difference possible between two subsets.

So,  expression would be as

min(S/2 - i) where T[n][i] = True and i>=0 and i<=S/2

Why we took i >=0 and i<S/2? Because, we want to be balanced, so i cannot be more than half of the total sum in any case.

Balanced partition problem: implementation

package com.company;

/**
 * Created by sangar on 25.11.18.
 */
public class BalancedPartition {
    public int findBalancePartition(int[] a){

        // Calculate sum of all the elements in set 
        int S = 0;
        for (int i=0; i<a.length; i++)
            S += a[i];

        boolean T[][] = new boolean[a.length + 1][S + 1];

        /* Initialize first column as true. 
            0 sum is possible with all elements. 
        */
        for (int i=0; i<=a.length; i++)
            T[i][0] = true;

        /*  Initialize top row, except dp[0][0], 
            as false. With 0 elements, no other 
            sum except 0 is possible
        */
        for (int i=1; i<=S; i++)
            T[0][i] = false;

        
        for (int i = 1; i <= a.length; i++) {
            for (int j = 1; j <= S; j++) {
                // If ith element is excluded 
                T[i][j] = T[i - 1][j];

                // If ith element is included 
                if (a[i - 1] <= j)
                    T[i][j] |= T[i - 1][j - a[i - 1]];
            }
        }

        // Initialize difference of two sums. 
        int diff = Integer.MAX_VALUE;

        for (int j = S/2; j >= 0; j--) {
            // Find the 
            if (T[a.length][j] == true)
            {
                diff = S - 2 * j;
                break;
            }
        }
        return diff;
    }
}

Once, we get the nearest sum, we can always backtrack the table and find elements of the subset itself. Actually, this problem is now reduced to 0/1 knapsack problem, where maximum value we can get is j from the set of integers.

Complexity to split set into two balanced partitions is O(n * S) with a space complexity of O(n * S), where S will be the max value array can have.

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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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