Count ways to split array into two subsets having difference between their sum equal to K

Given an array A[] of size N and an integer diff, the task is to count the number of ways to split the array into two subsets (non-empty subset is possible) such that the difference between their sums is equal to diff.

Examples:

Input: A[] = {1, 1, 2, 3}, diff = 1  
Output: 3  
Explanation: All possible combinations are as follows: 

• {1, 1, 2} and {3}
• {1, 3} and {1, 2}
• {1, 2} and {1, 3}

All partitions have difference between their sums equal to 1. Therefore, the count of ways is 3.

Input: A[] = {1, 6, 11, 5}, diff=1
Output: 2

Naive Approach: The simplest approach to solve the problem is based on the following observations:

Let the sum of elements in the partition subsets S1 and S2 be sum1 and sum2 respectively.
Let sum of the array A[] be X. 
Given, sum1 – sum2 = diff – (1)
Also, sum1 + sum2 = X – (2)

From equations (1) and (2), 
sum1 = (X + diff)/2

Therefore, the task is reduced to finding the number of subsets with a given sum. 
Therefore, the simplest approach is to solve this problem is by generating all the possible subsets and checking whether the subset has the required sum. 

Time Complexity: O(2^N)
Auxiliary Space: O(N)

Efficient Approach: To optimize the above approach, the idea is to use Dynamic Programming. Initialize a dp[][] table of size N*X, where dp[i][C] stores the number of subsets of the sub-array A[i…N-1] such that their sum is equal to C. Thus, the recurrence is very trivial as there are only two choices i.e. either consider the i^th element in the subset or don’t. So the recurrence relation will be:

dp[i][C] = dp[i – 1][C – A[i]] + dp[i-1][C]

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to count the number of ways to divide
// the array into two subsets and such that the
// difference between their sums is equal to diff
int countSubset(int arr[], int n, int diff)
{
    // Store the sum of the set S1
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
    sum += diff;
    sum = sum / 2;

    // Initializing the matrix
    int t[n + 1][sum + 1];

    // Number of ways to get sum
    // using 0 elements is 0
    for (int j = 0; j <= sum; j++)
        t[0][j] = 0;

    // Number of ways to get sum 0
    // using i elements is 1
    for (int i = 0; i <= n; i++)
        t[i][0] = 1;

    // Traverse the 2D array
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= sum; j++) {

            // If the value is greater
            // than the sum store the
            // value of previous state
            if (arr[i - 1] > j)
                t[i][j] = t[i - 1][j];

            else {
                t[i][j] = t[i - 1][j]
                          + t[i - 1][j - arr[i - 1]];
            }
        }
    }

    // Return the result
    return t[n][sum];
}

// Driver Code
int main()
{
    // Given Input
    int diff = 1, n = 4;
    int arr[] = { 1, 1, 2, 3 };

    // Function Call
    cout << countSubset(arr, n, diff);
}

Java

// Java program for the above approach
import java.io.*;
public class GFG
{

    // Function to count the number of ways to divide
    // the array into two subsets and such that the
    // difference between their sums is equal to diff
    static int countSubset(int []arr, int n, int diff)
    {

        // Store the sum of the set S1
        int sum = 0;
        for (int i = 0; i < n; i++)
            sum += arr[i];
        sum += diff;
        sum = sum / 2;

        // Initializing the matrix
        int t[][] = new int[n + 1][sum + 1];

        // Number of ways to get sum
        // using 0 elements is 0
        for (int j = 0; j <= sum; j++)
            t[0][j] = 0;

        // Number of ways to get sum 0
        // using i elements is 1
        for (int i = 0; i <= n; i++)
            t[i][0] = 1;

        // Traverse the 2D array
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= sum; j++) {

                // If the value is greater
                // than the sum store the
                // value of previous state
                if (arr[i - 1] > j)
                    t[i][j] = t[i - 1][j];

                else {
                    t[i][j] = t[i - 1][j]
                              + t[i - 1][j - arr[i - 1]];
                }
            }
        }

        // Return the result
        return t[n][sum];
    }

    // Driver Code
    public static void main(String[] args)
    {

        // Given Input
        int diff = 1, n = 4;
        int arr[] = { 1, 1, 2, 3 };

        // Function Call
        System.out.print(countSubset(arr, n, diff));
    }
}

// This code is contributed by AnkThon

Python3

# Python3 program for the above approach

# Function to count the number of ways to divide
# the array into two subsets and such that the
# difference between their sums is equal to diff
def countSubset(arr, n, diff):

    # Store the sum of the set S1
    sum = 0
    for i in range(n):
        sum += arr[i]

    sum += diff
    sum = sum // 2

    # Initializing the matrix
    t = [[0 for i in range(sum + 1)]
            for i in range(n + 1)]

    # Number of ways to get sum
    # using 0 elements is 0
    for j in range(sum + 1):
        t[0][j] = 0

    # Number of ways to get sum 0
    # using i elements is 1
    for i in range(n + 1):
        t[i][0] = 1

    # Traverse the 2D array
    for i in range(1, n + 1):
        for j in range(1, sum + 1):

            # If the value is greater
            # than the sum store the
            # value of previous state
            if (arr[i - 1] > j):
                t[i][j] = t[i - 1][j]
            else:
                t[i][j] = t[i - 1][j] + t[i - 1][j - arr[i - 1]]

    # Return the result
    return t[n][sum]

# Driver Code
if __name__ == '__main__':

    # Given Input
    diff, n = 1, 4
    arr = [ 1, 1, 2, 3 ]

    # Function Call
    print (countSubset(arr, n, diff))

# This code is contributed by mohit kumar 29

C#

// C# program for the above approach

using System;

public class GFG
{

    // Function to count the number of ways to divide
    // the array into two subsets and such that the
    // difference between their sums is equal to diff
    static int countSubset(int []arr, int n, int diff)
    {

        // Store the sum of the set S1
        int sum = 0;
        for (int i = 0; i < n; i++)
            sum += arr[i];
        sum += diff;
        sum = sum / 2;

        // Initializing the matrix
        int [,]t = new int[n + 1, sum + 1];

        // Number of ways to get sum
        // using 0 elements is 0
        for (int j = 0; j <= sum; j++)
            t[0,j] = 0;

        // Number of ways to get sum 0
        // using i elements is 1
        for (int i = 0; i <= n; i++)
            t[i,0] = 1;

        // Traverse the 2D array
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= sum; j++) {

                // If the value is greater
                // than the sum store the
                // value of previous state
                if (arr[i - 1] > j)
                    t[i,j] = t[i - 1,j];

                else {
                    t[i,j] = t[i - 1,j]
                              + t[i - 1,j - arr[i - 1]];
                }
            }
        }

        // Return the result
        return t[n,sum];
    }

    // Driver Code
    public static void Main(string[] args)
    {

        // Given Input
        int diff = 1, n = 4;
        int []arr = { 1, 1, 2, 3 };

        // Function Call
        Console.Write(countSubset(arr, n, diff));
    }
}

// This code is contributed by AnkThon

Javascript

<script>

// JavaScript program for the above approach

// Function to count the number of ways to divide
// the array into two subsets and such that the
// difference between their sums is equal to diff
function countSubset(arr, n, diff)
{
    // Store the sum of the set S1
    var sum = 0;
    for (var i = 0; i < n; i++){
        sum += arr[i];
    }
    sum += diff;
    sum = sum / 2;

    // Initializing the matrix
    //int t[n + 1][sum + 1];
    var t = new Array(n + 1);

    // Loop to create 2D array using 1D array
    for (var i = 0; i < t.length; i++) {
        t[i] = new Array(sum + 1);
    }

    // Loop to initialize 2D array elements.
    for (var i = 0; i < t.length; i++) {
        for (var j = 0; j < t[i].length; j++) {
            t[i][j] = 0;
        }
    }

    // Number of ways to get sum
    // using 0 elements is 0
    for (var j = 0; j <= sum; j++)
        t[0][j] = 0;

    // Number of ways to get sum 0
    // using i elements is 1
    for (var i = 0; i <= n; i++)
        t[i][0] = 1;

    // Traverse the 2D array
    for (var i = 1; i <= n; i++) {
        for (var j = 1; j <= sum; j++) {

            // If the value is greater
            // than the sum store the
            // value of previous state
            if (arr[i - 1] > j)
                t[i][j] = t[i - 1][j];

            else {
                t[i][j] = t[i - 1][j]
                          + t[i - 1][j - arr[i - 1]];
            }
        }
    }

    // Return the result
    return t[n][sum];
}

// Driver Code

// Given Input
var diff = 1;
var n = 4;
var arr = [ 1, 1, 2, 3 ];

// Function Call
document.write(countSubset(arr, n, diff));

</script>

Output:

3

Time Complexity: O(S*N), where S = sum of array elements + K/2 
Auxiliary Space: O(S*N)