# Count ways to split array into pair of subsets with difference between their sum equal to K

• Last Updated : 10 Nov, 2021

Given an array arr[] consisting of N integers and an integer K, the task is to find the number of ways to split the array into a pair of subsets such that the difference between their sum is K.

Examples:

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Input: arr[] = {1, 1, 2, 3}, K = 1
Output: 3
Explanation:
Following splits into a pair of subsets satisfies the given condition:

1. {1, 1, 2}, {3}, difference = (1 + 1 + 2) – 3 = 4 – 3 = 1.
2. {1, 3} {1, 2}, difference = (1 + 3) – (1 + 2) = 4 – 3 = 1.
3. {1, 3} {1, 2}, difference = (1 + 3) – (1 + 2) = 4 – 3 = 1.

Therefore, the count of ways to split is 3.

Input: arr[] = {1, 2, 3}, K = 2
Output: 1
Explanation:
The only possible split into a pair of subsets satisfying the given condition is {1, 3}, {2}, where the difference = (1 + 3) – 2 = 4 – 2 =2.
Therefore, the count of ways to split is 1.

Naive Approach: The simple approach to solve the given problem is to generate all the possible subsets and store the sum of each subset in an array say subset[]. Then, check if there exist any pair exists in the array subset[] whose difference is K. After checking for all the pairs, print the total count of such pairs as the result.
Time Complexity: O(2N)
Auxiliary Space: O(2N)

Efficient Approach: The above approach can be optimized using the following observations.

Let the sum of the first and second subsets be S1 and S2 respectively, and the sum of array elements be Y.

Now, the sum of both the subsets must be equal to the sum of the array elements.
Therefore, S1 + S2 = Y — (1)
To satisfy the given condition, their difference must be equal to K.
Therefore, S1 – S2 = K — (2)
Adding (1) & (2), the equation obtained is
S1 = (K + Y)/2 — (3)

Therefore, for a pair of subsets having sums S1 and S2, equation (3) must hold true, i.e., the sum of elements of the subset must be equal to (K + Y)/2. Now, the problem reduces to counting the number of subsets with a given sum. This problem can be solved using Dynamic Programming, whose recurrence relation is as follows:

dp[i][C] = dp[i + 1][C – arr[i]] + dp[i + 1][C]

Here, dp[i][C] stores the number of subsets of the subarray arr[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 ith array element in the subset or don’t.

Below is the implementation of the above approach :

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `#define maxN 20``#define maxSum 50``#define minSum 50``#define base 50` `// To store the states of DP``int` `dp[maxN][maxSum + minSum];``bool` `v[maxN][maxSum + minSum];` `// Function to find count of subsets``// with a given sum``int` `findCnt(``int``* arr, ``int` `i,``            ``int` `required_sum,``            ``int` `n)``{``    ``// Base case``    ``if` `(i == n) {``        ``if` `(required_sum == 0)``            ``return` `1;``        ``else``            ``return` `0;``    ``}` `    ``// If an already computed``    ``// subproblem occurs``    ``if` `(v[i][required_sum + base])``        ``return` `dp[i][required_sum + base];` `    ``// Set the state as solved``    ``v[i][required_sum + base] = 1;` `    ``// Recurrence relation``    ``dp[i][required_sum + base]``        ``= findCnt(arr, i + 1,``                  ``required_sum, n)``          ``+ findCnt(arr, i + 1,``                    ``required_sum - arr[i], n);``    ``return` `dp[i][required_sum + base];``}` `// Function to count ways to split array into``// pair of subsets with difference K``void` `countSubsets(``int``* arr, ``int` `K,``                  ``int` `n)``{` `    ``// Store the total sum of``    ``// element of the array``    ``int` `sum = 0;` `    ``// Traverse the array``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Calculate sum of array elements``        ``sum += arr[i];``    ``}` `    ``// Store the required sum``    ``int` `S1 = (sum + K) / 2;` `    ``// Print the number of subsets``    ``// with sum equal to S1``    ``cout << findCnt(arr, 0, S1, n);``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 1, 1, 2, 3 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(``int``);``    ``int` `K = 1;` `    ``// Function Call``    ``countSubsets(arr, K, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.io.*;``import` `java.util.*;` `class` `GFG``{``  ``static` `int` `maxN = ``20``;``  ``static` `int` `maxSum = ``50``;``  ``static` `int` `minSum = ``50``;``  ``static` `int` `Base = ``50``;` `  ``// To store the states of DP``  ``static` `int``[][] dp = ``new` `int``[maxN][maxSum + minSum];``  ``static` `boolean``[][] v = ``new` `boolean``[maxN][maxSum + minSum];` `  ``// Function to find count of subsets``  ``// with a given sum``  ``static` `int` `findCnt(``int``[] arr, ``int` `i,``                     ``int` `required_sum,``                     ``int` `n)``  ``{``    ``// Base case``    ``if` `(i == n) {``      ``if` `(required_sum == ``0``)``        ``return` `1``;``      ``else``        ``return` `0``;``    ``}` `    ``// If an already computed``    ``// subproblem occurs``    ``if` `(v[i][required_sum + Base])``      ``return` `dp[i][required_sum + Base];` `    ``// Set the state as solved``    ``v[i][required_sum + Base] = ``true``;` `    ``// Recurrence relation``    ``dp[i][required_sum + Base]``      ``= findCnt(arr, i + ``1``,``                ``required_sum, n)``      ``+ findCnt(arr, i + ``1``,``                ``required_sum - arr[i], n);``    ``return` `dp[i][required_sum + Base];``  ``}  ` `  ``// Function to count ways to split array into``  ``// pair of subsets with difference K``  ``static` `void` `countSubsets(``int``[] arr, ``int` `K,``                           ``int` `n)``  ``{``    ``// Store the total sum of``    ``// element of the array``    ``int` `sum = ``0``;` `    ``// Traverse the array``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{` `      ``// Calculate sum of array elements``      ``sum += arr[i];``    ``}` `    ``// Store the required sum``    ``int` `S1 = (sum + K) / ``2``;` `    ``// Print the number of subsets``    ``// with sum equal to S1``    ``System.out.print(findCnt(arr, ``0``, S1, n));``  ``} `  `  ``// Driver Code``  ``public` `static` `void` `main(String[] args)``  ``{``    ``int``[] arr = { ``1``, ``1``, ``2``, ``3` `};``    ``int` `N = arr.length;``    ``int` `K = ``1``;` `    ``// Function Call``    ``countSubsets(arr, K, N);``  ``}``}` `// This code is contributed by sanjoy_62.`

## Python3

 `# Python program for the above approach``maxN ``=` `20``;``maxSum ``=` `50``;``minSum ``=` `50``;``Base ``=` `50``;` `# To store the states of DP``dp ``=` `[[``0` `for` `i ``in` `range``(maxSum ``+` `minSum)] ``for` `j ``in` `range``(maxN)];``v ``=` `[[``False` `for` `i ``in` `range``(maxSum ``+` `minSum)] ``for` `j ``in` `range``(maxN)];` `# Function to find count of subsets``# with a given sum``def` `findCnt(arr, i, required_sum, n):``  ` `    ``# Base case``    ``if` `(i ``=``=` `n):``        ``if` `(required_sum ``=``=` `0``):``            ``return` `1``;``        ``else``:``            ``return` `0``;` `    ``# If an already computed``    ``# subproblem occurs``    ``if` `(v[i][required_sum ``+` `Base]):``        ``return` `dp[i][required_sum ``+` `Base];` `    ``# Set the state as solved``    ``v[i][required_sum ``+` `Base] ``=` `True``;` `    ``# Recurrence relation``    ``dp[i][required_sum ``+` `Base] ``=` `findCnt(arr, i ``+` `1``, required_sum, n)\``    ``+` `findCnt(arr, i ``+` `1``, required_sum ``-` `arr[i], n);``    ``return` `dp[i][required_sum ``+` `Base];` `# Function to count ways to split array into``# pair of subsets with difference K``def` `countSubsets(arr, K, n):``  ` `    ``# Store the total sum of``    ``# element of the array``    ``sum` `=` `0``;` `    ``# Traverse the array``    ``for` `i ``in` `range``(n):``      ` `        ``# Calculate sum of array elements``        ``sum` `+``=` `arr[i];` `    ``# Store the required sum``    ``S1 ``=` `(``sum` `+` `K) ``/``/` `2``;` `    ``# Print the number of subsets``    ``# with sum equal to S1``    ``print``(findCnt(arr, ``0``, S1, n));` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``1``, ``1``, ``2``, ``3``];``    ``N ``=` `len``(arr);``    ``K ``=` `1``;` `    ``# Function Call``    ``countSubsets(arr, K, N);` `    ``# This code is contributed by shikhasingrajput`

## C#

 `// C# program for the above approach``using` `System;``class` `GFG {``    ` `    ``static` `int` `maxN = 20;``    ``static` `int` `maxSum = 50;``    ``static` `int` `minSum = 50;``    ``static` `int` `Base = 50;``    ` `    ``// To store the states of DP``    ``static` `int``[,] dp = ``new` `int``[maxN, maxSum + minSum];``    ``static` `bool``[,] v = ``new` `bool``[maxN, maxSum + minSum];``      ` `    ``// Function to find count of subsets``    ``// with a given sum``    ``static` `int` `findCnt(``int``[] arr, ``int` `i,``                ``int` `required_sum,``                ``int` `n)``    ``{``        ``// Base case``        ``if` `(i == n) {``            ``if` `(required_sum == 0)``                ``return` `1;``            ``else``                ``return` `0;``        ``}``      ` `        ``// If an already computed``        ``// subproblem occurs``        ``if` `(v[i, required_sum + Base])``            ``return` `dp[i, required_sum + Base];``      ` `        ``// Set the state as solved``        ``v[i,required_sum + Base] = ``true``;``      ` `        ``// Recurrence relation``        ``dp[i,required_sum + Base]``            ``= findCnt(arr, i + 1,``                      ``required_sum, n)``              ``+ findCnt(arr, i + 1,``                        ``required_sum - arr[i], n);``        ``return` `dp[i,required_sum + Base];``    ``}``      ` `    ``// Function to count ways to split array into``    ``// pair of subsets with difference K``    ``static` `void` `countSubsets(``int``[] arr, ``int` `K,``                      ``int` `n)``    ``{``      ` `        ``// Store the total sum of``        ``// element of the array``        ``int` `sum = 0;``      ` `        ``// Traverse the array``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``      ` `            ``// Calculate sum of array elements``            ``sum += arr[i];``        ``}``      ` `        ``// Store the required sum``        ``int` `S1 = (sum + K) / 2;``      ` `        ``// Print the number of subsets``        ``// with sum equal to S1``        ``Console.Write(findCnt(arr, 0, S1, n));``    ``}  ` `  ``// Driver code``  ``static` `void` `Main()``  ``{``    ``int``[] arr = { 1, 1, 2, 3 };``    ``int` `N = arr.Length;``    ``int` `K = 1;``  ` `    ``// Function Call``    ``countSubsets(arr, K, N);``  ``}``}` `// This code is contributed by divyeshrabadiya07.`

## Javascript

 ``
Output:
`3`

Time Complexity: O(N*K)
Auxiliary Space: O(N*K)

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