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

• Difficulty Level : Medium
• 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:

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 â€” (1)
To satisfy the given condition, their difference must be equal to K.
Therefore, S1 – S2 = K â€” (2)
Adding (1) & (2), the equation obtained is
S1 = (K + Y)/2 â€” (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 â€“ arr[i]] + dp[i + 1][C]

Here, dp[i][C] stores the number of subsets of the subarray arr[i â€¦ 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 donâ€™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 DPint dp[maxN][maxSum + minSum];bool v[maxN][maxSum + minSum]; // Function to find count of subsets// with a given sumint 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 Kvoid 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 Codeint 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 approachimport 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 approachmaxN = 20;maxSum = 50;minSum = 50;Base = 50; # To store the states of DPdp = [[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 sumdef 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 Kdef 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 Codeif __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 approachusing 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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