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

Last Updated : 29 Apr, 2023

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:

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

Efficient approach: space optimization

To optimize space complexity we only need to keep track of the values of the previous row in the 2D array to compute the values of the current row. Hence, we can replace the 2D array t[n + 1][sum + 1] with a 1D array dp[sum + 1].

Implementation Steps :

Create vector Dp of size sum + 1 and initialize it with 0.
Now initialize DP with Base Case dp[0] =1.
To calculate answer iterate over subproblems with the help of nested loops and get the current value from previous computation.
At last return the answer stored in dp[sum].

Implementation:

C++

   // C++ program for 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)
{ 
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
    sum += diff;
    sum = sum / 2;
      
      // Initializing the vector Dp
    int dp[sum + 1] = {0};
   
      // Base Case
    dp[0] = 1;
      
      // iterate over subproblems to get the current computation
    for (int i = 0; i < n; i++) {
        for (int j = sum; j >= arr[i]; j--) {
              // update DP from previous values
            dp[j] += dp[j - arr[i]];
        }
    }
      
      // return answer
    return dp[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);
}
 
// this code is contributed by bhardwajji
 
  

Java

   // Java program for above approach
import java.util.*;
 
public class Main
{
   
    // 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
    public static int countSubset(int arr[], int n, int diff)
    {
        int sum = 0;
        for (int i = 0; i < n; i++)
            sum += arr[i];
        sum += diff;
        sum = sum / 2;
 
        // Initializing the vector Dp
        int dp[] = new int[sum + 1];
 
        // Base Case
        dp[0] = 1;
 
        // iterate over subproblems to get the current computation
        for (int i = 0; i < n; i++) {
            for (int j = sum; j >= arr[i]; j--) {
                // update DP from previous values
                dp[j] += dp[j - arr[i]];
            }
        }
 
        // return answer
        return dp[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.println(countSubset(arr, n, diff));
    }
}
 
  

Python3

   # 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):
    # Calculating the sum of all elements
    sum = 0
    for i in range(n):
        sum += arr[i]
    sum += diff
 
    # If sum is odd, then no such subset can exist
    if sum % 2 != 0:
        return 0
 
    # Initializing the vector Dp
    dp = [0] * (sum // 2 + 1)
 
    # Base Case
    dp[0] = 1
 
    # iterate over subproblems to get the current computation
    for i in range(n):
        for j in range(sum // 2, arr[i] - 1, -1):
            # update DP from previous values
            dp[j] += dp[j - arr[i]]
 
    # return answer
    return dp[sum // 2]
 
 
# Given Input
diff = 1
n = 4
arr = [1, 1, 2, 3]
 
# Function Call
print(countSubset(arr, n, diff))
 
  

C#

   using System;
 
class MainClass
{
   
  // 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)
  {
    int sum = 0;
    for (int i = 0; i < n; i++) {
      sum += arr[i];
    }
    sum += diff;
    sum = sum / 2;
 
    // Initializing the vector Dp
    int[] dp = new int[sum + 1];
 
    // Base Case
    dp[0] = 1;
 
    // Iterate over subproblems to get the current
    // computation
    for (int i = 0; i < n; i++) {
      for (int j = sum; j >= arr[i]; j--) {
        // Update DP from previous values
        dp[j] += dp[j - arr[i]];
      }
    }
 
    // Return answer
    return dp[sum];
  }
 
  // Driver Code
  public static void Main()
  {
    // Given Input
    int diff = 1, n = 4;
    int[] arr = { 1, 1, 2, 3 };
 
    // Function Call
    Console.WriteLine(countSubset(arr, n, diff));
  }
}
 
  

Javascript

   // 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) {
  let sum = 0;
  for (let i = 0; i < n; i++) {
    sum += arr[i];
  }
  sum += diff;
  sum = Math.floor(sum / 2);
 
  // Initializing the vector Dp
  let dp = new Array(sum + 1).fill(0);
 
  // Base Case
  dp[0] = 1;
 
  // Iterate over subproblems to get the current
  // computation
  for (let i = 0; i < n; i++) {
    for (let j = sum; j >= arr[i]; j--) {
      // Update DP from previous values
      dp[j] += dp[j - arr[i]];
    }
  }
 
  // Return answer
  return dp[sum];
}
 
// Driver Code
let diff = 1,
  n = 4;
let arr = [1, 1, 2, 3];
 
// Function Call
console.log(countSubset(arr, n, diff));
 
  

Output

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

Maximum subsequence sum with adjacent elements having atleast K difference in index

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

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