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

29 Jun, 2021
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=1Output: 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(2N)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 ith 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++
Java
Python3
C#
Javascript
// 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 diffint 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 Codeint main(){    // Given Input    int diff = 1, n = 4;    int arr[] = { 1, 1, 2, 3 };     // Function Call    cout << countSubset(arr, n, diff);}
// Java program for the above approachimport 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 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 diffdef 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 Codeif __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# 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
<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 difffunction 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 Inputvar diff = 1;var n = 4;var arr = [ 1, 1, 2, 3 ]; // Function Calldocument.write(countSubset(arr, n, diff)); </script>
3
 
Time Complexity: O(S*N), where S = sum of array elements + K/2 Auxiliary Space: O(S*N)
 
mohit kumar 29
ankthon
partition
subset
Arrays
Dynamic Programming
Mathematical
Arrays
Dynamic Programming
Mathematical
subset
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