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Minimize difference between sum of two K-length subsets
  • Difficulty Level : Medium
  • Last Updated : 23 Mar, 2021

Given an array arr[] consisting of N positive integers and a positive integer K, the task is to find the minimum difference between the sum of elements present in two subsets of size K, such that each array element can occur in at most 1 subset.

Examples:

Input: arr[] = {12, 3, 5, 6, 7, 17}, K = 2
Output: 1
Explanation: Considering two subsets {5, 6} and {3, 7}, the difference between their sum = (5 + 6) – (3 + 7) = (11 – 10) = 1, which is minimum possible.

Input: arr[] = {10, 10, 10, 10, 10, 2}, K = 3
Output: 8
Explanation: Considering two subsets {10, 10, 10} and {10, 10, 2}, the difference between their sum = (10 + 10 + 10) – (10 + 10 + 2) = (30 – 22) = 8.

Naive Approach: The simplest approach to solve the given problem is to generate all possible subsets of size K and choose every pair of subsets of size K such that each array element can occur in at most 1 subset. After finding all possible pairs of subsets, print the minimum difference between the sum of elements for each pair of subsets. 



Time Complexity: O(3N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized using Dynamic Programming. To implement this approach, the idea is to use 3D-array, say dp[][][], to store the states of each recursive call
Follow the steps below to solve the problem:

  • Initialize an auxiliary array, say dp[][][], where dp[i][S1][S2] stores the sum of the two subsets formed till every ith index.
  • Declare a recursive function, say minimumDifference(arr, N, K1, K2, S1, S2), that accepts an array, the current size of the array, the number of elements stored in the subsets(K1 and K2), and the sum of elements in each subset as parameters. 
    • Base Case: If the number of elements in the array is N, and the value of K1 and K2 is 0, then return the absolute difference between S1 and S2.
    • If the result of the current state is already computed, return the result.
    • Store the value returned by including the Nth element in the first subset and recursively call for the next iterations as minimumDifference(arr, N – 1, K1 – 1, K2, S1 + arr[N – 1], S2).
    • Store the value returned by including the Nth element in the first subset and recursively call for the next iterations as minimumDifference(arr, N – 1, K1, K2 – 1, S1, S2 + arr[N – 1]).
    • Store the value returned by not including the Nth element in either of the subsets and recursively call for the next iterations as minimumDifference(arr, N – 1, K1, K2, S1, S2).
    • Store the minimum of all the values returned by the above three recursive calls in the current state i.e., dp[N][S1][S2], and return its value.
  • After completing the above steps, print the value returned by the function minimumDifference(arr, N, K, K, 0, 0).

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Stores the values at recursive states
int dp[100][100][100];
 
// Function to find the minimum difference
// between sum of two K-length subsets
int minSumDifference(int* arr, int n,
                     int k1, int k2,
                     int sum1, int sum2)
{
    // Base Case
    if (n < 0) {
 
        // If k1 and k2 are 0, then
        // return the absolute difference
        // between sum1 and sum2
        if (k1 == 0 && k2 == 0) {
            return abs(sum1 - sum2);
        }
 
        // Otherwise, return INT_MAX
        else {
            return INT_MAX;
        }
    }
 
    // If the result is already
    // computed, return the result
    if (dp[n][sum1][sum2] != -1) {
        return dp[n][sum1][sum2];
    }
 
    // Store the 3 options
    int op1 = INT_MAX;
    int op2 = INT_MAX;
    int op3 = INT_MAX;
 
    // Including the element in first subset
    if (k1 > 0) {
        op1 = minSumDifference(arr, n - 1,
                               k1 - 1, k2,
                               sum1 + arr[n],
                               sum2);
    }
 
    // Including the element in second subset
    if (k2 > 0) {
        op2 = minSumDifference(arr, n - 1,
                               k1, k2 - 1, sum1,
                               sum2 + arr[n]);
    }
 
    // Not including the current element
    // in both the subsets
    op3 = minSumDifference(arr, n - 1,
                           k1, k2,
                           sum1, sum2);
 
    // Store minimum of 3 values obtained
    dp[n][sum1][sum2] = min(op1,
                            min(op2,
                                op3));
 
    // Return the value for
    // the current states
    return dp[n][sum1][sum2];
}
 
// Driver Code
int main()
{
    int arr[] = { 12, 3, 5, 6, 7, 17 };
    int K = 2;
    int N = sizeof(arr) / sizeof(arr[0]);
 
    memset(dp, -1, sizeof(dp));
 
    cout << minSumDifference(arr, N - 1,
                             K, K, 0, 0);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
     
// Stores the values at recursive states
static int[][][] dp = new int[100][100][100];
 
// Function to find the minimum difference
// between sum of two K-length subsets
static int minSumDifference(int[] arr, int n,
                            int k1, int k2,
                            int sum1, int sum2)
{
     
    // Base Case
    if (n < 0)
    {
         
        // If k1 and k2 are 0, then
        // return the absolute difference
        // between sum1 and sum2
        if (k1 == 0 && k2 == 0)
        {
            return Math.abs(sum1 - sum2);
        }
  
        // Otherwise, return INT_MAX
        else
        {
            return Integer.MAX_VALUE;
        }
    }
  
    // If the result is already
    // computed, return the result
    if (dp[n][sum1][sum2] != -1)
    {
        return dp[n][sum1][sum2];
    }
  
    // Store the 3 options
    int op1 = Integer.MAX_VALUE;
    int op2 = Integer.MAX_VALUE;
    int op3 = Integer.MAX_VALUE;
  
    // Including the element in first subset
    if (k1 > 0)
    {
        op1 = minSumDifference(arr, n - 1,
                               k1 - 1, k2,
                               sum1 + arr[n],
                               sum2);
    }
  
    // Including the element in second subset
    if (k2 > 0)
    {
        op2 = minSumDifference(arr, n - 1,
                               k1, k2 - 1, sum1,
                               sum2 + arr[n]);
    }
  
    // Not including the current element
    // in both the subsets
    op3 = minSumDifference(arr, n - 1,
                           k1, k2,
                           sum1, sum2);
  
    // Store minimum of 3 values obtained
    dp[n][sum1][sum2] = Math.min(op1,
                        Math.min(op2, op3));
  
    // Return the value for
    // the current states
    return dp[n][sum1][sum2];
}
  
// Driver Code
public static void main (String[] args)
{
    int arr[] = { 12, 3, 5, 6, 7, 17 };
    int K = 2;
    int N = arr.length;
     
    for(int[][] row:dp)
    {
        for(int[] innerrow : row)
        {
            Arrays.fill(innerrow,-1);
        }
    }
     
    System.out.println(minSumDifference(arr, N - 1, K,
                                        K, 0, 0));
}
}
 
// This code is contributed by avanitrachhadiya2155

Python3




# Python3 program for the above approach
import sys
 
# Stores the values at recursive states
dp = [[[ -1 for i in range(100)]
            for i in range(100)]
            for i in range(100)]
 
# Function to find the minimum difference
# between sum of two K-length subsets
def minSumDifference(arr, n, k1, k2, sum1, sum2):
     
    global dp
     
    # Base Case
    if (n < 0):
 
        # If k1 and k2 are 0, then
        # return the absolute difference
        # between sum1 and sum2
        if (k1 == 0 and k2 == 0):
            return abs(sum1 - sum2)
             
        # Otherwise, return INT_MAX
        else:
            return sys.maxsize + 1
 
    # If the result is already
    # computed, return the result
    if (dp[n][sum1][sum2] != -1):
        return dp[n][sum1][sum2]
 
    # Store the 3 options
    op1 = sys.maxsize + 1
    op2 = sys.maxsize + 1
    op3 = sys.maxsize + 1
 
    # Including the element in first subset
    if (k1 > 0):
        op1 = minSumDifference(arr, n - 1, k1 - 1,
                             k2, sum1 + arr[n], sum2)
 
    # Including the element in second subset
    if (k2 > 0):
        op2 = minSumDifference(arr, n - 1, k1, k2 - 1,
                           sum1, sum2 + arr[n])
 
    # Not including the current element
    # in both the subsets
    op3 = minSumDifference(arr, n - 1, k1,
                           k2, sum1, sum2)
 
    # Store minimum of 3 values obtained
    dp[n][sum1][sum2] = min(op1, min(op2, op3))
 
    # Return the value for
    # the current states
    return dp[n][sum1][sum2]
 
# Driver Code
if __name__ == '__main__':
     
    arr = [12, 3, 5, 6, 7, 17]
    K = 2
    N = len(arr)
 
    #memset(dp, -1, sizeof(dp))
 
    print (minSumDifference(arr, N - 1, K, K, 0, 0))
 
# This code is contributed by mohit kumar 29

C#




// C# program for the above approach
using System;
 
class GFG{
     
// Stores the values at recursive states
static int[,,] dp = new int[100, 100, 100];
  
// Function to find the minimum difference
// between sum of two K-length subsets
static int minSumDifference(int[] arr, int n,
                            int k1, int k2,
                            int sum1, int sum2)
{
     
    // Base Case
    if (n < 0)
    {
         
        // If k1 and k2 are 0, then
        // return the absolute difference
        // between sum1 and sum2
        if (k1 == 0 && k2 == 0)
        {
            return Math.Abs(sum1 - sum2);
        }
   
        // Otherwise, return INT_MAX
        else
        {
            return Int32.MaxValue;
        }
    }
   
    // If the result is already
    // computed, return the result
    if (dp[n, sum1, sum2] != -1)
    {
        return dp[n, sum1, sum2];
    }
   
    // Store the 3 options
    int op1 = Int32.MaxValue;
    int op2 = Int32.MaxValue;
    int op3 = Int32.MaxValue;
   
    // Including the element in first subset
    if (k1 > 0)
    {
        op1 = minSumDifference(arr, n - 1,
                               k1 - 1, k2,
                               sum1 + arr[n],
                               sum2);
    }
   
    // Including the element in second subset
    if (k2 > 0)
    {
        op2 = minSumDifference(arr, n - 1,
                               k1, k2 - 1, sum1,
                               sum2 + arr[n]);
    }
   
    // Not including the current element
    // in both the subsets
    op3 = minSumDifference(arr, n - 1,
                           k1, k2,
                           sum1, sum2);
   
    // Store minimum of 3 values obtained
    dp[n, sum1, sum2] = Math.Min(op1,
                        Math.Min(op2, op3));
   
    // Return the value for
    // the current states
    return dp[n, sum1, sum2];
}
   
// Driver Code
static public void Main()
{
    int[] arr = { 12, 3, 5, 6, 7, 17 };
    int K = 2;
    int N = arr.Length;
      
    for(int i = 0; i < 100; i++)
    {
        for(int j = 0; j < 100; j++)
        {
            for(int k = 0; k < 100; k++)
            {
                dp[i, j, k] = -1;
            }
        }
    }
     
    Console.WriteLine(minSumDifference(arr, N - 1, K,
                                        K, 0, 0));
}
}
 
// This code is contributed by rag2127
Output: 
1

 

Time Complexity: O(N*S2), where S is the sum of elements of the given array
Auxiliary Space: O(N*S2)

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