Minimize Subset sum difference after deleting K elements of Array

Last Updated : 26 Apr, 2022

Given an array arr[] containing N integers and an integer K, the task is to minimize the subset-sum difference after removing K elements from the given array.

Note: The subsets must be non-empty.

Examples:

Input: arr[] = {7, 9, 5, 8, 1, 3}, K = 2
Output: -23
Explanation: Remove 5 and 3 from the array and form subsets [1] and [7, 8, 9].
The subset difference of these subset is 1 - 24 = -23 that is the minimum possible difference.

Input: arr[] = {7, 9, 5, 8}, K = 3
Output: -1
Explanation: If 3 elements are removed it is not possible to form two non-empty subsets.

Approach: The problem can be solved using recursion based on the following idea:

For each element of the array there are 3 choices: either it can be deleted or be a part of any of the two subsets. Form all the possible subsets after deleting K elements and find the one where the difference of subset sums is the minimum.

Instead of considering the three options for each array element, only two options can be considered: either deleted or part of first subset (because if the sum of first subset elements are known then the other subset sum = array sum - first subset sum)

Follow the steps mentioned to solve the problem:

Calculate the sum of all the array elements.
Iterate recursively for each array element:
- Delete this if K elements are not already deleted and continue for the next index.
- Consider this as part of one subset.
- If the whole array is traversed, check the difference between two subset sums.
The minimum difference is the required answer.

Below is the implementation for the above-mentioned approach:

C++

// C++ code to implement the above approach
#include <bits/stdc++.h>
using namespace std;

int mini = INT_MAX;

// Function to form all possible subsets
void subset(vector<int> arr, int index, int sum, int total)
{
    if (index >= arr.size()) {
        if (sum != 0)
            mini = min(mini, sum - (total - sum));
        return;
    }
    subset(arr, index + 1, sum + arr[index], total);
    subset(arr, index + 1, sum, total);
}

// Function to get the minimum difference
static void print(int index, vector<int>& arr, int total,
                  vector<int>& ans)
{
    if (total == 0) {
        int sum = 0;
        for (int elements : ans) {
            sum += elements;
        }

        // Function call to form subset
        subset(ans, 0, 0, sum);
        return;
    }
    if (index >= arr.size()) {
        return;
    }
    ans.push_back(arr[index]);
    print(index + 1, arr, total - 1, ans);
    ans.pop_back();
    print(index + 1, arr, total, ans);
}

// Utility function to solve the problem
void solve(vector<int>& arr, int N)
{
    int selected = arr.size() - N;

    if (selected <= 1) {
        cout << -1;
        return;
    }
    vector<int> ans;
    print(0, arr, selected, ans);
}

// Driver code
int main()
{
    vector<int> arr = { 7, 9, 5, 8, 1, 3 };
    int N = 2;
    solve(arr, N);
    cout << (mini);
    return 0;
}

// This code is contributed by rakeshsahni

Java

// Java code to implement the above approach

import java.io.*;
import java.util.*;

class GFG {

    static int min = Integer.MAX_VALUE;

    // Function to form all possible subsets
    static void subset(List<Integer> arr,
                       int index,
                       int sum, int total)
    {
        if (index >= arr.size()) {
            if (sum != 0)
                min = Math.min(min,
                               sum - (total - sum));
            return;
        }
        subset(arr, index + 1,
               sum + arr.get(index), total);
        subset(arr, index + 1, sum, total);
    }

    // Function to get the minimum difference
    static void print(int index, int arr[],
                      int total,
                      List<Integer> ans)
    {
        if (total == 0) {
            int sum = 0;
            for (int elements : ans) {
                sum += elements;
            }

            // Function call to form subset
            subset(ans, 0, 0, sum);
            return;
        }
        if (index >= arr.length) {
            return;
        }
        ans.add(arr[index]);
        print(index + 1, arr, total - 1, ans);
        ans.remove(ans.size() - 1);
        print(index + 1, arr, total, ans);
    }

    // Utility function to solve the problem
    public static void solve(int arr[], int N)
    {
        int selected = arr.length - N;

        if (selected <= 1) {
            System.out.println(-1);
            return;
        }
        List<Integer> ans = new ArrayList<>();
        print(0, arr, selected, ans);
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 7, 9, 5, 8, 1, 3 };
        int N = 2;

        solve(arr, N);
        System.out.println(min);
    }
}

Python3

 # Python code to implement the above approach
import sys

# Function to form all possible subsets
def subset(arr,index,sum,total):
    global mini

    if (index >= len(arr)):
        if (sum != 0):
            mini = min(mini, sum - (total - sum))
        return

    subset(arr, index + 1, sum + arr[index], total)
    subset(arr, index + 1, sum, total)

# Function to get the minimum difference
def Print(index,arr,total,ans):

    if (total == 0):
        sum = 0
        for elements in ans:
            sum += elements

        # Function call to form subset
        subset(ans, 0, 0, sum)
        return

    if (index >= len(arr)):
        return

    ans.append(arr[index])
    Print(index + 1, arr, total - 1, ans)
    ans.pop()
    Print(index + 1, arr, total, ans)

# Utility function to solve the problem
def solve(arr,N):

    selected = len(arr) - N

    if (selected <= 1):
        print(-1)
        return

    ans = []
    Print(0, arr, selected, ans)

# Driver code

if __name__=="__main__":

    mini = sys.maxsize
    arr = [ 7, 9, 5, 8, 1, 3 ]
    N = 2
    solve(arr, N)
    print(mini)
    
# This code is contributed by shinjanpatra

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;

class GFG
{

  static int min = Int32.MaxValue;

  // Function to form all possible subsets
  static void subset(List<int> arr,
                     int index,
                     int sum, int total)
  {
    if (index >= arr.Count) {
      if (sum != 0)
        min = Math.Min(min,
                       sum - (total - sum));
      return;
    }
    subset(arr, index + 1,
           sum + arr[index], total);
    subset(arr, index + 1, sum, total);
  }

  // Function to get the minimum difference
  static void print(int index, int[] arr,
                    int total,
                    List<int> ans)
  {
    if (total == 0) {
      int sum = 0;
      foreach (int elements in ans) {
        sum += elements;
      }

      // Function call to form subset
      subset(ans, 0, 0, sum);
      return;
    }
    if (index >= arr.Length) {
      return;
    }
    ans.Add(arr[index]);
    print(index + 1, arr, total - 1, ans);
    ans.RemoveAt(ans.Count - 1);
    print(index + 1, arr, total, ans);
  }

  // Utility function to solve the problem
  public static void solve(int[] arr, int N)
  {
    int selected = arr.Length - N;

    if (selected <= 1) {
      Console.Write(-1);
      return;
    }
    List<int> ans = new List<int>();
    print(0, arr, selected, ans);
  }

  // Driver Code
  public static void Main()
  {
    int[] arr = { 7, 9, 5, 8, 1, 3 };
    int N = 2;

    solve(arr, N);
    Console.Write(min);
  }
}

// This code is contributed by sanjoy_62.

JavaScript

<script>

 // JavaScript code to implement the above approach
let mini = Number.MAX_VALUE;

// Function to form all possible subsets
function subset(arr,index,sum,total)
{
    if (index >= arr.length) {
        if (sum != 0)
            mini = Math.min(mini, sum - (total - sum));
        return;
    }
    subset(arr, index + 1, sum + arr[index], total);
    subset(arr, index + 1, sum, total);
}

// Function to get the minimum difference
function print(index,arr,total,ans)
{
    if (total == 0) {
        sum = 0;
        for (let elements of ans) {
            sum += elements;
        }

        // Function call to form subset
        subset(ans, 0, 0, sum);
        return;
    }
    if (index >= arr.length) {
        return;
    }
    ans.push(arr[index]);
    print(index + 1, arr, total - 1, ans);
    ans.pop();
    print(index + 1, arr, total, ans);
}

// Utility function to solve the problem
function solve(arr,N)
{
    let selected = arr.length - N;

    if (selected <= 1) {
        document.write(-1);
        return;
    }
    let ans = [];
    print(0, arr, selected, ans);
}

// Driver code

let arr = [ 7, 9, 5, 8, 1, 3 ];
let N = 2;
solve(arr, N);
document.write(mini);

// This code is contributed by shinjanpatra

</script>

Output

-23

Time Complexity: O(2^N)
Auxiliary Space: O(N)

Maximum possible middle element of the array after deleting exactly k elements

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