Minimize difference with 0 after adding or subtracting any element of the given Array

Given an array arr[] of N integers, the task is to find the minimum difference from 0 after adding or subtracting any element of the array from it. 

Examples:

Input: N = 4, arr[] = {1, 2, 1, 3}
Output: 1
Explanation: Add 1 and 2 with 0 and subtract 1 and 3.
So total sum = (1 + 2 – 1 – 3) = -1. Difference is 1.

Input: N = 3, arr[] = {3, 3, 3}
Output: 3
Explanation: No matter which order is chosen, the minimum possible difference is 3.

Input: N = 1, arr[] = {100}
Output: 100
Explanation: There is only one element so difference will be 100

Approach: The problem focusses on forming an Exponential Recursive Tree where there are two possibilities each time. One to add element and one to subtract it. Follow the given steps:

• Create a recursive function minDist having arguments original array arr, iterating index r starting from 0, length of the array N, and integer to calculate current distance k.
• If r becomes equal to n means all elements are traversed so return current distance k.
• Else return minimum of minDist(arr, r+1, N, k+arr[r]) and minDist(arr, r+1, N, k-arr[r]).

Below is the implementation of the above approach.

C++

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

// Function to find the minimum difference
long long minDist(int* arr, int r, int N,
                  long long k)
{
    if (r == N)
        return k;
    else
        return min(abs(minDist(arr, r + 1,
                               N, k
                                      + arr[r])),
                   abs(minDist(arr, r + 1,
                               N, k
                                      - arr[r])));
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 1, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << minDist(arr, 0, N, 0);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG
{

  // Function to find the minimum difference
  static long minDist(int arr[ ], int r, int N, long k)
  {
    if (r == N)
      return k;
    else
      return Math.min(Math.abs(minDist(arr, r + 1, N, k + arr[r])), Math.abs(minDist(arr, r + 1, N, k - arr[r])));
  }

  public static void main (String[] args)
  {
    int arr[] = { 1, 2, 1, 3 };
    int N = arr.length;
    System.out.print(minDist(arr, 0, N, 0));
  }
}

// This code is contributed by hrithikgarg03188

Python3

# Python code to implement above approach

# Function to find the minimum difference
def minDist(arr, r, N, k):
    if (r == N):
        return k
    else:
        return min(abs(minDist(arr, r + 1, N, k + arr[r])), abs(minDist(arr, r + 1, N, k - arr[r])))

# Driver code
arr = [1, 2, 1, 3]
N = len(arr)
print(minDist(arr, 0, N, 0))

# This code is contributed by gfgking

C#

// C# program for the above approach
using System;
class GFG {

    // Function to find the minimum difference
    static long minDist(int[] arr, int r, int N, long k)
    {
        if (r == N)
            return k;
        else
            return Math.Min(Math.Abs(minDist(arr, r + 1, N,
                                             k + arr[r])),
                            Math.Abs(minDist(arr, r + 1, N,
                                             k - arr[r])));
    }

    public static void Main()
    {
        int[] arr = { 1, 2, 1, 3 };
        int N = arr.Length;
        Console.Write(minDist(arr, 0, N, 0));
    }
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
    // JavaScript code to implement above approach

    // Function to find the minimum difference
    const minDist = (arr, r, N, k) => {
        if (r == N)
            return k;
        else
            return Math.min(Math.abs(minDist(arr, r + 1,
                N, k
            + arr[r])),
                Math.abs(minDist(arr, r + 1,
                    N, k
                - arr[r])));
    }

    // Driver code

    let arr = [1, 2, 1, 3];
    let N = arr.length;
    document.write(minDist(arr, 0, N, 0));

// This code is contributed by rakeshsahni

</script>

Output

1

Time Complexity: O(2^N)
Auxiliary Space: O(1)