Minimum value to add to arr[i] so that an array can be split at index i with equal sum

Given an array arr[] of integers, the task is to find the minimum non-negative integer k such that there exists an index j in the given array such that when arr[j] is updated as arr[j] + k, the sum of elements of an array from index arr[0] to arr[j] is equal to the sum of elements from arr[j + 1] to arr[n – 1] i.e. 

arr[0] + arr[1] + … + arr[j] = arr[j + 1] + arr[j + 2] + … + arr[n – 1]

If no such k exists then print -1.
Examples:

Input: arr[] = {6, 7, 1, 3, 8, 2, 4} 
Output: 3 
If 3 is added to 1 sum of elements from index 0 to 2 and 3 to 6 will be equal to 17.
Input: arr[] = {7, 3} 
Output: -1 

A simple approach is to run two loops. For every element, find the difference between sums of elements on the left and right. Finally, return the minimum difference between the two sums.
An efficient approach: is to first calculate the prefix sum and store in an array pre[] where pre[i] stores the sum of array elements from arr[0] to arr[i]. For each index, if the sum of elements left to it (including the element itself i.e. pre[i]) is less than or equal to the sum of right elements (pre[n – 1] – pre[i]) then update the value of k as min(k, (pre[n – 1] – pre[i]) – pre[i])
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the minimum value k to be added
int FindMinNum(int arr[], int n)
{

    // Array to store prefix sum
    int pre[n];

    // Initialize the prefix value for first index
    // as the first element of the array
    pre[0] = arr[0];

    // Compute the prefix sum for rest of the indices
    for (int i = 1; i < n; i++)
        pre[i] = pre[i - 1] + arr[i];

    int k = INT_MAX;

    for (int i = 0; i < n - 1; i++) {

        // Sum of elements from arr[i + 1] to arr[n - 1]
        int rightSum = pre[n - 1] - pre[i];

        // If sum on the right side of the ith element
        // is greater than or equal to the sum on the
        // left side then update the value of k
        if (rightSum >= pre[i])
            k = min(k, rightSum - pre[i]);
    }

    if (k != INT_MAX)
        return k;

    return -1;
}

// Driver code
int main()
{
    int arr[] = { 6, 7, 1, 3, 8, 2, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << FindMinNum(arr, n);
    return 0;
}

Java

// Java implementation of the approach
class GfG
{

// Function to return the minimum value k to be added
static int FindMinNum(int arr[], int n)
{

    // Array to store prefix sum
    int pre[] = new int[n];

    // Initialize the prefix value for first index
    // as the first element of the array
    pre[0] = arr[0];

    // Compute the prefix sum for rest of the indices
    for (int i = 1; i < n; i++)
        pre[i] = pre[i - 1] + arr[i];

    int k = Integer.MAX_VALUE;

    for (int i = 0; i < n - 1; i++)
    {

        // Sum of elements from arr[i + 1] to arr[n - 1]
        int rightSum = pre[n - 1] - pre[i];

        // If sum on the right side of the ith element
        // is greater than or equal to the sum on the
        // left side then update the value of k
        if (rightSum >= pre[i])
            k = Math.min(k, rightSum - pre[i]);
    }

    if (k != Integer.MAX_VALUE)
        return k;

    return -1;
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 6, 7, 1, 3, 8, 2, 4 };
    int n = arr.length;
    System.out.println(FindMinNum(arr, n));
}
}

// This code is contributed by Prerna Saini

Python3

# Python 3 implementation of the approach
import sys

# Function to return the minimum
# value k to be added
def FindMinNum(arr, n):

    # Array to store prefix sum
    pre = [0 for i in range(n)]

    # Initialize the prefix value for first
    # index as the first element of the array
    pre[0] = arr[0]

    # Compute the prefix sum for rest
    # of the indices
    for i in range(1, n, 1):
        pre[i] = pre[i - 1] + arr[i]

    k = sys.maxsize

    for i in range(n - 1):

        # Sum of elements from arr[i + 1] to arr[n - 1]
        rightSum = pre[n - 1] - pre[i]

        # If sum on the right side of the ith element
        # is greater than or equal to the sum on the
        # left side then update the value of k
        if (rightSum >= pre[i]):
            k = min(k, rightSum - pre[i])

    if (k != sys.maxsize):
        return k

    return -1

# Driver code
if __name__ == '__main__':
    arr = [6, 7, 1, 3, 8, 2, 4]
    n = len(arr)
    print(FindMinNum(arr, n))

# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach
using System;

class GfG
{

    // Function to return the minimum value k to be added
    static int FindMinNum(int []arr, int n)
    {

        // Array to store prefix sum
        int []pre = new int[n];

        // Initialize the prefix value for first index
        // as the first element of the array
        pre[0] = arr[0];

        // Compute the prefix sum for rest of the indices
        for (int i = 1; i < n; i++)
            pre[i] = pre[i - 1] + arr[i];

        int k = int.MaxValue;

        for (int i = 0; i < n - 1; i++)
        {

            // Sum of elements from arr[i + 1] to arr[n - 1]
            int rightSum = pre[n - 1] - pre[i];

            // If sum on the right side of the ith element
            // is greater than or equal to the sum on the
            // left side then update the value of k
            if (rightSum >= pre[i])
                k = Math.Min(k, rightSum - pre[i]);
        }

        if (k != int.MaxValue)
            return k;

        return -1;
    }

    // Driver code
    public static void Main()
    {
        int []arr = { 6, 7, 1, 3, 8, 2, 4 };
        int n = arr.Length;

        Console.WriteLine(FindMinNum(arr, n));
    }
}

// This code is contributed by Ryuga

PHP

<?php
// PHP implementation of the approach

// Function to return the minimum
// value k to be added
function FindMinNum($arr, $n)
{

    // Array to store prefix sum
    $pre = array();

    // Initialize the prefix value for first index
    // as the first element of the array
    $pre[0] = $arr[0];

    // Compute the prefix sum for
    // rest of the indices
    for ($i = 1; $i < $n; $i++)
        $pre[$i] = $pre[$i - 1] + $arr[$i];

    $k = PHP_INT_MAX;

    for ($i = 0; $i < $n - 1; $i++)
    {

        // Sum of elements from arr[i + 1] to arr[n - 1]
        $rightSum = $pre[$n - 1] - $pre[$i];

        // If sum on the right side of the ith element
        // is greater than or equal to the sum on the
        // left side then update the value of k
        if ($rightSum >= $pre[$i])
            $k = min($k, $rightSum - $pre[$i]);
    }

    if ($k != PHP_INT_MAX)
        return $k;

    return -1;
}

// Driver code
$arr = array(6, 7, 1, 3, 8, 2, 4);
$n = sizeof($arr);
echo FindMinNum($arr, $n);

// This code is contributed by Akanksha Rai
?>

Javascript

<script>
//Javascript Implementation

// Function to return the minimum value k to be added
function FindMinNum(arr, n)
{

    // Array to store prefix sum
    var pre = new Array(n);

    // Initialize the prefix value for first index
    // as the first element of the array
    pre[0] = arr[0];

    // Compute the prefix sum for rest of the indices
    for (var i = 1; i < n; i++)
        pre[i] = pre[i - 1] + arr[i];

    var k = Number.MAX_VALUE;

    for (var i = 0; i < n - 1; i++) {

        // Sum of elements from arr[i + 1] to arr[n - 1]
        var rightSum = pre[n - 1] - pre[i];

        // If sum on the right side of the ith element
        // is greater than or equal to the sum on the
        // left side then update the value of k
        if (rightSum >= pre[i])
            k = Math.min(k, rightSum - pre[i]);
    }

    if (k != Number.MAX_VALUE)
        return k;

    return -1;
}

/* Driver code*/
var arr = [6, 7, 1, 3, 8, 2, 4];
var n = arr.length;
document.write(FindMinNum(arr, n))
// This code is contributed by shubhamsingh10
</script>

Output

3

Time Complexity : O(n) 
Auxiliary Space : O(n)
Further optimization: We can avoid the use of extra space using the below steps. 

1. First, compute the sum of all the elements and store it in a variable sum. 
2. Iterate a loop for every index where the left sum at any index can be computed by keep on adding the current elements to the leftsum variable. 
3. The rightsum can be calculated by keep on subtracting the elements at every index from the sum variable.
4. For any ith index if we find that the rightsum is greater than the leftsum then we update the value of k.

Below is the code for the above approach.

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the minimum value k to be added
int FindMinNum(int arr[], int n)
{
    int sum = 0; // initialize sum of whole array
    int leftsum = 0; // initialize leftsum

    int k = INT_MAX;

    /* Find sum of the whole array */
    for (int i = 0; i < n; ++i)
        sum += arr[i];

    for (int i = 0; i < n; ++i) {
        sum -= arr[i]; // sum is now right sum for index i
        leftsum += arr[i]; // add current element to leftsum

        // If sum on the right side of the ith element
        // is greater than or equal to the sum on the
        // left side then update the value of k
        if (sum >= leftsum)
            k = min(k, sum - leftsum);
    }

    if (k != INT_MAX)
        return k;

    return -1;
}

// Driver code
int main()
{
    int arr[] = { 6, 7, 1, 3, 8, 2, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << FindMinNum(arr, n);
    return 0;
}

// This code is contributed by Pushpesh raj

Java

// Java implementation of the approach
class GfG {

    // Function to return the minimum value k to be added
    static int FindMinNum(int arr[], int n)
    {
        int sum = 0; // initialize sum of whole array
        int leftsum = 0; // initialize leftsum

        int k = Integer.MAX_VALUE;

        /* Find sum of the whole array */
        for (int i = 0; i < n; ++i)
            sum += arr[i];

        for (int i = 0; i < n; ++i) {
            sum -= arr[i]; // sum is now right sum for index
                           // i
            leftsum
                += arr[i]; // add current element to leftsum

            // If sum on the right side of the ith element
            // is greater than or equal to the sum on the
            // left side then update the value of k
            if (sum >= leftsum)
                k = Math.min(k, sum - leftsum);
        }

        if (k != Integer.MAX_VALUE)
            return k;

        return -1;
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 6, 7, 1, 3, 8, 2, 4 };
        int n = arr.length;
        System.out.println(FindMinNum(arr, n));
    }
}

// This code is contributed by Pushpesh Raj.

Python3

import sys
import math

# Python 3 implementation of the approach
class GfG :

    # Function to return the minimum value k to be added
    @staticmethod
    def  FindMinNum( arr,  n) :
        sum = 0

        # initialize sum of whole array
        leftsum = 0

        # initialize leftsum
        k = sys.maxsize

        # Find sum of the whole array
        i = 0
        while (i < n) :
            sum += arr[i]
            i += 1
        i = 0
        while (i < n) :
            sum -= arr[i]

            # sum is now right sum for index
            # i
            leftsum += arr[i]

            # add current element to leftsum
            # If sum on the right side of the ith element
            # is greater than or equal to the sum on the
            # left side then update the value of k
            if (sum >= leftsum) :
                k = min(k,sum - leftsum)
            i += 1
        if (k != sys.maxsize) :
            return k
        return -1

    # Driver code
    @staticmethod
    def main( args) :
        arr = [6, 7, 1, 3, 8, 2, 4]
        n = len(arr)
        print(GfG.FindMinNum(arr, n))

if __name__=="__main__":
    GfG.main([])

   # This code is contributed by utkarshshirode02.

C#

// C# implementation of the approach
using System;

class GfG {

    // Function to return the minimum value k to be added
    static int FindMinNum(int[] arr, int n)
    {
        int sum = 0; // initialize sum of whole array
        int leftsum = 0; // initialize leftsum

        int k = int.MaxValue;

        /* Find sum of the whole array */
        for (int i = 0; i < n; ++i)
            sum += arr[i];

        for (int i = 0; i < n; ++i) {
            sum -= arr[i]; // sum is now right sum for index
                           // i
            leftsum
                += arr[i]; // add current element to leftsum

            // If sum on the right side of the ith element
            // is greater than or equal to the sum on the
            // left side then update the value of k
            if (sum >= leftsum)
                k = Math.Min(k, sum - leftsum);
        }

        if (k != int.MaxValue)
            return k;

        return -1;
    }

    // Driver code
    public static void Main()
    {
        int[] arr = { 6, 7, 1, 3, 8, 2, 4 };
        int n = arr.Length;

        Console.WriteLine(FindMinNum(arr, n));
    }
}

// This code is contributed by Pushpesh Raj

Javascript

<script>
//Javascript Implementation

// Function to return the minimum value k to be added
function FindMinNum(arr, n)
{

    var sum = 0; // initialize sum of whole array
    var leftsum = 0; // initialize leftsum

    var k = Number.MAX_VALUE;

    /* Find sum of the whole array */
    for (var i = 0; i < n; ++i)
        sum += arr[i];

    for (var i = 0; i < n; ++i)
    {
        sum -= arr[i]; // sum is now right sum for index i
        leftsum+=arr[i]; // add current element to leftsum

        // If sum on the right side of the ith element
        // is greater than or equal to the sum on the
        // left side then update the value of k
        if (sum >= leftsum)
            k = Math.min(k, sum - leftsum);
    }

    if (k != Number.MAX_VALUE)
        return k;

    return -1;
}

/* Driver code*/
var arr = [6, 7, 1, 3, 8, 2, 4];
var n = arr.length;
document.write(FindMinNum(arr, n))
// This code is contributed by Pushpesh raj
</script>

Output

3

The idea is similar to optimized solution of equilibrium index problem.
Time Complexity : O(n) 
Auxiliary Space : O(1)