# 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:
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 ``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

 `= ``\$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

 ``

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 ``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

 ``

Output
`3`

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

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