# Find the minimum value to be added so that array becomes balanced

Last Updated : 19 Sep, 2023

Given an array of even size, task is to find minimum value that can be added to an element so that array become balanced. An array is balanced if the sum of the left half of the array elements is equal to the sum of right half. Suppose, we have an array 1 3 1 2 4 3. The Sum of first three elements is 1 + 3 + 1 = 5 and sum of last three elements is 2 + 4 + 3 = 9
So this is unbalanced, to make it balanced the minimum number we can add is 4 to any element in first half.

Examples :

```Input : 1 2 1 2 1 3
Output : 2
Sum of first 3 elements is 1 + 2 + 1 = 4,
sum of last three elements is 2 + 1 + 3 = 6
To make the array balanced you can add 2.

Input : 20 10
Output : 10```
Recommended Practice

The idea is simple, we compute sums of first and second halves. Once these sums are computed, we return absolute difference of these two values.

Implementation:

## C++

 `#include ` `using` `namespace` `std; ` ` `  `// Returns minimum value that need to be added ` `// to make array balanced. ` `int` `minValueToBalance(``int` `a[], ``int` `n) ` `{ ` `    ``// Calculating sum of first half elements ` `    ``// of an array ` `    ``int` `sum1 = 0; ` `    ``for` `(``int` `i = 0; i < n/2; i++) ` `        ``sum1 += a[i]; ` ` `  `    ``// Calculating sum of other half elements ` `    ``// of an array ` `    ``int` `sum2 = 0; ` `    ``for` `(``int` `i = n/2; i < n; i++) ` `        ``sum2 += a[i]; ` ` `  `    ``// calculating difference ` `    ``return` `abs``(sum1 - sum2); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = {1, 7, 1, 1, 3, 1}; ` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]); ` `    ``cout << minValueToBalance(arr, n)<

## Java

 `// Java program to Find the minimum value ` `// to be added so that array becomes balanced ` ` `  `class` `Minimum ` `{ ` `    ``// Returns minimum value that need to  ` `    ``// be added to make array balanced. ` `    ``public` `static` `int` `minValueToBalance(``int` `a[],  ` `                                        ``int` `n) ` `    ``{ ` `        ``// Calculating sum of first half ` `        ``// elements of an array ` `        ``int` `sum1 = ``0``; ` `        ``for` `(``int` `i = ``0``; i < n / ``2``; i++) ` `            ``sum1 += a[i]; ` ` `  `        ``// Calculating sum of other half ` `        ``// elements of an array ` `        ``int` `sum2 = ``0``; ` `        ``for` `(``int` `i = n/``2``; i < n; i++) ` `            ``sum2 += a[i]; ` ` `  `        ``// calculating difference ` `        ``return` `Math.abs(sum1 - sum2); ` `    ``} ` `     `  `    ``// driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `arr[] = {``1``, ``7``, ``1``, ``1``, ``3``, ``1``}; ` `        ``int` `n = ``6``; ` `        ``System.out.print(minValueToBalance(arr, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by rishabh_jain `

## Python3

 `# Python3 program to Find the  ` `# minimum value to be added so that  ` `# array becomes balanced ` ` `  `# Returns minimum value that need to  ` `# be added to make array balanced. ` `def` `minValueToBalance(a, n): ` `     `  `    ``# Calculating sum of first  ` `    ``# half elements of an array ` `    ``sum1 ``=` `0` `    ``for` `i ``in` `range``( ``int``(n ``/` `2``)): ` `        ``sum1 ``+``=` `a[i] ` `         `  `    ``# Calculating sum of other ` `    ``# half elements of an array ` `    ``sum2 ``=` `0``; ` `    ``i ``=` `int``(n ``/` `2``) ` `    ``while` `i < n: ` `        ``sum2 ``+``=` `a[i] ` `        ``i ``=` `i ``+` `1` `     `  `    ``# calculating difference ` `    ``return` `abs``(sum1 ``-` `sum2) ` ` `  `# Driver code ` `arr ``=` `[``1``, ``7``, ``1``, ``1``, ``3``, ``1``] ` `n ``=` `len``(arr) ` `print``(minValueToBalance(arr, n)) ` ` `  `# This code is contributed by "Sharad_Bhardwaj".  `

## C#

 `// C# program to Find the minimum value ` `// to be added so that array becomes balanced ` `using` `System; ` ` `  `class` `Minimum { ` `     `  `    ``// Returns minimum value that need to  ` `    ``// be added to make array balanced. ` `    ``public` `static` `int` `minValueToBalance(``int` `[]a,  ` `                                        ``int` `n) ` `    ``{ ` `         `  `        ``// Calculating sum of first half ` `        ``// elements of an array ` `        ``int` `sum1 = 0; ` `        ``for` `(``int` `i = 0; i < n / 2; i++) ` `            ``sum1 += a[i]; ` ` `  `        ``// Calculating sum of other half ` `        ``// elements of an array ` `        ``int` `sum2 = 0; ` `        ``for` `(``int` `i = n / 2; i < n; i++) ` `            ``sum2 += a[i]; ` ` `  `        ``// calculating difference ` `        ``return` `Math.Abs(sum1 - sum2); ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `[]arr = {1, 7, 1, 1, 3, 1}; ` `        ``int` `n = 6; ` `        ``Console.Write(minValueToBalance(arr, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by Nitin Mittal. `

## PHP

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## Javascript

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Output

`4`

Time complexity: O(n), where n is the size of the input array.

Space complexity: O(1), as the space used is constant irrespective of the input size.

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