# Minimum operations to make sum of neighbouring elements <= X

Given an array arr[] of N elements and an integer X, the task is to find the minimum number of operations required to make sum of neighbouring elements less than the given number X. In a single operation, you can choose an element arr[i] and decrease its value by 1.

Examples:

Input: arr[] = {2, 2, 2}, X = 3
Output: 1
Decrement arr[1] by 1 and the array becomes {2, 1, 2}.
Now, 2 + 1 = 3 and 1 + 2 = 3

Input: arr[] = {1, 6, 1, 2, 0, 4}, X = 1
Output: 11

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Suppose the elements of the array are a1, a2, …, an. Let’s assume a case when all a[i] are greater than X.
First we need to make all the a[i] equal to x. We will calculate the number of operations required for it.
Now, all elements will be of the form of x, x, x, x…, x N times. Here we can observe that the maximum neighbouring sum is equal to 2 * X.
Now, traverse the array from left to right, and for each i, if sum of two left neighbours that is a[i] + a[i – 1] > X then change a[i] to such a value that their net sum becomes equal to X.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the minimum ` `// number of operations required ` `int` `MinOperations(``int` `n, ``int` `x, ``int``* arr) ` `{ ` ` `  `    ``// To store total operations required ` `    ``int` `total = 0; ` `    ``for` `(``int` `i = 0; i < n; ++i) { ` ` `  `        ``// First make all elements equal to x ` `        ``// which are currenctly greater ` `        ``if` `(arr[i] > x) { ` `            ``int` `difference = arr[i] - x; ` `            ``total = total + difference; ` `            ``arr[i] = x; ` `        ``} ` `    ``} ` ` `  `    ``// Left scan the array ` `    ``for` `(``int` `i = 1; i < n; ++i) { ` `        ``int` `LeftNeigbouringSum = arr[i] + arr[i - 1]; ` ` `  `        ``// Update the current element such that ` `        ``// neighbouring sum is < x ` `        ``if` `(LeftNeigbouringSum > x) { ` `            ``int` `current_diff = LeftNeigbouringSum - x; ` `            ``arr[i] = max(0, arr[i] - current_diff); ` `            ``total = total + current_diff; ` `        ``} ` `    ``} ` `    ``return` `total; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `X = 1; ` `    ``int` `arr[] = { 1, 6, 1, 2, 0, 4 }; ` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(``int``); ` `    ``cout << MinOperations(N, X, arr); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG  ` `{ ` ` `  `// Function to return the minimum ` `// number of operations required ` `static` `int` `MinOperations(``int` `n, ``int` `x, ``int``[] arr) ` `{ ` ` `  `    ``// To store total operations required ` `    ``int` `total = ``0``; ` `    ``for` `(``int` `i = ``0``; i < n; ++i)  ` `    ``{ ` ` `  `        ``// First make all elements equal to x ` `        ``// which are currenctly greater ` `        ``if` `(arr[i] > x) ` `        ``{ ` `            ``int` `difference = arr[i] - x; ` `            ``total = total + difference; ` `            ``arr[i] = x; ` `        ``} ` `    ``} ` ` `  `    ``// Left scan the array ` `    ``for` `(``int` `i = ``1``; i < n; ++i) ` `    ``{ ` `        ``int` `LeftNeigbouringSum = arr[i] + arr[i - ``1``]; ` ` `  `        ``// Update the current element such that ` `        ``// neighbouring sum is < x ` `        ``if` `(LeftNeigbouringSum > x)  ` `        ``{ ` `            ``int` `current_diff = LeftNeigbouringSum - x; ` `            ``arr[i] = Math.max(``0``, arr[i] - current_diff); ` `            ``total = total + current_diff; ` `        ``} ` `    ``} ` `    ``return` `total; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String args[])  ` `{ ` `    ``int` `X = ``1``; ` `    ``int` `arr[] = { ``1``, ``6``, ``1``, ``2``, ``0``, ``4` `}; ` `    ``int` `N = arr.length; ` `    ``System.out.println(MinOperations(N, X, arr)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python

 `# Python3 implementation of the approach ` ` `  `# Function to return the minimum ` `# number of operations required ` `def` `MinOperations(n, x, arr): ` ` `  `    ``# To store total operations required ` `    ``total ``=` `0` `    ``for` `i ``in` `range``(n): ` ` `  `        ``# First make all elements equal to x ` `        ``# which are currenctly greater ` `        ``if` `(arr[i] > x): ` `            ``difference ``=` `arr[i] ``-` `x ` `            ``total ``=` `total ``+` `difference ` `            ``arr[i] ``=` `x ` ` `  ` `  `    ``# Left scan the array ` `    ``for` `i ``in` `range``(n): ` `        ``LeftNeigbouringSum ``=` `arr[i] ``+` `arr[i ``-` `1``] ` ` `  `        ``# Update the current element such that ` `        ``# neighbouring sum is < x ` `        ``if` `(LeftNeigbouringSum > x): ` `            ``current_diff ``=` `LeftNeigbouringSum ``-` `x ` `            ``arr[i] ``=` `max``(``0``, arr[i] ``-` `current_diff) ` `            ``total ``=` `total ``+` `current_diff ` ` `  `    ``return` `total ` ` `  ` `  `# Driver code ` `X ``=` `1` `arr``=``[``1``, ``6``, ``1``, ``2``, ``0``, ``4` `] ` `N ``=` `len``(arr) ` `print``(MinOperations(N, X, arr)) ` ` `  `# This code is contributed by mohit kumar 29 `

## C#

 `// C# implementation of the approach  ` `using` `System; ` `     `  `class` `GFG  ` `{ ` ` `  `// Function to return the minimum ` `// number of operations required ` `static` `int` `MinOperations(``int` `n, ``int` `x, ``int``[] arr) ` `{ ` ` `  `    ``// To store total operations required ` `    ``int` `total = 0; ` `    ``for` `(``int` `i = 0; i < n; ++i)  ` `    ``{ ` ` `  `        ``// First make all elements equal to x ` `        ``// which are currenctly greater ` `        ``if` `(arr[i] > x) ` `        ``{ ` `            ``int` `difference = arr[i] - x; ` `            ``total = total + difference; ` `            ``arr[i] = x; ` `        ``} ` `    ``} ` ` `  `    ``// Left scan the array ` `    ``for` `(``int` `i = 1; i < n; ++i) ` `    ``{ ` `        ``int` `LeftNeigbouringSum = arr[i] + arr[i - 1]; ` ` `  `        ``// Update the current element such that ` `        ``// neighbouring sum is < x ` `        ``if` `(LeftNeigbouringSum > x)  ` `        ``{ ` `            ``int` `current_diff = LeftNeigbouringSum - x; ` `            ``arr[i] = Math.Max(0, arr[i] - current_diff); ` `            ``total = total + current_diff; ` `        ``} ` `    ``} ` `    ``return` `total; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String []args)  ` `{ ` `    ``int` `X = 1; ` `    ``int` `[]arr = { 1, 6, 1, 2, 0, 4 }; ` `    ``int` `N = arr.Length; ` `    ``Console.WriteLine(MinOperations(N, X, arr)); ` `} ` `} ` ` `  `/* This code is contributed by PrinciRaj1992 */`

Output:

```11
```

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