# Minimum operations required to change the array such that |arr[i] – M| <= 1

Given an array[] of integers, the task is find the minimum number of operations required to change the array elements such that for any positive integer M, |arr[i] – M| ≤ 1 for all valid i.

In a single operation, any element of the array can either be incremented or decremented by 1.

Examples:

Input: arr[] = {10, 1, 4}
Output: 7
If we change 1 into 2 and 10 into 4 with count of operations being |1 – 2| + |10 – 4| = 7
After changing, array becomes {4, 2, 4} where every element’s absolute difference with M = 3 is ≤ 1

Input: arr[] = {5, 7, 4, 1, 4}
Output: 4

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

Approach: Starting from the minimum element of the array to the maximum element of the array say num, calculate the count of operations required to change every element such that it’s absolute difference with num is ≤ 1. The minimum among all possible operations is the required answer.

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` `changeTheArray(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// Minimum and maximum elements from the array ` `    ``int` `minEle = *(std::min_element(arr, arr + n)); ` `    ``int` `maxEle = *(std::max_element(arr, arr + n)); ` ` `  `    ``// To store the minimum number of ` `    ``// operations required ` `    ``int` `minOperations = INT_MAX; ` `    ``for` `(``int` `num = minEle; num <= maxEle; num++) { ` ` `  `        ``// To store the number of operations required ` `        ``// to change every element to either ` `        ``// (num - 1), num or (num + 1) ` `        ``int` `operations = 0; ` `        ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `            ``// If current element is not already num ` `            ``if` `(arr[i] != num) { ` ` `  `                ``// Add the count of operations ` `                ``// required to change arr[i] ` `                ``operations += (``abs``(num - arr[i]) - 1); ` `            ``} ` `        ``} ` ` `  `        ``// Update the minimum operations so far ` `        ``minOperations = min(minOperations, operations); ` `    ``} ` ` `  `    ``return` `minOperations; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 10, 1, 4 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``cout << changeTheArray(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG { ` ` `  `    ``// Function to return the minimum ` `    ``// number of operations required ` `    ``static` `int` `changeTheArray(``int` `arr[], ``int` `n) ` `    ``{ ` ` `  `        ``// Minimum and maximum elements from the array ` `        ``int` `minEle = Arrays.stream(arr).min().getAsInt(); ` `        ``int` `maxEle = Arrays.stream(arr).max().getAsInt(); ` ` `  `        ``// To store the minimum number of ` `        ``// operations required ` `        ``int` `minOperations = Integer.MAX_VALUE; ` `        ``for` `(``int` `num = minEle; num <= maxEle; num++) { ` ` `  `            ``// To store the number of operations required ` `            ``// to change every element to either ` `            ``// (num - 1), num or (num + 1) ` `            ``int` `operations = ``0``; ` `            ``for` `(``int` `i = ``0``; i < n; i++) { ` ` `  `                ``// If current element is not already num ` `                ``if` `(arr[i] != num) { ` ` `  `                    ``// Add the count of operations ` `                    ``// required to change arr[i] ` `                    ``operations += (Math.abs(num - arr[i]) - ``1``); ` `                ``} ` `            ``} ` ` `  `            ``// Update the minimum operations so far ` `            ``minOperations = Math.min(minOperations, operations); ` `        ``} ` ` `  `        ``return` `minOperations; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `arr[] = { ``10``, ``1``, ``4` `}; ` `        ``int` `n = arr.length; ` `        ``System.out.println(changeTheArray(arr, n)); ` `    ``} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

## Python3

 `# Python3 implementation of the approach  ` `import` `math ` `import` `sys ` ` `  `# Function to return the minimum  ` `# number of operations required  ` `def` `changeTheArray(arr, n): ` `     `  `    ``# Minimum and maximum elements ` `    ``# from the array  ` `    ``minEle ``=` `min``(arr) ` `    ``maxEle ``=` `max``(arr) ` ` `  `    ``# To store the minimum number of  ` `    ``# operations required  ` `    ``minOperations ``=` `sys.maxsize ` ` `  `    ``for` `num ``in` `range``(minEle, maxEle ``+` `1``): ` `         `  `        ``# To store the number of operations required  ` `        ``# to change every element to either  ` `        ``# (num - 1), num or (num + 1)  ` `        ``operations ``=` `0` `        ``for` `i ``in` `range``(n): ` ` `  `                ``# If current element is not already num ` `                ``if` `arr[i] !``=` `num: ` `                        ``operations ``+``=` `(``abs``(num ``-` `arr[i]) ``-` `1``) ` `         `  `        ``# Update the minimum operations so far  ` `        ``minOperations ``=` `min``(minOperations, operations) ` `    ``return` `minOperations ` ` `  `# Driver code ` `if` `__name__``=``=``'__main__'``: ` `    ``arr ``=` `[``10``, ``1``, ``4``] ` `    ``n ``=` `len``(arr) ` `    ``print``(changeTheArray(arr, n)) ` ` `  `# This code is contributed by Vikash Kumar 37 `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Linq; ` ` `  `class` `GFG  ` `{  ` ` `  `    ``// Function to return the minimum  ` `    ``// number of operations required  ` `    ``static` `int` `changeTheArray(``int` `[]arr, ``int` `n)  ` `    ``{  ` ` `  `        ``// Minimum and maximum elements from the array  ` `        ``int` `minEle = arr.Min();  ` `        ``int` `maxEle = arr.Max();  ` ` `  `        ``// To store the minimum number of  ` `        ``// operations required  ` `        ``int` `minOperations = ``int``.MaxValue;  ` `        ``for` `(``int` `num = minEle; num <= maxEle; num++)  ` `        ``{  ` ` `  `            ``// To store the number of operations required  ` `            ``// to change every element to either  ` `            ``// (num - 1), num or (num + 1)  ` `            ``int` `operations = 0;  ` `            ``for` `(``int` `i = 0; i < n; i++)  ` `            ``{  ` ` `  `                ``// If current element is not already num  ` `                ``if` `(arr[i] != num) ` `                ``{  ` ` `  `                    ``// Add the count of operations  ` `                    ``// required to change arr[i]  ` `                    ``operations += (Math.Abs(num - arr[i]) - 1);  ` `                ``}  ` `            ``}  ` ` `  `            ``// Update the minimum operations so far  ` `            ``minOperations = Math.Min(minOperations, operations);  ` `        ``}  ` ` `  `        ``return` `minOperations;  ` `    ``}  ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `Main(String []args)  ` `    ``{  ` `        ``int` `[]arr = { 10, 1, 4 };  ` `        ``int` `n = arr.Length;  ` `        ``Console.WriteLine(changeTheArray(arr, n));  ` `    ``}  ` `}  ` ` `  `// This code is contributed by Rajput-Ji `

Output:

```7
```

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