# Minimize increment/decrement of Array elements to make each modulo K equal

Given an array arr[] of length N and an integer K. In each operation any element(say arr[i]) can be selected from the array and can be changed to arr[i] + 1 or arr[i] – 1. The task is to find the minimum number of operation required to perform on array such that each value of the array modulo K remains same.
Examples:

Input: arr[] = {4, 5, 8, 3, 12},  k =5
Output: 4
Explanation:
Operation 1: { 3, 5, 8, 3, 12 }, decrease 4 at index 0 by 1.
Operation 2: { 3, 4, 8, 3, 12 }, decrease 5 at index 1 by 1.
Operation 3: { 3, 3, 8, 3, 12 }, decrease 4 at index 1 by 1.
Operation 4: { 3, 3, 8, 3, 13 }, increase 12 at index 4 by 1.
The modulo of each number is equal to 3 and minimum steps required were 4.

Input: arr[] = {2, 35, 48, 23, 52},  k =3
Output: 2
Explanation:
Minimum number of steps required to make modulo of each number equal is 2.

Approach: The idea is to use Hashing that keeps the count of each modulo that has been obtained.

• Now iterate for each value of i, in range 0 <= i < k, and find the number of operation required to make the modulo of all numbers equal.
• If it is less than the value obtained than the currently stored value then update it.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the minimum operations ` `// required to make the modulo of each ` `// element of the array equal to each other ` `int` `Find_min(set<``int``>& diff_mod, ` `             ``map<``int``, ``int``> count_mod, ``int` `k) ` `{ ` `    ``// Variable to store minimum ` `    ``// operation required ` `    ``int` `min_oprn = INT_MAX; ` ` `  `    ``// To store operation required to ` `    ``// make all modulo equal ` `    ``int` `oprn = 0; ` ` `  `    ``// Iterating through all ` `    ``// possible modulo value ` `    ``for` `(``int` `x = 0; x < k; x++) { ` `        ``oprn = 0; ` ` `  `        ``// Iterating through all different ` `        ``// modulo obtained so far ` `        ``for` `(``auto` `w : diff_mod) { ` ` `  `            ``// Caculating oprn required ` `            ``// to make all modulos equal ` `            ``// to x ` `            ``if` `(w != x) { ` ` `  `                ``if` `(w == 0) { ` ` `  `                    ``// Checking the operations ` `                    ``// that will cost less ` `                    ``oprn += min(x, k - x) ` `                            ``* count_mod[w]; ` `                ``} ` ` `  `                ``else` `{ ` ` `  `                    ``// Check operation that ` `                    ``// will cost less ` `                    ``oprn += min( ` `                                ``abs``(x - w), ` `                                ``k + x - w) ` `                            ``* count_mod[w]; ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``// Update the minimum ` `        ``// number of operations ` `        ``if` `(oprn < min_oprn) ` `            ``min_oprn = oprn; ` `    ``} ` ` `  `    ``// Returing the answer ` `    ``return` `min_oprn; ` `} ` ` `  `// Function to store different modulos ` `int` `Cal_min(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``// Set to store all ` `    ``// different modulo ` `    ``set<``int``> diff_mod; ` ` `  `    ``// Map to store count ` `    ``// of all different  modulo ` `    ``// obtained ` `    ``map<``int``, ``int``> count_mod; ` ` `  `    ``// Storing all different ` `    ``// modulo count ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// Insert into the set ` `        ``diff_mod.insert(arr[i] % k); ` ` `  `        ``// Increment count ` `        ``count_mod[arr[i] % k]++; ` `    ``} ` ` `  `    ``// Function call to return value of ` `    ``// min oprn required ` `    ``return` `Find_min(diff_mod, count_mod, k); ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 35, 48, 23, 52 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `k = 3; ` `    ``cout << Cal_min(arr, n, k); ` `    ``return` `0; ` `}`

## Java

 `// Java program for the above approach ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `     `  `// Function to find the minimum operations ` `// required to make the modulo of each ` `// element of the array equal to each other ` `static` `int` `Find_min(HashSet diff_mod, ` `                    ``HashMap count_mod, ` `                    ``int` `k) ` `{ ` `     `  `    ``// Variable to store minimum ` `    ``// operation required ` `    ``int` `min_oprn = Integer.MAX_VALUE; ` ` `  `    ``// To store operation required to ` `    ``// make all modulo equal ` `    ``int` `oprn = ``0``; ` ` `  `    ``// Iterating through all ` `    ``// possible modulo value ` `    ``for``(``int` `x = ``0``; x < k; x++) ` `    ``{ ` `        ``oprn = ``0``; ` ` `  `        ``// Iterating through all different ` `        ``// modulo obtained so far ` `        ``for``(``int` `w : diff_mod)  ` `        ``{ ` ` `  `            ``// Caculating oprn required ` `            ``// to make all modulos equal ` `            ``// to x ` `            ``if` `(w != x)  ` `            ``{ ` `                ``if` `(w == ``0``) ` `                ``{ ` `                     `  `                    ``// Checking the operations ` `                    ``// that will cost less ` `                    ``oprn += Math.min(x, k - x) * ` `                            ``count_mod.get(w); ` `                ``} ` `                ``else`  `                ``{ ` `                     `  `                    ``// Check operation that ` `                    ``// will cost less ` `                    ``oprn += Math.min(Math.abs(x - w),  ` `                                     ``k + x - w) *  ` `                                     ``count_mod.get(w); ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``// Update the minimum ` `        ``// number of operations ` `        ``if` `(oprn < min_oprn) ` `            ``min_oprn = oprn; ` `    ``} ` ` `  `    ``// Returing the answer ` `    ``return` `min_oprn; ` `} ` ` `  `// Function to store different modulos ` `static` `int` `Cal_min(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `     `  `    ``// Set to store all ` `    ``// different modulo ` `    ``HashSet diff_mod = ``new` `HashSet<>(); ` ` `  `    ``// Map to store count ` `    ``// of all different modulo ` `    ``// obtained ` `    ``HashMap count_mod = ``new` `HashMap<>(); ` ` `  `    ``// Storing all different ` `    ``// modulo count ` `    ``for``(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` `         `  `        ``// Insert into the set ` `        ``diff_mod.add(arr[i] % k); ` ` `  `        ``// Increment count ` `        ``count_mod.put(arr[i] % k,  ` `        ``count_mod.getOrDefault(arr[i] % k, ``0``) + ``1``); ` `    ``} ` ` `  `    ``// Function call to return value of ` `    ``// min oprn required ` `    ``return` `Find_min(diff_mod, count_mod, k); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``2``, ``35``, ``48``, ``23``, ``52` `}; ` `    ``int` `n = arr.length; ` `    ``int` `k = ``3``; ` `     `  `    ``System.out.print(Cal_min(arr, n, k)); ` `} ` `} ` ` `  `// This code is contributed by jrishabh99 `

Output:

```2
```

Time Complexity: O(N*K) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : jrishabh99