# Make array elements equal with minimum cost

Given an array of integers **arr[]**, the task is to find the minimum number of steps to make array elements equal by following two operations –

- Add or subtract 2 from the element with
**0**cost - Add or subtract 1 from the element with
**1**cost

**Examples:**

Input:arr[] = {4, 3, 2, 1}Output:2Explanation:

As in the above example all the array elements can be equal to the 4 or 3 with 2 as minimum cost

Cost of Making Array Elements equal to 4Index 0:As index 0 is already equal to 4, the cost is0Index 1:4 – 3 = 1, Adding 1 to the element will cost1Index 2:4 – 2 = 2, Adding 2 to the element will cost0Index 3:4 – 1 = 3, Add 2 and 1 once in the array element, which will cost1, because 2 can be added any number of times with cost 0.

Total Cost = 1 + 1 = 2

Input:arr[] = {3, 2, 1}Output:1Explanation:

As in the above example all the array elements can be made equal to 3 or 1 with 1 as minimum cost

Cost of Making Array Elements equal to 3Index 0:As index 0 is already equal to 3, the cost is0Index 1:3 – 2 = 1, Adding 1 to the element will cost1Index 2:3 – 1 = 2, Adding 2 to the element will cost0

Total Cost = 1

**Approach**: The idea is to use the fact that adding 2 to any element of the array will cost 0. Also the second observation is that adding 2 to any odd number will result in an Odd number and similarly adding 2 to an even number will result in an even number only. Therefore, any element can be made equal to a nearest integer at zero cost but with the same property that it will remain odd if it is odd as initial value or it will remain even if it is even as initial. So, to find the minimum cost we can find the minimum count of the odd or even elements in the array.

**Algorithm:**

- Find the count of the odd or even elements in the array (say
**count**). - Find the minimum value between count and (length – count) where
**length**is the length of the array. - The minimum value will be desired cost to make all array elements equal.

**Explanation with Example:**

Given Array be - {4, 2, 3, 1} Count of Odd Elements - 2 ({3, 1}) Count of Even Elements - 2 ({4, 2}) Hence, the minimum cost required is 2 and the array elements can be made equal to any integer of even or odd value

Below is the implementation of the above approach:

## C++

`// C++ implementation to make` `// array elements equal with` `// minimum cost` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the minimum` `// cost required to make array` `// elements equal` `void` `makearrayequal(` `int` `arr[], ` `int` `n)` `{` ` ` `int` `x = 0;` ` ` ` ` `// Loop to find the` ` ` `// count of odd elements` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `x += arr[i] & 1;` ` ` `}` ` ` `cout << min(x, n - x) << endl;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `arr[] = { 4, 3, 2, 1 };` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `makearrayequal(arr, n);` ` ` `return` `0;` `}` |

## Java

`// Java implementation to make` `// array elements equal with` `// minimum cost` `class` `GFG {` ` ` `// Function to find the minimum` ` ` `// cost required to make array` ` ` `// elements equal` ` ` `static` `void` `makearrayequal(` `int` `arr[], ` `int` `n)` ` ` `{` ` ` `int` `x = ` `0` `;` ` ` ` ` `// Loop to find the` ` ` `// count of odd elements` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) {` ` ` `x += (arr[i] & ` `1` `);` ` ` `}` ` ` ` ` `System.out.println(Math.min(x, n - x));` ` ` `}` ` ` ` ` `// Driver Code` ` ` `public` `static` `void` `main (String[] args)` ` ` `{` ` ` `int` `arr[] = { ` `4` `, ` `3` `, ` `2` `, ` `1` `};` ` ` `int` `n = arr.length;` ` ` `makearrayequal(arr, n);` ` ` ` ` `}` `}` `// This code is contributed by Yash_R` |

## Python3

`# Python3 implementation to make` `# array elements equal with` `# minimum cost` `# Function to find the minimum` `# cost required to make array` `# elements equal` `def` `makearrayequal(arr, n) :` ` ` `x ` `=` `0` `;` ` ` ` ` `# Loop to find the` ` ` `# count of odd elements` ` ` `for` `i ` `in` `range` `(n) :` ` ` `x ` `+` `=` `arr[i] & ` `1` `;` ` ` `print` `(` `min` `(x, n ` `-` `x));` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `arr ` `=` `[ ` `4` `, ` `3` `, ` `2` `, ` `1` `];` ` ` `n ` `=` `len` `(arr);` ` ` `makearrayequal(arr, n);` ` ` `# This code is contributed by Yash_R` |

## C#

`// C# implementation to make` `// array elements equal with` `// minimum cost` `using` `System;` `class` `GFG {` ` ` `// Function to find the minimum` ` ` `// cost required to make array` ` ` `// elements equal` ` ` `static` `void` `makearrayequal(` `int` `[]arr, ` `int` `n)` ` ` `{` ` ` `int` `x = 0;` ` ` ` ` `// Loop to find the` ` ` `// count of odd elements` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `x += (arr[i] & 1);` ` ` `}` ` ` ` ` `Console.WriteLine(Math.Min(x, n - x));` ` ` `}` ` ` ` ` `// Driver Code` ` ` `public` `static` `void` `Main (` `string` `[] args)` ` ` `{` ` ` `int` `[]arr = { 4, 3, 2, 1 };` ` ` `int` `n = arr.Length;` ` ` `makearrayequal(arr, n);` ` ` ` ` `}` `}` `// This code is contributed by Yash_R` |

## Javascript

`<script>` `// javascript implementation to make` `// array elements equal with` `// minimum cost` `// Function to find the minimum` `// cost required to make array` `// elements equal` `function` `makearrayequal(arr , n)` `{` ` ` `var` `x = 0;` ` ` ` ` `// Loop to find the` ` ` `// count of odd elements` ` ` `for` `(i = 0; i < n; i++)` ` ` `{` ` ` `x += (arr[i] & 1);` ` ` `}` ` ` `document.write(Math.min(x, n - x));` `}` `// Driver Code` `var` `arr = [ 4, 3, 2, 1 ];` `var` `n = arr.length;` `makearrayequal(arr, n);` `// This code is contributed by Amit Katiyar` `</script>` |

**Output:**

2

**Performance Analysis:**

**Time Complexity:**O(N).**Auxiliary Space:**O(1)