# Minimum inversions required so that no two adjacent elements are same

Given a binary array arr[] of size N. The task is to find the minimum number of inversions required so that no two adjacent elements are same. After a single inversion, an element could change from 0 to 1 or from 1 to 0.

Examples:

Input: arr[] = {1, 1, 1}
Output: 1
Change arr from 1 to 0 and
the array becomes {1, 0, 1}.

Input: arr[] = {1, 0, 0, 1, 0, 0, 1, 0}
Output: 3

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

Approach: There are only two possibilities to make the array {1, 0, 1, 0, 1, 0, 1, …} or {0, 1, 0, 1, 0, 1, 0, …}. Let ans_a and ans_b be the count of changes required to get these arrays respectively. Now, the final answer will be min(ans_a, ans_b).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the minimum ` `// inversions required so that no ` `// two adjacent elements are same ` `int` `min_changes(``int` `a[], ``int` `n) ` `{ ` `    ``// To store the inversions required ` `    ``// to make the array {1, 0, 1, 0, 1, 0, 1, ...} ` `    ``// and {0, 1, 0, 1, 0, 1, 0, ...} respectively ` `    ``int` `ans_a = 0, ans_b = 0; ` ` `  `    ``// Find all the changes required ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(i % 2 == 0) { ` `            ``if` `(a[i] == 0) ` `                ``ans_a++; ` `            ``else` `                ``ans_b++; ` `        ``} ` `        ``else` `{ ` `            ``if` `(a[i] == 0) ` `                ``ans_b++; ` `            ``else` `                ``ans_a++; ` `        ``} ` `    ``} ` ` `  `    ``// Return the required answer ` `    ``return` `min(ans_a, ans_b); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a[] = { 1, 0, 0, 1, 0, 0, 1, 0 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` ` `  `    ``cout << min_changes(a, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` ` `  `// Function to return the minimum ` `// inversions required so that no ` `// two adjacent elements are same ` `static` `int` `min_changes(``int` `a[], ``int` `n) ` `{ ` `    ``// To store the inversions required ` `    ``// to make the array {1, 0, 1, 0, 1, 0, 1, ...} ` `    ``// and {0, 1, 0, 1, 0, 1, 0, ...} respectively ` `    ``int` `ans_a = ``0``, ans_b = ``0``; ` ` `  `    ``// Find all the changes required ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `        ``if` `(i % ``2` `== ``0``)  ` `        ``{ ` `            ``if` `(a[i] == ``0``) ` `                ``ans_a++; ` `            ``else` `                ``ans_b++; ` `        ``} ` `        ``else`  `        ``{ ` `            ``if` `(a[i] == ``0``) ` `                ``ans_b++; ` `            ``else` `                ``ans_a++; ` `        ``} ` `    ``} ` ` `  `    ``// Return the required answer ` `    ``return` `Math.min(ans_a, ans_b); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `a[] = { ``1``, ``0``, ``0``, ``1``, ``0``, ``0``, ``1``, ``0` `}; ` `    ``int` `n = a.length; ` ` `  `    ``System.out.println(min_changes(a, n)); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the minimum ` `# inversions required so that no ` `# two adjacent elements are same ` `def` `min_changes(a, n): ` ` `  `    ``# To store the inversions required ` `    ``# to make the array {1, 0, 1, 0, 1, 0, 1, ...} ` `    ``# and {0, 1, 0, 1, 0, 1, 0, ...} respectively ` `    ``ans_a ``=` `0``; ` `    ``ans_b ``=` `0``; ` ` `  `    ``# Find all the changes required ` `    ``for` `i ``in` `range``(n): ` `        ``if` `(i ``%` `2` `=``=` `0``): ` `            ``if` `(a[i] ``=``=` `0``): ` `                ``ans_a ``+``=` `1``; ` `            ``else``: ` `                ``ans_b ``+``=` `1``; ` ` `  `        ``else``: ` `            ``if` `(a[i] ``=``=` `0``): ` `                ``ans_b ``+``=` `1``; ` `            ``else``: ` `                ``ans_a ``+``=` `1``; ` ` `  `    ``# Return the required answer ` `    ``return` `min``(ans_a, ans_b); ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` ` `  `    ``a ``=` `[ ``1``, ``0``, ``0``, ``1``, ``0``, ``0``, ``1``, ``0` `]; ` `    ``n ``=` `len``(a); ` ` `  `    ``print``(min_changes(a, n)); ` ` `  `# This code is contributed by Rajput-Ji  `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to return the minimum ` `// inversions required so that no ` `// two adjacent elements are same ` `static` `int` `min_changes(``int` `[]a, ``int` `n) ` `{ ` `    ``// To store the inversions required ` `    ``// to make the array {1, 0, 1, 0, 1, 0, 1, ...} ` `    ``// and {0, 1, 0, 1, 0, 1, 0, ...} respectively ` `    ``int` `ans_a = 0, ans_b = 0; ` ` `  `    ``// Find all the changes required ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``if` `(i % 2 == 0)  ` `        ``{ ` `            ``if` `(a[i] == 0) ` `                ``ans_a++; ` `            ``else` `                ``ans_b++; ` `        ``} ` `        ``else` `        ``{ ` `            ``if` `(a[i] == 0) ` `                ``ans_b++; ` `            ``else` `                ``ans_a++; ` `        ``} ` `    ``} ` ` `  `    ``// Return the required answer ` `    ``return` `Math.Min(ans_a, ans_b); ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]a = { 1, 0, 0, 1, 0, 0, 1, 0 }; ` `    ``int` `n = a.Length; ` ` `  `    ``Console.WriteLine(min_changes(a, n)); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

Output:

```3
```

My Personal Notes arrow_drop_up pawanasipugmailcom

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : Rajput-Ji