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

• Difficulty Level : Medium
• Last Updated : 03 Jun, 2021

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

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[0]);` `    ``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`

## Javascript

 ``
Output:
`3`

My Personal Notes arrow_drop_up