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[1] from 1 to 0 and
the array becomes {1, 0, 1}.
Input: arr[] = {1, 0, 0, 1, 0, 0, 1, 0}
Output: 3
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 <bits/stdc++.h>using namespace std;// Function to return the minimum// inversions required so that no// two adjacent elements are sameint 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 codeint 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 approachclass GFG{// Function to return the minimum// inversions required so that no// two adjacent elements are samestatic 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 codepublic 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 samedef 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 codeif __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 approachusing System;class GFG{// Function to return the minimum// inversions required so that no// two adjacent elements are samestatic 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 codepublic 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
<script>// JavaScript implementation of the approach// Function to return the minimum// inversions required so that no// two adjacent elements are samefunction 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 let ans_a = 0, ans_b = 0; // Find all the changes required for (let 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 codelet a = [1, 0, 0, 1, 0, 0, 1, 0];let n = a.length;document.write(min_changes(a, n));</script> |
Output:
3
Time Complexity: O(n)
Auxiliary Space: O(1)
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