Given a binary array arr[] (containing only 0s and 1s) and an integer K. The task is to find the minimum number of moves to make the array K-periodic.
An array is said to be K-periodic if the sub-arrays [1 to K], [k+1 to 2K], [2k+1 to 3K], … are all exactly same.
In a single move any 1 can be changed to a 0 or any 0 can be changed into a 1.
Examples:
Input: arr[] = {1, 1, 0, 0, 1, 1}, K = 2
Output: 2
The new array can be {1, 1, 1, 1, 1, 1}Input: arr[] = {1, 0, 0, 0, 1, 0}, K = 2
Output: 1
The new array can be {1, 0, 1, 0, 1, 0}
Approach: For an array to be K-periodic. Consider indices i where i % K = X. All these indices must have the same value. So either the 1s can be converted to 0s or vice-versa. In order to reduce the number of moves, we choose the conversion which is minimum.
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 moves requiredint minMoves(int n, int a[], int k){ int ct1[k] = { 0 }, ct0[k] = { 0 }, moves = 0; // Count the number of 1s and 2s // at each X such that i % K = X for (int i = 0; i < n; i++) if (a[i] == 1) ct1[i % k]++; else ct0[i % k]++; // Choose the minimum elements to change for (int i = 0; i < k; i++) moves += min(ct1[i], ct0[i]); // Return the minimum moves required return moves;}// Driver codeint main(){ int k = 2; int a[] = { 1, 0, 0, 0, 1, 0 }; int n = sizeof(a) / sizeof(a[0]); cout << minMoves(n, a, k); return 0;} |
Java
// Java implementation of the approachimport java.io.*;class GFG{// Function to return the minimum// moves requiredstatic int minMoves(int n, int a[], int k){ int ct1[] = new int [k]; int ct0[] = new int[k]; int moves = 0; // Count the number of 1s and 2s // at each X such that i % K = X for (int i = 0; i < n; i++) if (a[i] == 1) ct1[i % k]++; else ct0[i % k]++; // Choose the minimum elements to change for (int i = 0; i < k; i++) moves += Math.min(ct1[i], ct0[i]); // Return the minimum moves required return moves;}// Driver codepublic static void main (String[] args){ int k = 2; int a[] = { 1, 0, 0, 0, 1, 0 }; int n = a.length; System.out.println(minMoves(n, a, k));}}// This is code contributed by inder_verma |
Python3
# Python3 implementation of the approach# Function to return the minimum # moves requireddef minMoves(n, a, k): ct1 = [0 for i in range(k)] ct0 = [0 for i in range(k)] moves = 0 # Count the number of 1s and 2s # at each X such that i % K = X for i in range(n): if (a[i] == 1): ct1[i % k] += 1 else: ct0[i % k] += 1 # Choose the minimum elements to change for i in range(k): moves += min(ct1[i], ct0[i]) # Return the minimum moves required return moves# Driver codeif __name__ == '__main__': k = 2 a = [1, 0, 0, 0, 1, 0] n = len(a) print(minMoves(n, a, k))# This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of the approachusing System;class GFG{// Function to return the minimum// moves requiredstatic int minMoves(int n, int[] a, int k){ int[] ct1 = new int [k]; int[] ct0 = new int[k]; int moves = 0; // Count the number of 1s and 2s // at each X such that i % K = X for (int i = 0; i < n; i++) if (a[i] == 1) ct1[i % k]++; else ct0[i % k]++; // Choose the minimum elements to change for (int i = 0; i < k; i++) moves += Math.Min(ct1[i], ct0[i]); // Return the minimum moves required return moves;}// Driver codepublic static void Main (){ int k = 2; int[] a = { 1, 0, 0, 0, 1, 0 }; int n = a.Length; Console.WriteLine(minMoves(n, a, k));}}// This is code contributed by Code_Mech |
PHP
<?php// PHP implementation of the approach// Function to return the minimum// moves requiredfunction minMoves($n, $a, $k){ $ct1 = array(); $ct0 = array(); $moves = 0; // Count the number of 1s and 2s // at each X such that i % K = X for ($i = 0; $i < $n; $i++) if ($a[$i] == 1) $ct1[$i % $k]++; else $ct0[$i % $k]++; // Choose the minimum elements to change for ($i = 0; $i < $k; $i++) $moves += min($ct1[$i], $ct0[$i]); // Return the minimum moves required return $moves;}// Driver code$k = 2;$a = array(1, 0, 0, 0, 1, 0);$n = sizeof($a);echo(minMoves($n, $a, $k));// This is code contributed by Code_Mech. |
Javascript
<script>// Javascript implementation of the approach// Function to return the minimum moves requiredfunction minMoves(n, a, k){ let ct1 = new Uint8Array(k), ct0 = new Uint8Array(k), moves = 0; // Count the number of 1s and 2s // at each X such that i % K = X for(let i = 0; i < n; i++) if (a[i] == 1) ct1[i % k]++; else ct0[i % k]++; // Choose the minimum elements to change for(let i = 0; i < k; i++) moves += Math.min(ct1[i], ct0[i]); // Return the minimum moves required return moves;}// Driver codelet k = 2;let a = [ 1, 0, 0, 0, 1, 0 ];let n = a.length;document.write(minMoves(n, a, k)); // This code is contributed by Mayank Tyagi </script> |
Output:
1
Time Complexity: O(n + k), where n is the size of the given array and k is the given input.
Auxiliary Space: O(k), where k is the given input.
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