Minimum number of moves to make a binary array K periodic

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:

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the minimum moves required
int 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 code
int 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;
}

chevron_right


Output:

1

Time Complexity: O(N)



My Personal Notes arrow_drop_up


Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.