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++ 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;
} |
// Java implementation of the approach import java.io.*;
class GFG
{ // Function to return the minimum // moves required static 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 code public 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 implementation of the approach # Function to return the minimum # moves required def 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 code if __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# implementation of the approach using System;
class GFG
{ // Function to return the minimum // moves required static 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 code public 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 implementation of the approach // Function to return the minimum // moves required function 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. |
<script> // Javascript implementation of the approach // Function to return the minimum moves required function 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 code let 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> |
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.