# 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}

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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

 `// 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

 `# Python 3 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#

 `// 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

 `

Output:

```1
```

Time Complexity: O(N)

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