# Minimum flips in a Binary array such that XOR of consecutive subarrays of size K have different parity

Given a binary array arr[] of length N, the task is to find the minimum flips required in the array such that XOR of a consecutive sub-arrays of size K have different parity.

Examples:

Input: arr[] = {0, 1, 0, 1, 1}, K = 2
Output: 2
Explanation:
For the above given array XOR of consective sub-array of size 2 is: {(0, 1): 1, (1, 0): 1, (0, 1): 1, (1, 1): 0}
There are two flips required which can be done on the following indices:
Index 0: It is required to flip the bit of the 0th index, such that XOR of first sub-array remains 1.
Index 1: It is required to flip the bit of 1st index, such that XOR of second sub-array becomes 0.

Input: arr[]={0, 0, 1, 1, 0, 0}, K = 3
Output: 1
Explanation:
For the above given array XOR of consective sub-array of size 2 is: {(0, 0, 1): 1, (0, 1, 1): 0, (1, 1, 0): 0, (1, 0, 0): 1}
There is one flip required which can be done on the following indices:
Index 4: It is required to flip the bit of the 4th index, such that XOR of third sub-array becomes 1 and XOR of fourth subarray becomes 0.

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

Approach: To make the different parity of consecutive subarrays, the total array is dependent upon the first subarray of size K. That is every next subarray of size K should be the negation of the previous subarray.
For Example: For an array of size 4, such that consecutive subarray of size 2 have different parity:

```Let the first subarray of size 2 be {1, 1}
Then the next subarray can be {0, 0}

Consecutive subarrays of size 2 in this array:
{(1, 1): 0, (1, 0): 1, (0, 0): 0}
```

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find the ` `// minimum flips required such that ` `// alternate subarrays have  ` `// different parity ` ` `  `#include ` `#include ` `using` `namespace` `std; ` ` `  `     `  `// Function to find the minimum ` `// flips required in binary array ` `int` `count_flips(``int` `a[], ``int` `n, ``int` `k) ` `{ ` `    ``// Boolean value to indicate ` `    ``// odd or even value of 1's ` `    ``bool` `set = ``false``; ` `    ``int` `ans = 0, min_diff = INT_MAX; ` `     `  `    ``// Loop to iterate over  ` `    ``// the subarrays of size K ` `    ``for` `(``int` `i = 0; i < k; i++) { ` `         `  `        ``// curr_index is used to iterate ` `        ``// over all the subarrays ` `        ``int` `curr_index = i, segment = 0,  ` `          ``count_zero = 0, count_one = 0; ` `         `  `        ``// Loop to iterate over the array ` `        ``// at the jump of K to  ` `        ``// consider every parity         ` `        ``while` `(curr_index < n) { ` ` `  `            ``// Condition to check if the  ` `            ``// subarray is at even position ` `            ``if` `(segment % 2 == 0) { ` ` `  `                ``// The value needs to be  ` `                ``// same as the first subarray ` `                ``if` `(a[curr_index] == 1) ` `                    ``count_zero++; ` `                ``else` `                    ``count_one++; ` `            ``} ` `            ``else` `{ ` `                ``// The value needs to be  ` `                ``// opposite of the first subarray ` `                ``if` `(a[curr_index] == 0) ` `                    ``count_zero++; ` `                ``else` `                    ``count_one++; ` `            ``} ` `            ``curr_index = curr_index + k; ` `            ``segment++; ` `        ``} ` `        ``ans += min(count_one, count_zero); ` `        ``if` `(count_one < count_zero) ` `            ``set = !set; ` `        ``// Update the minimum difference ` `        ``min_diff = min(min_diff,  ` `         ``abs``(count_zero - count_one)); ` `    ``} ` ` `  `    ``// Condition to check if the 1s ` `    ``// in the subarray is odd ` `    ``if` `(set) ` `        ``return` `ans; ` `    ``else` `        ``return` `ans + min_diff; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `n = 6, k = 3; ` `    ``int` `a[] = { 0, 0, 1, 1, 0, 0 }; ` `    ``cout << count_flips(a, n, k); ` `} `

## Java

 `// Java implementation to find the minimum flips ` `// required such that alternate subarrays ` `// have different parity ` ` `  `import` `java.util.*; ` ` `  `class` `Count_Flips { ` `     `  `    ``// Function to find the minimum ` `    ``// flips required in binary array ` `    ``public` `static` `int` `count_flips( ` `              ``int` `a[], ``int` `n, ``int` `k){ ` `         `  `        ``// Boolean value to indicate ` `        ``// odd or even value of 1's ` `        ``boolean` `set = ``false``; ` `        ``int` `ans = ``0``,  ` `           ``min_diff = Integer.MAX_VALUE; ` `         `  `        ``// Loop to iterate over  ` `        ``// the subarrays of size K ` `        ``for` `(``int` `i = ``0``; i < k; i++) { ` `             `  `            ``// curr_index is used to iterate ` `            ``// over all the subarrays ` `            ``int` `curr_index = i, segment = ``0``,  ` `               ``count_zero = ``0``, count_one = ``0``; ` `             `  `            ``// Loop to iterate over the array ` `            ``// at the jump of K to  ` `            ``// consider every parity ` `            ``while` `(curr_index < n) { ` `                 `  `                ``// Condition to check if the  ` `                ``// subarray is at even position ` `                ``if` `(segment % ``2` `== ``0``) { ` `                     `  `                    ``// The value needs to be  ` `                    ``// same as the first subarray ` `                    ``if` `(a[curr_index] == ``1``) ` `                        ``count_zero++; ` `                    ``else` `                        ``count_one++; ` `                ``} ` `                ``else` `{ ` `                    ``// The value needs to be  ` `                    ``// opposite of the first subarray ` `                    ``if` `(a[curr_index] == ``0``) ` `                        ``count_zero++; ` `                    ``else` `                        ``count_one++; ` `                ``} ` `                ``curr_index = curr_index + k; ` `                ``segment++; ` `            ``} ` `            ``ans += Math.min(count_one, count_zero); ` `            ``if` `(count_one < count_zero) ` `                ``set = !set; ` `            ``// Update the minimum difference ` `            ``min_diff = Math.min(min_diff,  ` `             ``Math.abs(count_zero - count_one)); ` `        ``} ` `        ``// Condition to check if the 1s ` `        ``// in the subarray is odd ` `        ``if` `(set) ` `            ``return` `ans; ` `        ``else` `            ``return` `ans + min_diff; ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `n = ``6``, k = ``3``; ` `        ``int` `a[] = { ``0``, ``0``, ``1``, ``1``, ``0``, ``0` `}; ` `        ``System.out.println(count_flips(a, n, k)); ` `    ``} ` `} `

## Python3

 `# Python implementation to find the ` `# minimum flips required such that ` `# alternate subarrays have  ` `# different parity ` ` `  `# Function to find the minimum ` `# flips required in binary array ` `def` `count_flips(a, n, k): ` `    ``min_diff, ans, ``set` `=` `n, ``0``, ``False` `     `  `    ``# Loop to iterate over  ` `    ``# the subarrays of size K ` `    ``for` `i ``in` `range``(k): ` `        ``# curr_index is used to iterate ` `        ``# over all the subarrays ` `        ``curr_index, segment,\ ` `        ``count_zero, count_one ``=``\ ` `                   ``i, ``0``, ``0``, ``0` `         `  `        ``# Loop to iterate over the array ` `        ``# at the jump of K to  ` `        ``# consider every parity ` `        ``while` `curr_index < n: ` `             `  `            ``# Condition to check if the  ` `            ``# subarray is at even position ` `            ``if` `segment ``%` `2` `=``=` `0``: ` `                ``# The value needs to be  ` `                ``# same as the first subarray ` `                ``if` `a[curr_index] ``=``=` `1``: ` `                    ``count_zero ``+``=` `1` `                ``else``: ` `                    ``count_one ``+``=` `1` `            ``else``: ` `                ``# The value needs to be  ` `                ``# opposite of the first subarray ` `                ``if` `a[curr_index] ``=``=` `0``: ` `                    ``count_zero ``+``=` `1` `                ``else``: ` `                    ``count_one ``+``=` `1` `            ``curr_index ``+``=` `k ` `            ``segment``+``=` `1` `        ``ans ``+``=` `min``(count_zero, count_one) ` `        ``if` `count_one

## C#

 `// C# implementation to find the minimum flips ` `// required such that alternate subarrays ` `// have different parity ` `using` `System; ` ` `  `class` `Count_Flips ` `{ ` `     `  `    ``// Function to find the minimum ` `    ``// flips required in binary array ` `    ``static` `int` `count_flips(``int` `[]a, ``int` `n, ``int` `k) ` `    ``{ ` `         `  `        ``// Boolean value to indicate ` `        ``// odd or even value of 1's ` `        ``bool` `set` `= ``false``; ` `        ``int` `ans = 0, min_diff = ``int``.MaxValue; ` `         `  `        ``// Loop to iterate over  ` `        ``// the subarrays of size K ` `        ``for` `(``int` `i = 0; i < k; i++) { ` `             `  `            ``// curr_index is used to iterate ` `            ``// over all the subarrays ` `            ``int` `curr_index = i, segment = 0,  ` `            ``count_zero = 0, count_one = 0; ` `             `  `            ``// Loop to iterate over the array ` `            ``// at the jump of K to  ` `            ``// consider every parity ` `            ``while` `(curr_index < n) { ` `                 `  `                ``// Condition to check if the  ` `                ``// subarray is at even position ` `                ``if` `(segment % 2 == 0) { ` `                     `  `                    ``// The value needs to be  ` `                    ``// same as the first subarray ` `                    ``if` `(a[curr_index] == 1) ` `                        ``count_zero++; ` `                    ``else` `                        ``count_one++; ` `                ``} ` `                ``else` `{ ` `                    ``// The value needs to be  ` `                    ``// opposite of the first subarray ` `                    ``if` `(a[curr_index] == 0) ` `                        ``count_zero++; ` `                    ``else` `                        ``count_one++; ` `                ``} ` `                ``curr_index = curr_index + k; ` `                ``segment++; ` `            ``} ` `            ``ans += Math.Min(count_one, count_zero); ` `            ``if` `(count_one < count_zero) ` `                ``set` `= !``set``; ` `                 `  `            ``// Update the minimum difference ` `            ``min_diff = Math.Min(min_diff,  ` `            ``Math.Abs(count_zero - count_one)); ` `        ``} ` `         `  `        ``// Condition to check if the 1s ` `        ``// in the subarray is odd ` `        ``if` `(``set``) ` `            ``return` `ans; ` `        ``else` `            ``return` `ans + min_diff; ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `Main(``string``[] args) ` `    ``{ ` `        ``int` `n = 6, k = 3; ` `        ``int` `[]a = { 0, 0, 1, 1, 0, 0 }; ` `        ``Console.WriteLine(count_flips(a, n, k)); ` `    ``} ` `} ` ` `  `// This code is contributed by Yash_R `

Output:

```1
```

Performance Analysis:

• Time Complexity: As in the above approach, There is only one loop which takes O(N) time in worst case. Hence the Time Complexity will be O(N).
• Auxiliary Space Complexity: As in the above approach, There is no extra space used. Hence the auxiliary space complexity will be O(1).

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Improved By : Yash_R