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

• Difficulty Level : Medium
• Last Updated : 10 Sep, 2021

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:
Explanation:
For the above given array XOR of consecutive 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:
Explanation:
For the above given array XOR of consecutive 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.

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`

## Javascript

 ``
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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