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 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: 1
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 <iostream>#include <limits.h>using namespace std; // Function to find the minimum// flips required in binary arrayint 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 Codeint 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 parityimport 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 arraydef 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<count_zero: set = not set 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 Codeif __name__ == "__main__": n, k = 6, 3 a =[0, 0, 1, 1, 0, 0] print(count_flips(a, n, k)) |
C#
// C# implementation to find the minimum flips// required such that alternate subarrays// have different parityusing 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
<script>// Javascript implementation to find the minimum flips// required such that alternate subarrays// have different parity // Function to find the minimum // flips required in binary array function count_flips(a , n , k) { // Boolean value to indicate // odd or even value of 1's var set = false; var ans = 0, min_diff = Number.MAX_VALUE; // Loop to iterate over // the subarrays of size K for (i = 0; i < k; i++) { // curr_index is used to iterate // over all the subarrays var 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 var n = 6, k = 3; var a = [ 0, 0, 1, 1, 0, 0 ]; document.write(count_flips(a, n, k));// This code contributed by Rajput-Ji </script> |
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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