Check if Binary Array can be split into X Subarray with same bitwise XOR

Last Updated : 10 Nov, 2022

Given a binary array A[] and an integer X, the task is to check whether it is possible to divide A[] into exactly X non-empty and non-overlapping subarrays such that each A_ibelongs to exactly one subarray and the bitwise XOR of each subarray is the same.

Examples:

Input: A[] = {0, 1, 1, 0, 0}, X = 3
Output: Yes
?Explanation: One of the possible ways of dividing A is {0}, {1, 1} & {0, 0}. Here XOR of each subarray is 0.

Input: A[] = {1, 1, 1}, X = 2
?Output: No

Approach: The problem can be solved based on the following observation:

The bitwise XOR of any binary array is either 0 or 1. Therefore if an answer exists then it is either X non-overlapping subarrays having XOR equal to 1 or X non - overlapping subarrays having XOR equal to 0. We can iterate over the binary array and check whether we can divide the array into X non-overlapping subarrays having XOR equal to 0 or X non-overlapping substrings having XOR equal to 1.

Follow the steps mentioned below to implement the above idea:

First set xor = 0, count0 = 0 and count1 = 0 .
Iterate a loop to count the number of times the xor of the prefix element of the array is 0. Let's say the count is count0.
After that check count0 ? X and xor != 1, if it is true then return "Yes".
- If it is not true set the xor = 0.
- Iterate another loop to count the number of times the xor of the prefix element of the array is 1 and reset xor = 0. Let's say the count is count1.
- After that check count1 ? X and (count1 - X) % 2 == 0, if it is true then return "Yes".
- Otherwise, return "No".

Below is the implementation of the above approach.

C++

// C++ code to implement the approach #include <iostream> #include <vector> using namespace std;  // Function to find check whether // array can be divided into exactly // X non-empty subarrays string check(vector<int> &arr, int n, int x) {     int xor_ = 0;     int count0 = 0, count1 = 0;          for (int i = 0; i < n; i++) {         xor_ ^= arr[i];         if (xor_ == 0)             count0++;     }     if (count0 >= x && xor_ != 1) {         return "Yes";     }     xor_ = 0;     for (int i = 0; i < n; i++) {         xor_ ^= arr[i];         if (xor_ == 1) {             count1++;             xor_ = 0;         }     }     if (count1 >= x && (count1 - x) % 2 == 0) {         return "Yes";     }     return "No"; }  // Driver Code int main() {     vector<int> A = { 0, 1, 1, 0, 0 };     int N = A.size();     int X = 3;          // Function Call     cout << check(A, N, X) << endl;     return 0; }  // This code is contributed Tapesh(tapeshdua420)

Java

// Java code to implement the approach  import java.io.*; import java.util.*;  public class GFG {      // Function to find check whether     // array can be divided into exactly     // X non-empty subarrays     public static String check(int arr[], int n, int x)     {         int xor = 0;         int count0 = 0, count1 = 0;          for (int i = 0; i < n; i++) {             xor ^= arr[i];             if (xor == 0)                 count0++;         }         if (count0 >= x && xor != 1) {             return "Yes";         }         xor = 0;         for (int i = 0; i < n; i++) {             xor ^= arr[i];             if (xor == 1) {                 count1++;                 xor = 0;             }         }         if (count1 >= x && (count1 - x) % 2 == 0) {             return "Yes";         }         return "No";     }      // Driver Code     public static void main(String[] args)     {         int[] A = { 0, 1, 1, 0, 0 };         int N = A.length;         int X = 3;          // Function Call         System.out.println(check(A, N, X));     } }

Python3

# Python code to implement the approach  # Function to find check whether # array can be divided into exactly # X non-empty subarrays def check(arr, n, x):     xor = 0     count0 = 0     count1 = 0          for i in range(n):         xor ^= arr[i]         if xor == 0:             count0 += 1                  if count0 >= x and xor != 1:         return "Yes"          xor = 0     for i in range(n):         xor ^= arr[i]         if xor == 1:             count1 += 1             xor = 0                  if count1 >= x and (count1 - x) % 2 == 0:         return "Yes"      return "No"  # Driver Code if __name__ == '__main__':     A = [0, 1, 1, 0, 0]       N = len(A)       X = 3           # Function Call     print(check(A, N, X))      # This code is contributed Tapesh(tapeshdua420)

// C# code to implement the approach using System;  class Program {     // Driver Code     static void Main(string[] args)     {         int[] A = { 0, 1, 1, 0, 0 };         int N = A.Length;         int X = 3;          // Function Call         Console.WriteLine(check(A, N, X));     }      // Function to find check whether     // array can be divided into exactly     // X non-empty subarrays     public static string check(int[] arr, int n, int x)     {         int xor = 0;         int count0 = 0, count1 = 0;          for (int i = 0; i < n; i++) {             xor ^= arr[i];             if (xor == 0)                 count0++;         }         if (count0 >= x && xor != 1) {             return "Yes";         }         xor = 0;         for (int i = 0; i < n; i++) {             xor ^= arr[i];             if (xor == 1) {                 count1++;                 xor = 0;             }         }          if (count1 >= x && ((count1 - x) % 2 == 0)) {             return "Yes";         }         return "No";     } }  // This code is contributed by Tapesh(tapeshdua420)

JavaScript

  // JavaScript code to implement the approach    // Function to find check whether   // array can be divided into exactly   // X non-empty subarrays   function check(arr, n, x) {       let xor = 0       let count0 = 0, count1 = 0        for (let i = 0; i < n; i++) {           xor ^= arr[i]           if (xor == 0)               count0++       }       if (count0 >= x && xor != 1) {           return "Yes"       }       xor = 0       for (let i = 0; i < n; i++) {           xor ^= arr[i]           if (xor == 1) {               count1++               xor = 0           }       }       if (count1 >= x && (count1 - x) % 2 == 0) {           return "Yes"       }       return "No"   }    // Driver Code   var A = [ 0, 1, 1, 0, 0 ]   var N = A.length   var X = 3    // Function Call   console.log(check(A, N, X))    // This code is contributed Tapesh(tapeshdua420).