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  1. import java.util.*;
  2.  
  3. public class Main {
  4.     public static void main(String[] args) {
  5.         int[] arr = new int[]{4, 2, 2, 6, 4};
  6.         int k = 6;
  7.  
  8.         System.out.println(countSubarraysWithXOR(arr, k));
  9.     }
  10.  
  11.     public static int countSubarraysWithXOR(int[] arr, int k) {
  12.  
  13.         Map<Integer, Integer> map = new HashMap<>();
  14.  
  15.         // Putting a prefix XOR of 0 has occurred once initially
  16.         map.put(0, 1);
  17.  
  18.         int currentXOR = 0;
  19.         int count = 0;
  20.  
  21.         for (int num : arr) {
  22.             // Update the running prefix XOR
  23.             currentXOR ^= num;
  24.  
  25.             // Looking for the required prefix XOR value that satisfies the property
  26.             int targetXOR = currentXOR ^ k;
  27.  
  28.             // If the target prefix XOR exists, add its frequency to our count
  29.             if (map.containsKey(targetXOR)) {
  30.                 count += map.get(targetXOR);
  31.             }
  32.  
  33.             // Record the current prefix XOR frequency in the map
  34.             map.put(currentXOR, map.getOrDefault(currentXOR, 0) + 1);
  35.         }
  36.  
  37.         return count;
  38.     }
  39. }
  40.  
Success #stdin #stdout 0.07s 54552KB
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https://ideone.com/W61oEw
language:
Java (HotSpot 12)
created:
9 hours ago
visibility:
public

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