Count Triplets That Can Form Two Arrays of Equal XOR
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I have successfully developed and executed detailed test plans, test cases, and scripts for Android and web applications, ensuring high-quality standards and user satisfaction. I have also demonstrated my proficiency in manual REST API testing with Postman, as well as in end-to-end performance and automation testing using Jmeter and selenium with Java, TestNG and Maven. Additionally, I have utilized Azure DevOps for bug tracking and issue management.
Given an array of integers arr.
We want to select three indices i, j and k where (0 <= i < j <= k < arr.length).
Let's define a and b as follows:
a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k]
Note that ^ denotes the bitwise-xor operation.
Return the number of triplets (i, j and k) Where a == b.
LeetCode Problem - 1442
class Solution {
public int countTriplets(int[] arr) {
int length = arr.length; // Get the length of the input array
int answer = 0; // Initialize a counter to keep track of the number of valid triplets
// Loop through each possible starting point i
for(int i = 0; i < length; i++) {
int a = 0; // Initialize a variable to store the XOR of elements from i to j-1
// Loop through each possible middle point j, starting from i+1
for(int j = i + 1; j < length; j++) {
a ^= arr[j - 1]; // Update the XOR value a by XORing with the element at j-1
int b = 0; // Initialize a variable to store the XOR of elements from j to k
// Loop through each possible end point k, starting from j
for(int k = j; k < length; k++) {
b ^= arr[k]; // Update the XOR value b by XORing with the element at k
if(a == b) { // If a and b are equal, it means the condition is satisfied
answer++; // Increment the counter for valid triplets
}
}
}
}
return answer; // Return the total count of valid triplets
}
}
