LeetCode 1310: XOR Queries of a Subarray Solution
Master LeetCode problem 1310 (XOR Queries of a Subarray), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
1310. XOR Queries of a Subarray
Problem Explanation
Explanation
To efficiently compute the XOR of elements within a subarray, we can precompute the XOR prefix array. The XOR prefix array at index i will store the XOR of all elements from index 0 to i in the original array. Then, to answer a query [left, right], we can XOR the prefix values at index right and left-1 if left is not 0.
Algorithm:
- Initialize an XOR prefix array of the same size as the original array.
- Compute the XOR prefix array by iterating through the original array.
- Answer each query by XORing the prefix values at the query indices.
Time Complexity: O(n) where n is the size of the original array. Space Complexity: O(n) for the XOR prefix array.
Solution Code
class Solution {
public int[] xorQueries(int[] arr, int[][] queries) {
int n = arr.length;
int[] xorPrefix = new int[n];
xorPrefix[0] = arr[0];
for (int i = 1; i < n; i++) {
xorPrefix[i] = xorPrefix[i - 1] ^ arr[i];
}
int[] result = new int[queries.length];
for (int i = 0; i < queries.length; i++) {
int left = queries[i][0];
int right = queries[i][1];
result[i] = (left == 0) ? xorPrefix[right] : xorPrefix[right] ^ xorPrefix[left - 1];
}
return result;
}
}Try It Yourself
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