LeetCode #1310 — MEDIUM

XOR Queries of a Subarray

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an array arr of positive integers. You are also given the array queries where queries[i] = [left_i,right_i].

For each query i compute the XOR of elements from left_i to right_i (that is, arr[left_i] XOR arr[left_i + 1] XOR ... XOR arr[right_i] ).

Return an array answer where answer[i] is the answer to the i^th query.

Example 1:

Input: arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]
Output: [2,7,14,8] 
Explanation: 
The binary representation of the elements in the array are:
1 = 0001 
3 = 0011 
4 = 0100 
8 = 1000 
The XOR values for queries are:
[0,1] = 1 xor 3 = 2 
[1,2] = 3 xor 4 = 7 
[0,3] = 1 xor 3 xor 4 xor 8 = 14 
[3,3] = 8

Example 2:

Input: arr = [4,8,2,10], queries = [[2,3],[1,3],[0,0],[0,3]]
Output: [8,0,4,4]

Constraints:

1 <= arr.length, queries.length <= 3 * 10⁴
1 <= arr[i] <= 10⁹
queries[i].length == 2
0 <= left_i <= right_i < arr.length

Step 01

Brute Force Baseline

Problem summary: You are given an array arr of positive integers. You are also given the array queries where queries[i] = [lefti, righti]. For each query i compute the XOR of elements from lefti to righti (that is, arr[lefti] XOR arr[lefti + 1] XOR ... XOR arr[righti] ). Return an array answer where answer[i] is the answer to the ith query.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Bit Manipulation

Example 1

[1,3,4,8]
[[0,1],[1,2],[0,3],[3,3]]

Example 2

[4,8,2,10]
[[2,3],[1,3],[0,0],[0,3]]

Step 02

Core Insight

What unlocks the optimal approach

What is the result of x ^ y ^ x ?
Compute the prefix sum for XOR.
Process the queries with the prefix sum values.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1310: XOR Queries of a Subarray
class Solution {
    public int[] xorQueries(int[] arr, int[][] queries) {
        int n = arr.length;
        int[] s = new int[n + 1];
        for (int i = 1; i <= n; ++i) {
            s[i] = s[i - 1] ^ arr[i - 1];
        }
        int m = queries.length;
        int[] ans = new int[m];
        for (int i = 0; i < m; ++i) {
            int l = queries[i][0], r = queries[i][1];
            ans[i] = s[r + 1] ^ s[l];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1310: XOR Queries of a Subarray
func xorQueries(arr []int, queries [][]int) (ans []int) {
	n := len(arr)
	s := make([]int, n+1)
	for i, x := range arr {
		s[i+1] = s[i] ^ x
	}
	for _, q := range queries {
		l, r := q[0], q[1]
		ans = append(ans, s[r+1]^s[l])
	}
	return
}

# Accepted solution for LeetCode #1310: XOR Queries of a Subarray
class Solution:
    def xorQueries(self, arr: List[int], queries: List[List[int]]) -> List[int]:
        s = list(accumulate(arr, xor, initial=0))
        return [s[r + 1] ^ s[l] for l, r in queries]

// Accepted solution for LeetCode #1310: XOR Queries of a Subarray
struct Solution;

impl Solution {
    fn xor_queries(mut arr: Vec<i32>, queries: Vec<Vec<i32>>) -> Vec<i32> {
        let n = arr.len();
        for i in 1..n {
            arr[i] ^= arr[i - 1];
        }
        let mut res = vec![];
        for query in queries {
            let l = query[0] as usize;
            let r = query[1] as usize;
            let x = if l > 0 { arr[r] ^ arr[l - 1] } else { arr[r] };
            res.push(x);
        }
        res
    }
}

#[test]
fn test() {
    let arr = vec![1, 3, 4, 8];
    let queries = vec_vec_i32![[0, 1], [1, 2], [0, 3], [3, 3]];
    let res = vec![2, 7, 14, 8];
    assert_eq!(Solution::xor_queries(arr, queries), res);
    let arr = vec![4, 8, 2, 10];
    let queries = vec_vec_i32![[2, 3], [1, 3], [0, 0], [0, 3]];
    let res = vec![8, 0, 4, 4];
    assert_eq!(Solution::xor_queries(arr, queries), res);
}

// Accepted solution for LeetCode #1310: XOR Queries of a Subarray
function xorQueries(arr: number[], queries: number[][]): number[] {
    const n = arr.length;
    const s: number[] = Array(n + 1).fill(0);
    for (let i = 0; i < n; ++i) {
        s[i + 1] = s[i] ^ arr[i];
    }
    return queries.map(([l, r]) => s[r + 1] ^ s[l]);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n+m)

Space

O(n)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.