LeetCode #2564 — MEDIUM

Substring XOR Queries

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

The Problem

Problem Statement

You are given a binary string s, and a 2D integer array queries where queries[i] = [first_i, second_i].

For the i^th query, find the shortest substring of s whose decimal value, val, yields second_i when bitwise XORed with first_i. In other words, val ^ first_i == second_i.

The answer to the i^th query is the endpoints (0-indexed) of the substring [left_i, right_i] or [-1, -1] if no such substring exists. If there are multiple answers, choose the one with the minimum left_i.

Return an array ans where ans[i] = [left_i, right_i] is the answer to the i^th query.

A substring is a contiguous non-empty sequence of characters within a string.

Example 1:

Input: s = "101101", queries = [[0,5],[1,2]]
Output: [[0,2],[2,3]]
Explanation: For the first query the substring in range [0,2] is "101" which has a decimal value of 5, and 5 ^ 0 = 5, hence the answer to the first query is [0,2]. In the second query, the substring in range [2,3] is "11", and has a decimal value of 3, and 3 ^ 1 = 2. So, [2,3] is returned for the second query.

Example 2:

Input: s = "0101", queries = [[12,8]]
Output: [[-1,-1]]
Explanation: In this example there is no substring that answers the query, hence [-1,-1] is returned.

Example 3:

Input: s = "1", queries = [[4,5]]
Output: [[0,0]]
Explanation: For this example, the substring in range [0,0] has a decimal value of 1, and 1 ^ 4 = 5. So, the answer is [0,0].

Constraints:

1 <= s.length <= 10⁴
s[i] is either '0' or '1'.
1 <= queries.length <= 10⁵
0 <= first_i, second_i <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given a binary string s, and a 2D integer array queries where queries[i] = [firsti, secondi]. For the ith query, find the shortest substring of s whose decimal value, val, yields secondi when bitwise XORed with firsti. In other words, val ^ firsti == secondi. The answer to the ith query is the endpoints (0-indexed) of the substring [lefti, righti] or [-1, -1] if no such substring exists. If there are multiple answers, choose the one with the minimum lefti. Return an array ans where ans[i] = [lefti, righti] is the answer to the ith query. A substring is a contiguous non-empty sequence of characters within a string.

Baseline thinking

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

Pattern signal: Array · Hash Map · Bit Manipulation

Example 1

"101101"
[[0,5],[1,2]]

Example 2

"0101"
[[12,8]]

Example 3

"1"
[[4,5]]

Core Insight

What unlocks the optimal approach

You do not need to consider substrings having lengths greater than 30.
Pre-process all substrings with lengths not greater than 30, and add the best endpoints to a dictionary.

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 #2564: Substring XOR Queries
class Solution {
    public int[][] substringXorQueries(String s, int[][] queries) {
        Map<Integer, int[]> d = new HashMap<>();
        int n = s.length();
        for (int i = 0; i < n; ++i) {
            int x = 0;
            for (int j = 0; j < 32 && i + j < n; ++j) {
                x = x << 1 | (s.charAt(i + j) - '0');
                d.putIfAbsent(x, new int[] {i, i + j});
                if (x == 0) {
                    break;
                }
            }
        }
        int m = queries.length;
        int[][] ans = new int[m][2];
        for (int i = 0; i < m; ++i) {
            int first = queries[i][0], second = queries[i][1];
            int val = first ^ second;
            ans[i] = d.getOrDefault(val, new int[] {-1, -1});
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2564: Substring XOR Queries
func substringXorQueries(s string, queries [][]int) (ans [][]int) {
	d := map[int][]int{}
	for i := range s {
		x := 0
		for j := 0; j < 32 && i+j < len(s); j++ {
			x = x<<1 | int(s[i+j]-'0')
			if _, ok := d[x]; !ok {
				d[x] = []int{i, i + j}
			}
			if x == 0 {
				break
			}
		}
	}
	for _, q := range queries {
		first, second := q[0], q[1]
		val := first ^ second
		if v, ok := d[val]; ok {
			ans = append(ans, v)
		} else {
			ans = append(ans, []int{-1, -1})
		}
	}
	return
}

# Accepted solution for LeetCode #2564: Substring XOR Queries
class Solution:
    def substringXorQueries(self, s: str, queries: List[List[int]]) -> List[List[int]]:
        d = {}
        n = len(s)
        for i in range(n):
            x = 0
            for j in range(32):
                if i + j >= n:
                    break
                x = x << 1 | int(s[i + j])
                if x not in d:
                    d[x] = [i, i + j]
                if x == 0:
                    break
        return [d.get(first ^ second, [-1, -1]) for first, second in queries]

// Accepted solution for LeetCode #2564: Substring XOR Queries
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2564: Substring XOR Queries
// class Solution {
//     public int[][] substringXorQueries(String s, int[][] queries) {
//         Map<Integer, int[]> d = new HashMap<>();
//         int n = s.length();
//         for (int i = 0; i < n; ++i) {
//             int x = 0;
//             for (int j = 0; j < 32 && i + j < n; ++j) {
//                 x = x << 1 | (s.charAt(i + j) - '0');
//                 d.putIfAbsent(x, new int[] {i, i + j});
//                 if (x == 0) {
//                     break;
//                 }
//             }
//         }
//         int m = queries.length;
//         int[][] ans = new int[m][2];
//         for (int i = 0; i < m; ++i) {
//             int first = queries[i][0], second = queries[i][1];
//             int val = first ^ second;
//             ans[i] = d.getOrDefault(val, new int[] {-1, -1});
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2564: Substring XOR Queries
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2564: Substring XOR Queries
// class Solution {
//     public int[][] substringXorQueries(String s, int[][] queries) {
//         Map<Integer, int[]> d = new HashMap<>();
//         int n = s.length();
//         for (int i = 0; i < n; ++i) {
//             int x = 0;
//             for (int j = 0; j < 32 && i + j < n; ++j) {
//                 x = x << 1 | (s.charAt(i + j) - '0');
//                 d.putIfAbsent(x, new int[] {i, i + j});
//                 if (x == 0) {
//                     break;
//                 }
//             }
//         }
//         int m = queries.length;
//         int[][] ans = new int[m][2];
//         for (int i = 0; i < m; ++i) {
//             int first = queries[i][0], second = queries[i][1];
//             int val = first ^ second;
//             ans[i] = d.getOrDefault(val, new int[] {-1, -1});
//         }
//         return ans;
//     }
// }

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 × log M + m)

Space

O(n × log M)

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.