Leetcode #2564: Substring XOR Queries
In this guide, we solve Leetcode #2564 Substring XOR Queries in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Bit Manipulation, Array, Hash Table, String
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
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]
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.