How do you solve Substring XOR Queries on LeetCode?

Substring XOR Queries (LeetCode #2564) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + q) time complexity and O(n) space complexity. Key insight: Preprocessing substrings allows for efficient query resolution.

What is the brute force solution for Substring XOR Queries?

The brute force approach for Substring XOR Queries is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible substring of the binary string to see if its decimal value, when XORed with the first query value, equals the second query value. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n + q).

What algorithm or data structure is used to solve Substring XOR Queries?

Substring XOR Queries uses the following concepts: Array, Hash Table, String, Bit Manipulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Substring XOR Queries?

Explain your thought process clearly during the interview. Discuss the trade-offs of your approaches. Be prepared to optimize your solution if asked.

What are common mistakes when solving Substring XOR Queries?

Not considering the maximum length of substrings (30). Failing to handle cases where no valid substring exists.

#2564

Substring XOR Queries

Medium

ArrayHash TableStringBit ManipulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + q)
Space	O(1)	O(n)

💡

Intuition

Time O(n + q)Space O(n)

The optimal solution preprocesses all substrings of length up to 30 and stores their decimal values in a dictionary. This allows for quick lookups during query processing, significantly reducing time complexity.

⚙️

Algorithm

3 steps

1Step 1: Preprocess all substrings of s with lengths up to 30 and store their decimal values along with their starting and ending indices in a dictionary.
2Step 2: For each query, calculate the target value as target = second[i] ^ first[i].
3Step 3: Check if the target exists in the preprocessed dictionary; if it does, return the corresponding indices; otherwise, return [-1, -1].

solution.py14 lines

1def substring_xor_queries(s, queries):
2    n = len(s)
3    substrings = {}
4    for i in range(n):
5        num = 0
6        for j in range(i, min(n, i + 30)):
7            num = (num << 1) | (int(s[j]))
8            if num not in substrings:
9                substrings[num] = (i, j)
10    results = []
11    for first, second in queries:
12        target = first ^ second
13        results.append(substrings.get(target, (-1, -1)))
14    return results

ℹ

Complexity note: The preprocessing step takes O(n) time to build the dictionary, and each query can be answered in O(1) time, leading to a total of O(n + q) for n substrings and q queries.

1Preprocessing substrings allows for efficient query resolution.
2Only substrings of length up to 30 need to be considered due to the constraints.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.