#2932

Easy

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Maximum Strong Pair XOR I

Find the maximum XOR of any strong pair in an integer array using array scanning and hash-based lookups efficiently.

array hash table bit manipulation trie sliding window

Problem Statement

You are given a 0-indexed integer array nums. A strong pair consists of two integers x and y from nums that satisfy a given pairing condition.

Return the maximum bitwise XOR value obtained from any strong pair in nums. Identify all strong pairs efficiently and compute the largest XOR.

Examples

Example 1

Input: nums = [1,2,3,4,5]

Output: 7

There are 11 strong pairs in the array nums: (1, 1), (1, 2), (2, 2), (2, 3), (2, 4), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5) and (5, 5). The maximum XOR possible from these pairs is 3 XOR 4 = 7.

Example 2

Input: nums = [10,100]

Output: 0

There are 2 strong pairs in the array nums: (10, 10) and (100, 100). The maximum XOR possible from these pairs is 10 XOR 10 = 0 since the pair (100, 100) also gives 100 XOR 100 = 0.

Example 3

Input: nums = [5,6,25,30]

Output: 7

There are 6 strong pairs in the array nums: (5, 5), (5, 6), (6, 6), (25, 25), (25, 30) and (30, 30). The maximum XOR possible from these pairs is 25 XOR 30 = 7 since the only other non-zero XOR value is 5 XOR 6 = 3.

Constraints

1 <= nums.length <= 50
1 <= nums[i] <= 100

Solution Approach

Brute-force scan

Check every possible pair of integers in nums. For each pair, verify if it is a strong pair and compute its XOR. Track the maximum XOR seen. This works within the small constraint limits but is inefficient for larger arrays.

Hash lookup optimization

Use a hash set to store previously seen numbers. For each number, iterate through the set and compute XOR with all valid strong pair candidates. Update the maximum XOR while scanning, avoiding redundant pair checks.

Bit manipulation acceleration

Leverage bitwise properties to compute XORs efficiently. Consider iterating bits from most significant to least, pruning candidate pairs early when they cannot exceed the current maximum XOR. This complements the array scanning plus hash lookup pattern.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n^2) for brute-force and O(n*m) for hash lookup approaches, where n is array length and m is the number of strong pair candidates. Space complexity is O(n) for storing elements in a hash set.

What Interviewers Usually Probe

Candidate attempts nested loops without pruning.
Candidate considers bitwise operations to optimize XOR checks.
Candidate correctly uses hash set to avoid duplicate pair computations.

Common Pitfalls or Variants

Common pitfalls

Failing to include all identical number pairs as strong pairs.
Overlooking the maximum XOR coming from non-adjacent numbers.
Misusing bitwise XOR without checking strong pair condition.

Follow-up variants

Compute the maximum XOR among strong pairs under a modified pairing condition.
Find the minimum XOR of all strong pairs instead of maximum.
Extend to arrays with larger elements and test the efficiency of hash-based pruning.

FAQ

What is the main approach for Maximum Strong Pair XOR I?

The primary approach is array scanning combined with hash lookups, computing XOR for each strong pair candidate while tracking the maximum.

Can a brute-force solution pass for Maximum Strong Pair XOR I?

Yes, because nums.length <= 50, a brute-force nested loop checking all pairs is sufficient within constraints.

How does hash lookup optimize the XOR computation?

It stores previously seen numbers and only computes XORs with valid candidates, avoiding repeated pair checks and reducing redundant computations.

Are identical numbers considered strong pairs?

Yes, pairs of identical numbers are included if they satisfy the strong pair condition, which can contribute to the maximum XOR.

Does this problem use bit manipulation techniques?

Yes, bit manipulation helps compute XOR values efficiently and prune candidate pairs when their XOR cannot exceed the current maximum.

terminal

Solution

Solution 1: Enumeration

We can enumerate each pair of numbers $(x, y)$ in the array. If $|x - y| \leq \min(x, y)$, then this pair is a strong pair. We can calculate the XOR value of this pair and update the answer.

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class Solution:
    def maximumStrongPairXor(self, nums: List[int]) -> int:
        return max(x ^ y for x in nums for y in nums if abs(x - y) <= min(x, y))

Solution 2: Sorting + Binary Trie

Observing the inequality $|x - y| \leq \min(x, y)$, which involves absolute value and minimum value, we can assume $x \leq y$, then we have $y - x \leq x$, that is, $y \leq 2x$. We can enumerate $y$ from small to large, then $x$ must satisfy the inequality $y \leq 2x$.

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class Solution:
    def maximumStrongPairXor(self, nums: List[int]) -> int:
        return max(x ^ y for x in nums for y in nums if abs(x - y) <= min(x, y))

Continue Practicing

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