LeetCode #2932 — EASY

Maximum Strong Pair XOR I

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given a 0-indexed integer array nums. A pair of integers x and y is called a strong pair if it satisfies the condition:

|x - y| <= min(x, y)

You need to select two integers from nums such that they form a strong pair and their bitwise XOR is the maximum among all strong pairs in the array.

Return the maximum XOR value out of all possible strong pairs in the array nums.

Note that you can pick the same integer twice to form a pair.

Example 1:

Input: nums = [1,2,3,4,5]
Output: 7
Explanation: 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
Explanation: 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
Explanation: 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

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array nums. A pair of integers x and y is called a strong pair if it satisfies the condition: |x - y| <= min(x, y) You need to select two integers from nums such that they form a strong pair and their bitwise XOR is the maximum among all strong pairs in the array. Return the maximum XOR value out of all possible strong pairs in the array nums. Note that you can pick the same integer twice to form a pair.

Baseline thinking

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

Pattern signal: Array · Hash Map · Bit Manipulation · Trie · Sliding Window

Example 1

[1,2,3,4,5]

Example 2

[10,100]

Example 3

[5,6,25,30]

Core Insight

What unlocks the optimal approach

The constraints are small enough to make brute-force solutions pass.

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 #2932: Maximum Strong Pair XOR I
class Solution {
    public int maximumStrongPairXor(int[] nums) {
        int ans = 0;
        for (int x : nums) {
            for (int y : nums) {
                if (Math.abs(x - y) <= Math.min(x, y)) {
                    ans = Math.max(ans, x ^ y);
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2932: Maximum Strong Pair XOR I
func maximumStrongPairXor(nums []int) (ans int) {
	for _, x := range nums {
		for _, y := range nums {
			if abs(x-y) <= min(x, y) {
				ans = max(ans, x^y)
			}
		}
	}
	return
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #2932: Maximum Strong Pair XOR I
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))

// Accepted solution for LeetCode #2932: Maximum Strong Pair XOR I
fn maximum_strong_pair_xor(nums: Vec<i32>) -> i32 {
    let mut ret = 0;
    let len = nums.len();

    for i in 0..len {
        for j in i..len {
            if (nums[i] - nums[j]).abs() <= std::cmp::min(nums[i], nums[j]) {
                ret = std::cmp::max(ret, nums[i] ^ nums[j]);
            }
        }
    }

    ret
}

fn main() {
    let nums = vec![1, 2, 3, 4, 5];
    let ret = maximum_strong_pair_xor(nums);
    println!("ret={ret}");
}

#[test]
fn test() {
    {
        let nums = vec![1, 2, 3, 4, 5];
        let ret = maximum_strong_pair_xor(nums);
        assert_eq!(ret, 7);
    }
    {
        let nums = vec![10, 100];
        let ret = maximum_strong_pair_xor(nums);
        assert_eq!(ret, 0);
    }
    {
        let nums = vec![5, 6, 25, 30];
        let ret = maximum_strong_pair_xor(nums);
        assert_eq!(ret, 7);
    }
}

// Accepted solution for LeetCode #2932: Maximum Strong Pair XOR I
function maximumStrongPairXor(nums: number[]): number {
    let ans = 0;
    for (const x of nums) {
        for (const y of nums) {
            if (Math.abs(x - y) <= Math.min(x, y)) {
                ans = Math.max(ans, x ^ y);
            }
        }
    }
    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)

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.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.