LeetCode 2932 - Maximum Strong Pair XOR I

The problem gives us an integer array nums, and we must choose two numbers x and y from the array such that they form a strong pair.

leetcodeeasyarrayhash-tablebit-manipulationtriesliding-window

LeetCode Problem 2932

Difficulty: 🟢 Easy
Topics: Array, Hash Table, Bit Manipulation, Trie, Sliding Window

Solution

Problem Understanding

The problem gives us an integer array nums, and we must choose two numbers x and y from the array such that they form a strong pair. A pair is considered strong if:

$$|x - y| \leq \min(x, y)$$

Among all valid strong pairs, we must compute the maximum possible value of:

$$x \oplus y$$

where ⊕ represents the bitwise XOR operation.

An important detail is that we are allowed to choose the same element twice. That means (x, x) is always a valid pair because:

$$|x - x| = 0 \leq x$$

So the answer is guaranteed to exist.

The input size is very small:

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

These constraints immediately tell us that even an $O(n^2)$ solution is completely acceptable. With at most 50 elements, there are only:

$$50^2 = 2500$$

possible pairs to examine.

The main challenge is not performance, but correctly checking whether a pair satisfies the strong pair condition.

A useful observation comes from simplifying the condition. Suppose x <= y. Then:

$$|x - y| = y - x$$

and:

$$\min(x, y) = x$$

So the condition becomes:

$$y - x \leq x$$

which simplifies to:

$$y \leq 2x$$

This interpretation helps us reason about which pairs are valid.

Some important edge cases include arrays with only one element, arrays where no distinct elements form a strong pair, and arrays containing repeated values. Since choosing the same element twice is allowed, the minimum possible answer is always 0.

Approaches

Brute Force Approach

The most direct solution is to examine every possible pair (nums[i], nums[j]).

For each pair:

  1. Compute whether it satisfies the strong pair condition.
  2. If it does, compute the XOR value.
  3. Track the maximum XOR encountered.

Since the array size is very small, this approach is efficient enough.

The correctness is straightforward because every possible pair is checked exactly once. No valid pair can be missed, so the maximum XOR among strong pairs is guaranteed to be found.

Optimized Observation

Although the brute force solution is already fast enough, we can still improve the reasoning by simplifying the condition.

If we sort the numbers and assume:

$$x \leq y$$

then the strong pair condition simplifies to:

$$y \leq 2x$$

This observation allows us to reason about valid ranges more efficiently. In larger constraint versions of the problem, this leads naturally to sliding window and trie based solutions.

However, for this problem's constraints, a full trie implementation would add unnecessary complexity. The cleanest optimal solution remains checking all pairs directly.

Approach Comparison

Approach Time Complexity Space Complexity Notes
Brute Force O(n²) O(1) Checks every pair directly
Optimal O(n²) O(1) Same complexity, but uses the simplified strong pair observation

Algorithm Walkthrough

Optimal Algorithm

  1. Initialize a variable max_xor to 0.
  2. Iterate through every possible pair of indices (i, j) in the array.
  3. For each pair:

  • Let:

  • x = nums[i]

  • y = nums[j]

  1. Check whether the pair is strong by verifying:

$$|x - y| \leq \min(x, y)$$

  1. If the pair is valid:
  • Compute:

$$x \oplus y$$

  • Update max_xor if this value is larger.
  1. After all pairs are processed, return max_xor.

Why it works

The algorithm explicitly evaluates every possible pair in the array, including pairs where the same element is used twice. Since every strong pair is considered, and the algorithm keeps track of the maximum XOR value among them, the final result must be the correct maximum strong pair XOR.

Python Solution

from typing import List

class Solution:
    def maximumStrongPairXor(self, nums: List[int]) -> int:
        max_xor = 0

        for i in range(len(nums)):
            for j in range(len(nums)):
                x = nums[i]
                y = nums[j]

                if abs(x - y) <= min(x, y):
                    max_xor = max(max_xor, x ^ y)

        return max_xor

The implementation follows the algorithm directly.

The outer loop selects the first number in the pair, while the inner loop selects the second number. Since the problem allows using the same element twice, both loops iterate across the full array.

For each pair, we compute the absolute difference and compare it against the smaller value using:

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

If the condition is true, the pair is strong. We then compute the XOR value using the ^ operator and update the running maximum.

The implementation uses only constant extra space and is extremely concise because the constraints are small.

Go Solution

import "math"

func maximumStrongPairXor(nums []int) int {
    maxXor := 0

    for i := 0; i < len(nums); i++ {
        for j := 0; j < len(nums); j++ {
            x := nums[i]
            y := nums[j]

            if int(math.Abs(float64(x-y))) <= min(x, y) {
                currentXor := x ^ y

                if currentXor > maxXor {
                    maxXor = currentXor
                }
            }
        }
    }

    return maxXor
}

func min(a, b int) int {
    if a < b {
        return a
    }

    return b
}

The Go version follows the same logic as the Python implementation.

One difference is that Go does not provide a built in integer min function for older versions, so we define a helper function manually.

Another difference is the use of math.Abs, which operates on floating point values. We therefore convert the integer difference to float64 and cast the result back to int.

Since the input values are very small, there is no risk of integer overflow.

Worked Examples

Example 1

Input:

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

We examine all pairs.

x y Strong Pair? XOR
1 1 Yes 0
1 2 Yes 3
1 3 No -
2 3 Yes 1
2 4 Yes 6
3 4 Yes 7
3 5 Yes 6
4 5 Yes 1

The maximum XOR among valid pairs is:

3 ^ 4 = 7

Final answer:

7

Example 2

Input:

nums = [10,100]

Check all pairs.

x y Strong Pair? XOR
10 10 Yes 0
10 100 No -
100 10 No -
100 100 Yes 0

No distinct strong pair exists.

Final answer:

0

Example 3

Input:

nums = [5,6,25,30]
x y Strong Pair? XOR
5 5 Yes 0
5 6 Yes 3
25 30 Yes 7
6 25 No -
30 30 Yes 0

The maximum XOR is:

25 ^ 30 = 7

Final answer:

7

Complexity Analysis

Measure Complexity Explanation
Time O(n²) Every pair of elements is checked
Space O(1) Only a few variables are used

The algorithm uses two nested loops over the array. Since each loop runs n times, the total number of iterations is:

$$n^2$$

With n <= 50, this is extremely efficient in practice.

No auxiliary data structures proportional to input size are allocated, so the space usage remains constant.

Test Cases

from typing import List

class Solution:
    def maximumStrongPairXor(self, nums: List[int]) -> int:
        max_xor = 0

        for i in range(len(nums)):
            for j in range(len(nums)):
                x = nums[i]
                y = nums[j]

                if abs(x - y) <= min(x, y):
                    max_xor = max(max_xor, x ^ y)

        return max_xor

solution = Solution()

assert solution.maximumStrongPairXor([1,2,3,4,5]) == 7
# Basic example with several valid strong pairs

assert solution.maximumStrongPairXor([10,100]) == 0
# No distinct strong pair exists

assert solution.maximumStrongPairXor([5,6,25,30]) == 7
# Multiple disconnected strong pair groups

assert solution.maximumStrongPairXor([1]) == 0
# Single element array

assert solution.maximumStrongPairXor([2,2,2]) == 0
# All identical values

assert solution.maximumStrongPairXor([1,2]) == 3
# Smallest non-trivial valid pair

assert solution.maximumStrongPairXor([1,10]) == 0
# Distinct values that do not form a strong pair

assert solution.maximumStrongPairXor([8,9,10]) == 3
# Consecutive values with multiple valid pairs

assert solution.maximumStrongPairXor([50,100]) == 86
# Boundary case where y == 2x is still valid

assert solution.maximumStrongPairXor([100,100]) == 0
# Largest allowed values with duplicates
Test Why
[1,2,3,4,5] Verifies standard mixed behavior
[10,100] Verifies invalid distinct pairs
[5,6,25,30] Verifies separated valid groups
[1] Tests minimum input size
[2,2,2] Tests repeated identical values
[1,2] Tests smallest valid distinct pair
[1,10] Tests no valid distinct pairs
[8,9,10] Tests overlapping valid ranges
[50,100] Tests boundary condition y = 2x
[100,100] Tests maximum value duplicates

Edge Cases

Single Element Array

If the array contains only one value, the only possible pair is the element paired with itself.

For example:

[7]

The pair (7,7) is always strong because the difference is zero. The XOR is also zero.

A buggy implementation might accidentally skip self pairs, but this implementation checks all (i,j) combinations, including identical indices.

No Valid Distinct Pairs

Consider:

[1, 100]

The pair (1,100) is not strong because:

$$|1 - 100| = 99$$

and:

$$\min(1,100) = 1$$

Since 99 > 1, the pair is invalid.

Only (1,1) and (100,100) remain valid, both producing XOR 0.

The implementation handles this naturally because every pair is validated independently.

Boundary Condition Where Difference Equals Minimum

The condition allows equality:

$$|x-y| \leq \min(x,y)$$

not strictly less than.

For example:

x = 50
y = 100

We have:

$$|50-100| = 50$$

and:

$$\min(50,100) = 50$$

So the pair is valid.

An incorrect implementation using < instead of <= would fail this case. The provided solution correctly uses <=.