How do you solve Maximum Strong Pair XOR II on LeetCode?

Maximum Strong Pair XOR II (LeetCode #2935) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Sorting the array helps in efficiently finding valid pairs that satisfy the strong pair condition.

What is the brute force solution for Maximum Strong Pair XOR II?

The brute force approach for Maximum Strong Pair XOR II is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check all possible pairs in the array to see if they form a strong pair and calculate their XOR. This method is straightforward but inefficient for large arrays. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Maximum Strong Pair XOR II?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases, such as arrays with duplicate values or very small sizes. Discuss your thought process and approach with the interviewer to ensure you're on the right track.

What are common mistakes when solving Maximum Strong Pair XOR II?

Not considering the condition for strong pairs correctly, leading to incorrect pairs being checked. Overlooking the need to sort the array before checking pairs, which can lead to missed valid pairs.

#2935

Maximum Strong Pair XOR II

Hard

ArrayHash TableBit ManipulationTrieSliding WindowHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(1)

By sorting the array and leveraging the condition for strong pairs, we can significantly reduce the number of pairs we need to check. This allows us to focus on pairs that are likely to yield a high XOR.

⚙️

Algorithm

4 steps

1Step 1: Sort the array nums.
2Step 2: Initialize max_xor to 0.
3Step 3: Iterate through the sorted array and for each element, find valid pairs that satisfy the strong pair condition y <= 2 * x.
4Step 4: For each valid pair, calculate the XOR and update max_xor if this XOR is greater.

solution.py11 lines

1# Full working Python code
2
3def max_strong_pair_xor(nums):
4    nums.sort()
5    max_xor = 0
6    n = len(nums)
7    for i in range(n):
8        for j in range(i, n):
9            if nums[j] <= 2 * nums[i]:
10                max_xor = max(max_xor, nums[i] ^ nums[j])
11    return max_xor

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, and the nested loop runs in O(n) in the worst case, leading to an overall complexity of O(n log n). The space complexity is O(1) since we are not using additional data structures.

1Sorting the array helps in efficiently finding valid pairs that satisfy the strong pair condition.
2The condition for strong pairs can be simplified to y <= 2 * x, which reduces the number of pairs we need to check.

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