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

Maximum Strong Pair XOR I (LeetCode #2932) 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: Strong pairs are defined by the relationship between the two numbers, which can be leveraged to limit the pairs we check.

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

The brute force approach for Maximum Strong Pair XOR I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible pairs of integers in the array to see if they form a strong pair. If they do, we calculate their XOR and keep track of the maximum XOR found. The optimal Optimal Solution approach improves this to O(n log n).

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

Always clarify the problem statement and constraints before diving into coding. Consider edge cases, such as arrays with duplicate values or very small sizes. Explain your thought process while coding to demonstrate your understanding of the problem.

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

Not checking the condition for strong pairs correctly, leading to incorrect maximum XOR calculations. Overlooking the possibility of selecting the same integer twice.

#2932

Maximum Strong Pair XOR I

Easy

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)

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Intuition

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

The optimal solution leverages the properties of XOR and the strong pair condition to reduce the number of pairs we need to check. By sorting the array and only checking adjacent elements, we can efficiently find the maximum XOR.

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Algorithm

5 steps

1Step 1: Sort the array nums.
2Step 2: Initialize max_xor to 0.
3Step 3: Iterate through the sorted array and check adjacent pairs (nums[i], nums[i+1]).
4Step 4: For each adjacent pair, check if they satisfy the strong pair condition. If they do, calculate their XOR and update max_xor.
5Step 5: Return max_xor after checking all adjacent pairs.

solution.py14 lines

1# Full working Python code
2
3def max_strong_pair_xor_optimal(nums):
4    nums.sort()
5    max_xor = 0
6    n = len(nums)
7    for i in range(n - 1):
8        if abs(nums[i] - nums[i + 1]) <= min(nums[i], nums[i + 1]):
9            max_xor = max(max_xor, nums[i] ^ nums[i + 1])
10    return max_xor
11
12# Example usage
13print(max_strong_pair_xor_optimal([1, 2, 3, 4, 5]))  # Output: 7
14

ℹ

Complexity note: The time complexity is O(n log n) due to sorting the array, while the space complexity is O(1) since we are not using any additional data structures that grow with input size.

1Strong pairs are defined by the relationship between the two numbers, which can be leveraged to limit the pairs we check.
2Sorting the array allows us to efficiently find valid pairs by only checking adjacent elements.

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