Problem Description

You have an array of integers nums (0-indexed). Your task is to find two numbers from this array that form a "strong pair" and have the maximum possible XOR value.

A strong pair is defined as two integers x and y that satisfy the condition:

• |x - y| <= min(x, y)

This means the absolute difference between the two numbers must be less than or equal to the smaller of the two numbers.

Your goal is to:

1. Find all possible strong pairs in the array
2. Calculate the bitwise XOR for each strong pair
3. Return the maximum XOR value among all strong pairs

Important notes:

• You can use the same number twice to form a pair (e.g., if the array contains 5, you can pair 5 with itself)
• The XOR operation is the bitwise exclusive OR operation

For example, if you have two numbers x = 5 and y = 3:

• Check if they form a strong pair: |5 - 3| = 2 and min(5, 3) = 3, so 2 <= 3 ✓
• Calculate their XOR: 5 ^ 3 = 6
• This would be one candidate for the maximum XOR value

The solution iterates through all possible pairs, checks if each pair is strong, calculates the XOR for valid pairs, and returns the maximum XOR value found.

Intuition

When we look at this problem, we need to find the maximum XOR value among all valid strong pairs. The key insight is that the constraint for a strong pair |x - y| <= min(x, y) significantly limits which pairs we need to consider.

Let's think about what this constraint actually means. If we assume x >= y (without loss of generality), then:

• |x - y| = x - y (since x >= y)
• min(x, y) = y
• So the constraint becomes: x - y <= y
• Which simplifies to: x <= 2y

This tells us that for any pair to be strong, the larger number cannot be more than twice the smaller number. This is a relatively loose constraint that still allows many pairs.

Since the array size isn't specified as being particularly large in the problem, and we need to check all possible strong pairs anyway to find the maximum XOR, a brute force approach makes sense. We can:

1. Check every possible pair (x, y) from the array
2. Verify if it satisfies the strong pair condition
3. If it does, calculate x ^ y
4. Keep track of the maximum XOR value seen

The beauty of this approach is its simplicity - we don't need any complex data structures or algorithms. We simply enumerate all possibilities, filter by the strong pair condition, and find the maximum. The one-liner solution elegantly combines all these steps using Python's generator expression with max().

The reason we can use the same element twice (as mentioned in the problem) is that a number paired with itself always forms a strong pair: |x - x| = 0 <= min(x, x) = x (assuming x >= 0). However, x ^ x = 0, so same-element pairs won't contribute to the maximum XOR unless all valid pairs give 0.

Learn more about Trie and Sliding Window patterns.

Solution Approach

The solution uses a brute force enumeration approach, which directly implements the problem requirements in a single line of code.

Implementation Details:

The solution leverages Python's generator expression combined with the max() function to find the answer:

return max(x ^ y for x in nums for y in nums if abs(x - y) <= min(x, y))

Let's break down how this works:

1. Nested Iteration: The code uses two nested loops through nums:

   • for x in nums - iterates through each element as the first number
   • for y in nums - for each x, iterates through all elements as the second number
   • This generates all possible pairs (x, y), including pairs where x == y

2. Strong Pair Filtering: For each pair (x, y), we check the condition:

   • abs(x - y) <= min(x, y)
   • Only pairs satisfying this condition are considered for XOR calculation

3. XOR Calculation: For each valid strong pair, we calculate:

   • x ^ y - the bitwise XOR of the two numbers

4. Maximum Selection: The max() function:

   • Evaluates all XOR values from valid strong pairs
   • Returns the largest XOR value found

Time Complexity: O(n²) where n is the length of the array, as we check all possible pairs.

Space Complexity: O(1) as we only use a constant amount of extra space (the generator doesn't create a list but evaluates on-the-fly).

This approach is optimal for smaller arrays and when we need to guarantee finding the exact maximum. The simplicity of the implementation makes it easy to verify correctness - we're literally checking every possibility that meets the criteria and selecting the best one.

Example Walkthrough

Let's walk through the solution with a small example: nums = [1, 2, 3, 4, 5]

Step 1: Generate all possible pairs We'll check every combination of (x, y) from the array:

Step 2: Check each pair and calculate XOR if valid

Let me trace through several pairs:

• Pair (1, 1):

  • Check: |1 - 1| = 0 ≤ min(1, 1) = 1 ✓ Valid
  • XOR: 1 ^ 1 = 0

• Pair (1, 2):

  • Check: |1 - 2| = 1 ≤ min(1, 2) = 1 ✓ Valid
  • XOR: 1 ^ 2 = 3

• Pair (1, 3):

  • Check: |1 - 3| = 2 ≤ min(1, 3) = 1 ✗ Invalid (2 > 1)
  • Skip this pair

• Pair (2, 3):

  • Check: |2 - 3| = 1 ≤ min(2, 3) = 2 ✓ Valid
  • XOR: 2 ^ 3 = 1

• Pair (3, 4):

  • Check: |3 - 4| = 1 ≤ min(3, 4) = 3 ✓ Valid
  • XOR: 3 ^ 4 = 7

• Pair (4, 5):

  • Check: |4 - 5| = 1 ≤ min(4, 5) = 4 ✓ Valid
  • XOR: 4 ^ 5 = 1

• Pair (2, 5):

  • Check: |2 - 5| = 3 ≤ min(2, 5) = 2 ✗ Invalid (3 > 2)
  • Skip this pair

Step 3: Find maximum XOR From all valid pairs, we collected XOR values: 0, 3, 1, 7, 1, ...

The maximum XOR value is 7 (from the pair (3, 4)).

The solution max(x ^ y for x in nums for y in nums if abs(x - y) <= min(x, y)) automatically performs all these checks and returns 7.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n^2), where n is the length of the array nums.

The code uses a nested iteration through the array with two variables x and y, both iterating over all elements in nums. This creates n × n = n^2 pairs to check. For each pair, the operations performed (absolute difference abs(x - y), minimum min(x, y), comparison <=, and XOR ^) are all O(1) operations. The max() function processes all valid pairs generated by the nested loops, but doesn't add additional complexity beyond the O(n^2) pairs already being generated. Therefore, the overall time complexity is O(n^2).

Space Complexity: O(1).

The generator expression (x ^ y for x in nums for y in nums if abs(x - y) <= min(x, y)) doesn't create a list to store all results at once; instead, it generates values one at a time as needed by the max() function. The max() function only needs to keep track of the current maximum value as it iterates through the generator, requiring constant extra space. No additional data structures proportional to the input size are created, resulting in O(1) space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Misunderstanding the Strong Pair Condition

A frequent mistake is misinterpreting the strong pair condition |x - y| <= min(x, y). Developers often make these errors:

• Forgetting absolute value: Writing x - y <= min(x, y) instead of abs(x - y) <= min(x, y)
• Using max instead of min: Writing abs(x - y) <= max(x, y) which would always be true for non-negative numbers
• Not considering self-pairs: Assuming x and y must be different values

Solution: Always use the exact condition abs(x - y) <= min(x, y). Note that when x == y, the condition becomes 0 <= x, which is satisfied for all non-negative numbers.

2. Overlooking Edge Cases with Negative Numbers

The strong pair condition behaves unexpectedly with negative numbers. For example:

• If x = -5 and y = -3, then min(x, y) = -5 (negative!)
• The condition becomes |-5 - (-3)| <= -5, which is 2 <= -5 (false)
• Two negative numbers can never form a strong pair!

Solution: Be aware that negative numbers in the array will rarely form strong pairs with each other. The condition essentially requires at least one non-negative number in the pair for most cases.

3. Assuming Distinct Array Elements

Some developers optimize by only checking pairs where i < j in the indices, assuming each element appears once:

# INCORRECT if we can use the same value twice
for i in range(len(nums)):
    for j in range(i + 1, len(nums)):
        x, y = nums[i], nums[j]
        # Missing pairs like (nums[i], nums[i])

Solution: The problem allows pairing a number with itself. Use the full nested loop to check all combinations, including (nums[i], nums[i]).

4. Empty Array or Generator Expression

When using the one-liner with max():

return max(x ^ y for x in nums for y in nums if abs(x - y) <= min(x, y))

If no pairs satisfy the strong pair condition, the generator will be empty, causing max() to raise a ValueError.

Solution: Either use a default value with max() or check for valid pairs first:

# Option 1: Provide default value
return max((x ^ y for x in nums for y in nums if abs(x - y) <= min(x, y)), default=0)

# Option 2: Use the explicit loop approach with initialization
max_xor = 0
for x in nums:
    for y in nums:
        if abs(x - y) <= min(x, y):
            max_xor = max(max_xor, x ^ y)
return max_xor