0615MediumArrayHash TableBit ManipulationTrie

Largest Channel Spread

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Trains the technique from

LeetCode 421Maximum XOR of Two Numbers in an Array

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A lighting desk holds n fixtures. Fixture i is described by the non-negative integer nums[i], whose set bits say which channels that fixture drives.

Pair up two fixtures at positions i and j, where i and j may be the same position. The pair's spread is the number you get by combining the two descriptions with bitwise exclusive-or, so a bit of the spread is set exactly when one fixture of the pair drives that channel and the other does not.

Return the largest spread over all pairs. Pairing a fixture with itself gives a spread of 0, so a desk with a single fixture answers 0.

Examples

Example 1

Input: nums = [8, 10, 2]
Output: 10

Pairing the fixtures at positions 0 and 2 gives `8 XOR 2 = 10`, the largest spread this desk reaches.

Example 2

Input: nums = [1024, 2048, 4096]
Output: 6144

Each description has one bit set. Pairing 4096 with 2048 gives a spread of 6144, which is the largest here.

Example 3

Input: nums = [5, 5]
Output: 0

Both fixtures drive the same channels, so every pair, including the two distinct positions, spreads to 0.

Constraints

1 <= nums.length <= 2 * 10^5
0 <= nums[i] <= 2^31 - 1

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def find_maximum_x_o_r(nums: list[int]) -> int:

Java

public int findMaximumXOR(int[] nums)

The intended approach

Time O(31 * n) — one pass over all fixtures per bit position.Space O(n) for the set of truncated descriptions.

The answer is settled one bit at a time, starting from bit 30 and working down, keeping the best value achievable using only the bits decided so far. At each step the code asks whether the next bit can be a 1: it shifts every fixture description right so only the decided bits remain, stores those truncated values in a set, and looks for a pair whose exclusive-or equals the candidate. Rearranging that condition turns the search into one set lookup per truncated value. If a pair exists the bit is kept as 1, otherwise it is left at 0.

Pythonentry point: find_maximum_x_o_r

def find_maximum_x_o_r(nums):
    # O(31 * n) time, O(n) space. Build the answer one bit at a time from the
    # top. Suppose the top few bits of the answer are already fixed; ask
    # whether a 1 can be appended. Truncate every mask to the bits decided so
    # far and drop those truncations in a set: a 1 is reachable exactly when
    # some pair of truncations differs by the candidate value, which is a
    # single lookup per truncation because the exclusive-or of two of them
    # equalling the candidate means one is the candidate combined with the
    # other.
    best = 0
    for bit in range(30, -1, -1):
        best <<= 1
        candidate = best | 1
        prefixes = {value >> bit for value in nums}
        for p in prefixes:
            if (candidate ^ p) in prefixes:
                best = candidate
                break
    return best

The obvious version

Take every pair of positions, including a position with itself, combine the two descriptions with exclusive-or, and keep the largest result.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_find_maximum_x_o_r(nums):
    best = 0
    for i in range(len(nums)):
        for j in range(i, len(nums)):
            spread = nums[i] ^ nums[j]
            if spread > best:
                best = spread
    return best

How this problem was checked

Reference solution executed against 14 test cases — 14 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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