0925MediumArrayBit Manipulation

Longest Pick With a Non-Zero Combination

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Trains the technique from

LeetCode 3702Longest Subsequence With Non-Zero Bitwise XOR

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.

Register values are given as nums. A pick takes some of them, keeping their order, and its combination is the bitwise exclusive-or of everything picked.

Return how many values the longest pick with a non-zero combination holds, or 0 when no pick has one.

Examples

Example 1

Input: nums = [13, 27, 6, 41]
Output: 4

The four values combine to something other than zero, so the whole list is the longest pick.

Example 2

Input: nums = [6, 10, 12]
Output: 2

All three combine to zero, so dropping one is the best available, and dropping any of them leaves that value behind as the combination, which is not zero.

Example 3

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

Every value is zero, so every pick combines to zero and there is no answer.

Constraints

1 <= nums.length <= 10^5
0 <= nums[i] <= 10^9

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 longest_subsequence(nums: list[int]) -> int:

Java

public int longestSubsequence(int[] nums)

The intended approach

Time O(n) for n values, one pass.Space O(1).

Take everything first and see what happens. If the whole list combines to something non-zero, nothing beats it, so the answer is the full length. If it combines to zero, the full list is out, so the best possible is one short. Dropping a single value leaves that value as the combination, since removing it from a total of zero leaves exactly it, so dropping any non-zero value works and one short is reachable. That only fails when every value is zero, in which case every pick combines to zero and there is no answer at all. So the whole thing is one pass to combine and one check for a non-zero value.

Pythonentry point: longest_subsequence

def longest_subsequence(nums):
    total = 0
    for v in nums:
        total ^= v
    if total != 0:
        return len(nums)
    for v in nums:
        if v != 0:
            return len(nums) - 1
    return 0

The obvious version

Try every pick there is, combine each one, and keep the longest whose combination is not zero. Exponential in the value count and it makes no argument about dropping a single value.

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_longest_subsequence(nums):
    n = len(nums)
    best = 0
    for mask in range(1 << n):
        total = 0
        size = 0
        for i in range(n):
            if mask >> i & 1:
                total ^= nums[i]
                size += 1
        if total != 0 and size > best:
            best = size
    return best

How this problem was checked

Reference solution executed against 12 test cases — 12 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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