LeetCode #2980 — EASY

Check if Bitwise OR Has Trailing Zeros

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given an array of positive integers nums.

You have to check if it is possible to select two or more elements in the array such that the bitwise OR of the selected elements has at least one trailing zero in its binary representation.

For example, the binary representation of 5, which is "101", does not have any trailing zeros, whereas the binary representation of 4, which is "100", has two trailing zeros.

Return true if it is possible to select two or more elements whose bitwise OR has trailing zeros, return false otherwise.

Example 1:

Input: nums = [1,2,3,4,5]
Output: true
Explanation: If we select the elements 2 and 4, their bitwise OR is 6, which has the binary representation "110" with one trailing zero.

Example 2:

Input: nums = [2,4,8,16]
Output: true
Explanation: If we select the elements 2 and 4, their bitwise OR is 6, which has the binary representation "110" with one trailing zero.
Other possible ways to select elements to have trailing zeroes in the binary representation of their bitwise OR are: (2, 8), (2, 16), (4, 8), (4, 16), (8, 16), (2, 4, 8), (2, 4, 16), (2, 8, 16), (4, 8, 16), and (2, 4, 8, 16).

Example 3:

Input: nums = [1,3,5,7,9]
Output: false
Explanation: There is no possible way to select two or more elements to have trailing zeros in the binary representation of their bitwise OR.

Constraints:

2 <= nums.length <= 100
1 <= nums[i] <= 100

Step 01

Brute Force Baseline

Problem summary: You are given an array of positive integers nums. You have to check if it is possible to select two or more elements in the array such that the bitwise OR of the selected elements has at least one trailing zero in its binary representation. For example, the binary representation of 5, which is "101", does not have any trailing zeros, whereas the binary representation of 4, which is "100", has two trailing zeros. Return true if it is possible to select two or more elements whose bitwise OR has trailing zeros, return false otherwise.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Bit Manipulation

Example 1

[1,2,3,4,5]

Example 2

[2,4,8,16]

Example 3

[1,3,5,7,9]

Core Insight

What unlocks the optimal approach

Bitwise <code>OR</code> can never unset a bit. If there is a solution, there must be a solution with only a pair of elements.
We can brute force the solution: enumerate all the pairs.
As the least significant bit must stay unset, the question is whether the array has at least two even elements.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2980: Check if Bitwise OR Has Trailing Zeros
class Solution {
    public boolean hasTrailingZeros(int[] nums) {
        int cnt = 0;
        for (int x : nums) {
            cnt += (x & 1 ^ 1);
        }
        return cnt >= 2;
    }
}

// Accepted solution for LeetCode #2980: Check if Bitwise OR Has Trailing Zeros
func hasTrailingZeros(nums []int) bool {
	cnt := 0
	for _, x := range nums {
		cnt += (x&1 ^ 1)
	}
	return cnt >= 2
}

# Accepted solution for LeetCode #2980: Check if Bitwise OR Has Trailing Zeros
class Solution:
    def hasTrailingZeros(self, nums: List[int]) -> bool:
        return sum(x & 1 ^ 1 for x in nums) >= 2

// Accepted solution for LeetCode #2980: Check if Bitwise OR Has Trailing Zeros
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2980: Check if Bitwise OR Has Trailing Zeros
// class Solution {
//     public boolean hasTrailingZeros(int[] nums) {
//         int cnt = 0;
//         for (int x : nums) {
//             cnt += (x & 1 ^ 1);
//         }
//         return cnt >= 2;
//     }
// }

// Accepted solution for LeetCode #2980: Check if Bitwise OR Has Trailing Zeros
function hasTrailingZeros(nums: number[]): boolean {
    let cnt = 0;
    for (const x of nums) {
        cnt += (x & 1) ^ 1;
    }
    return cnt >= 2;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.