Check if Bitwise OR Has Trailing Zeros

Check if Bitwise OR Has Trailing Zeros - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: nums = [2,1,4]

› Output: true

💡 Note: We can select 2 and 4. Their OR is 2|4 = 6, which is 110 in binary (has one trailing zero).

Example 2 — All Odd Numbers

$ Input: nums = [1,3,5]

› Output: false

💡 Note: All numbers are odd. Any OR combination will be odd (no trailing zeros). 1|3=3, 1|5=5, 3|5=7, 1|3|5=7.

Example 3 — Multiple Even Numbers

$ Input: nums = [2,4,6,8]

› Output: true

💡 Note: Multiple even numbers exist. For example, 2|4 = 6 (binary: 110) has trailing zero.

Constraints

2 ≤ nums.length ≤ 100
1 ≤ nums[i] ≤ 100

Visualization

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Asked in

G Google 15 M Microsoft 12 a Amazon 8

The key insight is that a number has trailing zeros in binary if and only if it's even. The OR of any two even numbers is always even (has trailing zeros). Best approach is to count even numbers - if there are 2 or more, return true. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Check All Combinations

⏱️ Time: O(2ⁿ × n) Space: O(1)

Try every possible subset of size 2 or more, calculate the bitwise OR of each subset, and check if the result is even (has trailing zeros). A number has trailing zeros if and only if it's even.

Count Even Numbers

⏱️ Time: O(n) Space: O(1)

The key insight is that a number has trailing zeros in binary if and only if it's even. The OR of two even numbers is always even, so we just need to count even numbers.

Brute Force - Check All Combinations — Algorithm Steps

Generate all combinations of 2+ elements
For each combination, calculate bitwise OR
Check if OR result is even (last bit is 0)
Return true if any combination produces even OR

Visualization

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Step-by-Step Walkthrough

Generate Combinations

Create all subsets of size 2+

Calculate OR

For each subset, compute bitwise OR

Check Even

Return true if any OR is even

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int popcount(int x) {
    int count = 0;
    while (x) {
        count += x & 1;
        x >>= 1;
    }
    return count;
}

int solution(int* nums, int n) {
    // Try all possible combinations of 2 or more elements
    for (int mask = 3; mask < (1 << n); mask++) {
        if (popcount(mask) >= 2) { // At least 2 elements
            int orResult = 0;
            for (int i = 0; i < n; i++) {
                if (mask & (1 << i)) {
                    orResult |= nums[i];
                }
            }
            // Check if OR has trailing zeros (is even)
            if (orResult % 2 == 0) {
                return 1;
            }
        }
    }
    return 0;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    int nums[20], n = 0;
    char* token = strtok(line + 1, ",]");
    while (token) {
        nums[n++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    int result = solution(nums, n);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2ⁿ × n)

Generate 2^n subsets, each taking O(n) time to compute OR

✓ Linear Growth

Space Complexity

O(1)

Only use constant extra space for variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check All Combinations — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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