Check if Bitwise OR Has Trailing Zeros - Practice Coding Problems

Check if Bitwise OR Has Trailing Zeros - Problem

You're given an array of positive integers, and your task is to determine if it's possible to select two or more elements such that their bitwise OR operation results in a number with at least one trailing zero.

A trailing zero in binary representation means the number ends with one or more '0' bits. For example:

5 in binary is "101" - no trailing zeros
4 in binary is "100" - two trailing zeros
12 in binary is "1100" - two trailing zeros

Return true if such a selection is possible, false otherwise.

Key insight: A number has trailing zeros if and only if it's divisible by 2 (even). So we need to find if any combination of 2+ elements can produce an even OR result.

Input & Output

example_1.py — Basic Case with Even Numbers

$ Input: [2, 1, 3]

› Output: false

💡 Note: Only one even number (2) exists. We need at least 2 even numbers to create an OR result with trailing zeros. 2|1=3 (odd), 2|3=3 (odd), 1|3=3 (odd), 2|1|3=3 (odd).

example_2.py — Multiple Even Numbers

$ Input: [1, 2, 3, 4, 5]

› Output: true

💡 Note: We have two even numbers: 2 and 4. Their OR is 2|4 = 6 (binary: 110), which has one trailing zero. Since 6 is even, it has at least one trailing zero.

example_3.py — All Odd Numbers

$ Input: [1, 3, 5, 7]

› Output: false

💡 Note: All numbers are odd. Any OR combination of odd numbers will always result in an odd number, which cannot have trailing zeros.

Constraints

2 ≤ nums.length ≤ 100
1 ≤ nums[i] ≤ 100
All integers are positive

Visualization

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Understanding the Visualization

Observe Binary Patterns

Even numbers end in 0, odd numbers end in 1

Apply OR Logic

OR with any 1 bit produces 1 (odd result)

Find Valid Combinations

Only even-only combinations can produce trailing zeros

Key Takeaway

🎯 Key Insight: For bitwise OR to produce trailing zeros (even result), ALL selected numbers must be even. Count even numbers - if ≥2 exist, answer is true!

Asked in

G Google 23 a Amazon 18 ⊞ Microsoft 15 f Meta 12

The optimal solution uses a key mathematical insight: a bitwise OR result has trailing zeros (is even) only if ALL selected numbers are even. Since we need at least 2 elements, we simply count even numbers in the array. If there are 2 or more even numbers, return true; otherwise, return false. This gives us O(n) time complexity with O(1) space.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Check All Combinations)	O(2^n * n)	O(1)	Generate all possible combinations of 2+ elements and check their OR result
Optimal Mathematical Insight	O(n)	O(1)	Count even numbers - if there are 2+ even numbers, return true

Brute Force (Check All Combinations) — Algorithm Steps

Generate all combinations of 2 elements, then 3 elements, etc.
For each combination, compute the bitwise OR of all elements
Check if the OR result is even (has trailing zeros)
Return true as soon as we find one valid combination

Visualization

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Step-by-Step Walkthrough

Generate Combinations

Create all possible pairs and larger groups

Compute OR

Calculate bitwise OR for each combination

Check Trailing Zeros

Test if result is even (has trailing zeros)

Code -

solution.c — C

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

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

bool hasTrailingZeros(int* nums, int n) {
    // Check all combinations using bit mask
    for (int mask = 3; mask < (1 << n); mask++) {
        if (popcount(mask) < 2) continue;
        
        int orResult = 0;
        for (int i = 0; i < n; i++) {
            if (mask & (1 << i)) {
                orResult |= nums[i];
            }
        }
        
        if (orResult % 2 == 0) {
            return true;
        }
    }
    return false;
}

int main() {
    int nums[1000];
    int n = 0;
    char c;
    
    scanf("%c", &c); // Read '['
    while (scanf("%d", &nums[n]) == 1) {
        n++;
        scanf("%c", &c); // Read ',' or ']'
        if (c == ']') break;
    }
    
    printf("%s\n", hasTrailingZeros(nums, n) ? "true" : "false");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n * n)

We generate O(2^n) combinations and each combination takes O(n) time to compute OR

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Check All Combinations) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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