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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.

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!

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the array to count even numbers

✓ Linear Growth

Space Complexity

O(1)

Only storing a counter variable

✓ Linear Space

Constraints

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

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
✓ Optimal Mathematical Insight	O(n)	O(1)	Count even numbers - if there are 2+ even numbers, return true
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 — Algorithm Steps

Count how many even numbers exist in the array
If count >= 2, return true (we can select any 2 even numbers)
If count < 2, return false (impossible to get even OR)

Visualization

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Step-by-Step Walkthrough

Identify Pattern

OR result is even only if all inputs are even

Count Even Numbers

Scan array and count even numbers

Apply Logic

If 2+ even numbers exist, return true

Code -

solution.c — C

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

bool hasTrailingZeros(int* nums, int n) {
    // Key insight: OR result has trailing zeros only if ALL selected numbers are even
    int evenCount = 0;
    
    for (int i = 0; i < n; i++) {
        if (nums[i] % 2 == 0) {
            evenCount++;
            if (evenCount >= 2) {
                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(n)

Single pass through the array to count even numbers

✓ Linear Growth

Space Complexity

O(1)

Only storing a counter variable

✓ Linear Space

Constraints

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

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Optimal Mathematical Insight — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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