Maximum Number That Makes Result of Bitwise AND Zero - Problem

Given an integer n, return the maximum integer x such that x ≤ n, and the bitwise AND of all the numbers in the range [x, n] is 0.

The bitwise AND operation compares each bit position of two numbers and returns 1 only when both bits are 1, otherwise returns 0.

When we AND all numbers in a range, we need to find the largest starting point where this result becomes 0.

Input & Output

Example 1 — Basic Case

$ Input: n = 5

› Output: 3

💡 Note: We need to find the maximum x where AND of range [x,5] equals 0. For x=3, we get 3&4&5 = 0&5 = 0 (since 3&4=0). For x=4, we get 4&5=4≠0. So maximum x is 3.

Example 2 — Single Number

$ Input: n = 1

› Output: 0

💡 Note: For x=0, the range [0,1] computes 0&1 = 0. This gives AND result 0, so x=0 is the maximum answer.

Example 3 — Edge Case Zero

$ Input: n = 0

› Output: 0

💡 Note: For x=0, the range [0,0] contains only 0, and AND of 0 is 0. So x=0 is the answer.

Constraints

0 ≤ n ≤ 10⁹

Visualization

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Asked in

G Google 15 M Microsoft 12 a Amazon 8

The key insight is that any range starting from 0 will always have AND result 0, since 0 AND anything equals 0. The optimal approach uses this mathematical property to return 0 in constant time. Time: O(1), Space: O(1)

Common Approaches

✓ Bit Manipulation Analysis

⏱️ Time: O(log n) Space: O(1)

In consecutive integers, certain bit positions will flip. When we have enough consecutive numbers, every bit position will be covered by at least one 0, making the overall AND equal to 0.

Brute Force

⏱️ Time: O(n²) Space: O(1)

For each possible value of x from n down to 0, calculate the bitwise AND of all numbers in range [x,n]. Return the maximum x where the result is 0.

Mathematical Pattern Recognition

⏱️ Time: O(1) Space: O(1)

The key insight is that when we start from 0 and go to any n, we always get AND=0 because the range [0,n] will always include 0, and 0 AND anything equals 0. Since we want the maximum x, and x=0 always works, the answer is always 0.

Bit Manipulation Analysis — Algorithm Steps

Find the highest bit set in n
Calculate how many numbers we need to cover all bit positions
Determine the starting point that ensures AND equals 0

Visualization

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Step-by-Step Walkthrough

Binary Analysis

Examine bit patterns of consecutive numbers

Find Pattern

Identify when bits flip to create zeros

Calculate Result

Determine maximum x mathematically

Code -

solution.c — C

#include <stdio.h>

int solution(int n) {
    if (n == 0) return 0;
    if (n == 1) return 0;
    
    // Find the highest power of 2 that is <= n
    int power_of_2 = 1;
    while (power_of_2 <= n) {
        power_of_2 <<= 1;
    }
    power_of_2 >>= 1;
    
    // The answer is power_of_2 - 1
    return power_of_2 - 1;
}

int main() {
    int n;
    scanf("%d", &n);
    int result = solution(n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(log n)

Only need to examine the bits in n, which is at most log n bits

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables for bit calculations

✓ Linear Space

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Maximum Number That Makes Result of Bitwise AND Zero - Problem

Input & Output

Constraints

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Related Problems

Common Approaches

Bit Manipulation Analysis — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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