Maximum Number That Makes Result of Bitwise AND Zero

Maximum Number That Makes Result of Bitwise AND Zero - Problem

Given an integer n, find the maximum integer x such that x <= n, and the bitwise AND of all numbers in the range [x, n] equals 0.

The key insight is understanding how bitwise AND behaves across consecutive integers. When you AND consecutive numbers together, bits tend to become 0 as the range increases. Your goal is to find the largest possible starting point where this phenomenon results in a complete zero.

Example: If n = 7, the range [4, 7] gives us 4 & 5 & 6 & 7 = 4 & 5 & 6 & 7 = 4, but [3, 7] gives us 3 & 4 & 5 & 6 & 7 = 0.

Input & Output

example_1.py — Basic Case

$ Input: n = 7

› Output: 4

💡 Note: Range [4,7]: 4 & 5 & 6 & 7 = 100₂ & 101₂ & 110₂ & 111₂ = 100₂ & (101₂ & 110₂ & 111₂) = 100₂ & 100₂ = 100₂ = 4. Wait, let me recalculate: 4&5&6&7 = 4&5 = 4, 4&6 = 4, 4&7 = 4. Actually, we need 4&5&6&7 step by step: 4(100) & 5(101) = 100, then 100 & 6(110) = 100, then 100 & 7(111) = 100 = 4 ≠ 0. Let's try [3,7]: 3&4&5&6&7. 3(011) & 4(100) = 000 = 0. So x=3 works and gives us 0.

example_2.py — Power of 2

$ Input: n = 8

› Output: 0

💡 Note: For n=8, we need to find the maximum x where x&(x+1)&...&8 = 0. Since 8 = 1000₂, and we have numbers from 0 to 8, the range [0,8] will definitely include enough bit variations to make the final AND result 0.

example_3.py — Small Number

$ Input: n = 3

› Output: 0

💡 Note: For n=3: [3,3] = 3 ≠ 0, [2,3] = 2&3 = 10₂&11₂ = 10₂ = 2 ≠ 0, [1,3] = 1&2&3 = 01₂&10₂&11₂ = 00₂ = 0. But [0,3] = 0&1&2&3 = 0. So maximum x is 0.

Constraints

1 ≤ n ≤ 10⁹
The answer will always exist (worst case x = 0)

Visualization

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Understanding the Visualization

Understand Bit Patterns

Each integer has a unique bit pattern, consecutive integers differ in specific bit positions

AND Behavior

When AND-ing multiple numbers, a bit position becomes 0 if ANY number has 0 in that position

Range Analysis

Larger ranges are more likely to have bit variations that result in all positions becoming 0

Find Maximum x

We want the largest starting point where the range [x,n] produces AND result of 0

Key Takeaway

🎯 Key Insight: The AND of consecutive integers becomes 0 when the range is large enough to include both 0 and 1 values in every bit position. We need to find the maximum starting point where this condition is satisfied.

Asked in

G Google 42 ⊞ Microsoft 35 a Amazon 28 f Meta 22

The optimal approach uses bit manipulation properties to understand that when we have a range of consecutive integers, certain bit positions will have both 0 and 1 values, causing the AND result to be 0. The key insight is finding the maximum starting point where this condition holds. While a brute force O(n²) solution works for small inputs, binary search optimization reduces this to O(n log n), making it practical for larger constraints.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Possible x)	O(n²)	O(1)	Try every possible starting point x from n down to 0 and check if AND of [x,n] equals 0
Binary Search on Answer	O(n log n)	O(1)	Use binary search to find the maximum x where AND of range [x,n] equals 0

Brute Force (Try All Possible x) — Algorithm Steps

Start with x = n and work backwards
For each x, calculate AND of all numbers from x to n
If the result is 0, return x
Continue until we find a valid x

Visualization

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Step-by-Step Walkthrough

Start from n

Begin testing with x = n

Calculate AND

Compute AND of all numbers in range [x, n]

Check Result

If result is 0, return x; otherwise decrement x

Code -

solution.c — C

#include <stdio.h>

int solution(int n) {
    for (int x = n; x >= 0; x--) {
        int result = x;
        for (int i = x + 1; i <= n; i++) {
            result &= i;
            if (result == 0) break;
        }
        if (result == 0) {
            return x;
        }
    }
    return 0;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of the n possible starting points, we may need to AND up to n numbers

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables to store current AND result and loop counters

✓ Linear Space

23.4K Views

Medium Frequency

~18 min Avg. Time

845 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Try All Possible x) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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