Smallest Number With All Set Bits

Smallest Number With All Set Bits - Problem

You are given a positive number n. Return the smallest number x greater than or equal to n, such that the binary representation of x contains only set bits (all bits are 1).

A number with all set bits means every bit position in its binary representation is 1. For example:

1 (binary: 1) - all bits are 1
3 (binary: 11) - all bits are 1
7 (binary: 111) - all bits are 1
15 (binary: 1111) - all bits are 1

Input & Output

Example 1 — Basic Case

$ Input: n = 5

› Output: 7

💡 Note: 5 in binary is 101. The smallest number ≥ 5 with all set bits is 7 (111 in binary).

Example 2 — Already All Set Bits

$ Input: n = 3

› Output: 3

💡 Note: 3 in binary is 11. Since all bits are already set, return 3.

Example 3 — Single Bit

$ Input: n = 1

› Output: 1

💡 Note: 1 in binary is 1. Since the only bit is set, return 1.

Constraints

1 ≤ n ≤ 10⁹

Visualization

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Asked in

G Google 25 M Microsoft 18 A Amazon 15

The key insight is that numbers with all set bits follow the pattern 2^k - 1. We can find the bit length of n and calculate the next such number directly. Best approach uses bit manipulation with Time: O(1), Space: O(1).

Common Approaches

✓ Bit Length Calculation

⏱️ Time: O(log n) Space: O(1)

Calculate the number of bits needed to represent n. If n already has all bits set, return n. Otherwise, find the next number with all bits set using 2^(bits+1) - 1.

Brute Force Check

⏱️ Time: O(k × log k) Space: O(1)

Start from n and check each number to see if all its bits are 1. Keep incrementing until we find a number whose binary representation contains only 1s.

Single Pass Optimal

⏱️ Time: O(1) Space: O(1)

Find the highest bit set in n, then create a mask with all bits set up to the next position. This gives us the smallest number with all set bits greater than or equal to n.

Bit Length Calculation — Algorithm Steps

Find bit length of n
Check if n is already all set bits (2^k - 1)
If not, return 2^(bits) - 1 where bits is the next power

Visualization

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Step-by-Step Walkthrough

Check Input

See if n already has all bits set

Find Bit Length

Calculate bits needed for n

Calculate Result

Use 2^length - 1 formula

Code -

solution.c — C

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

int solution(int n) {
    // Check if n is already a number with all set bits
    if ((n & (n + 1)) == 0) {
        return n;
    }
    
    // Find the number of bits in n
    int bitLength = floor(log2(n)) + 1;
    
    // Return 2^bit_length - 1
    return (1 << bitLength) - 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)

Finding bit length takes O(log n) time

⚡ Linearithmic

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Input & Output

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Common Approaches

Bit Length Calculation — Algorithm Steps

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Code -

Time & Space Complexity

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