Problem Description

You are given a positive integer n. Your task is to find the smallest number x that satisfies two conditions:

1. x must be greater than or equal to n (i.e., x >= n)
2. The binary representation of x must contain only set bits (all 1s, no 0s)

In other words, you need to find the smallest number that is at least n and whose binary representation consists entirely of 1s.

For example:

• Numbers with only set bits in binary: 1 (binary: 1), 3 (binary: 11), 7 (binary: 111), 15 (binary: 1111), 31 (binary: 11111), etc.
• If n = 5, the answer would be 7 because:
  • 7 in binary is 111 (all set bits)
  • 7 >= 5
  • Numbers smaller than 7 with all set bits are 3 (binary: 11) and 1 (binary: 1), but both are less than 5

The pattern of numbers with all set bits follows the sequence: 2^k - 1 where k is a positive integer (1, 3, 7, 15, 31, 63, 127, ...).

Intuition

To understand the solution, let's first recognize what numbers have only set bits in their binary representation. These are numbers like 1 (binary: 1), 3 (binary: 11), 7 (binary: 111), 15 (binary: 1111), and so on. Notice a pattern here - these numbers are always one less than a power of 2:

• 1 = 2^1 - 1
• 3 = 2^2 - 1
• 7 = 2^3 - 1
• 15 = 2^4 - 1
• 31 = 2^5 - 1

Why does this pattern exist? When we have a power of 2 like 8 (binary: 1000), subtracting 1 gives us 7 (binary: 111). The subtraction causes a cascade effect where the single 1 bit becomes 0 and all positions to its right become 1.

Now, to find the smallest such number that is greater than or equal to n, we need to find the smallest power of 2, let's call it x, such that x - 1 >= n. This is equivalent to finding the smallest x where x > n.

The approach becomes clear: start with x = 1 (which is 2^0) and keep doubling it (left shifting by 1 is the same as multiplying by 2) until x - 1 becomes greater than or equal to n. At each step:

• x = 1 gives us x - 1 = 0
• x = 2 gives us x - 1 = 1
• x = 4 gives us x - 1 = 3
• x = 8 gives us x - 1 = 7
• x = 16 gives us x - 1 = 15

We keep going until we find the first x where x - 1 >= n, and that x - 1 is our answer.

Learn more about Math patterns.

Solution Approach

The implementation uses bit manipulation to efficiently find the answer. Here's how the algorithm works step by step:

1. Initialize x = 1: We start with the smallest power of 2, which is 2^0 = 1.

2. Loop with condition while x - 1 < n: We continue as long as x - 1 is still less than n. This means the current value of x - 1 (which has all set bits) is not yet large enough to be our answer.

3. Left shift operation x <<= 1: In each iteration, we left shift x by 1 position. This is equivalent to multiplying x by 2, moving us to the next power of 2:

   • First iteration: x = 1 becomes x = 2
   • Second iteration: x = 2 becomes x = 4
   • Third iteration: x = 4 becomes x = 8
   • And so on...

4. Return x - 1: Once the loop exits, we have found the smallest power of 2 where x - 1 >= n. Since x is a power of 2, x - 1 will have all bits set in its binary representation.

Example walkthrough with n = 5:

• Start: x = 1, check x - 1 = 0 < 5, continue
• After shift: x = 2, check x - 1 = 1 < 5, continue
• After shift: x = 4, check x - 1 = 3 < 5, continue
• After shift: x = 8, check x - 1 = 7 >= 5, exit loop
• Return 7

The time complexity is O(log n) because we need at most log₂(n) + 1 iterations to find the answer. The space complexity is O(1) as we only use a single variable x.

Example Walkthrough

Let's walk through the solution with n = 10 to find the smallest number with all set bits that is greater than or equal to 10.

Step-by-step process:

1. Initialize: Start with x = 1

   • Check: x - 1 = 0, is 0 >= 10? No, continue.

2. First iteration: Left shift x → x = 2

   • Binary: x = 10₂
   • Check: x - 1 = 1, is 1 >= 10? No, continue.
   • Note: 1 in binary is 1₂ (all set bits, but too small)

3. Second iteration: Left shift x → x = 4

   • Binary: x = 100₂
   • Check: x - 1 = 3, is 3 >= 10? No, continue.
   • Note: 3 in binary is 11₂ (all set bits, but still too small)

4. Third iteration: Left shift x → x = 8

   • Binary: x = 1000₂
   • Check: x - 1 = 7, is 7 >= 10? No, continue.
   • Note: 7 in binary is 111₂ (all set bits, but still not enough)

5. Fourth iteration: Left shift x → x = 16

   • Binary: x = 10000₂
   • Check: x - 1 = 15, is 15 >= 10? Yes, stop!
   • Note: 15 in binary is 1111₂ (all set bits, and it's >= 10)

Result: Return x - 1 = 15

Verification:

• 15 in binary is 1111₂ ✓ (all bits are set)
• 15 >= 10 ✓ (satisfies the minimum requirement)
• The previous candidate 7 (binary: 111₂) was less than 10, so 15 is indeed the smallest valid answer

The algorithm efficiently finds that we need 4 set bits to create a number at least as large as 10, resulting in the answer 15.

Solution Implementation

Time and Space Complexity

The time complexity is O(log n), and the space complexity is O(1).

Time Complexity Analysis: The algorithm starts with x = 1 and repeatedly doubles it using left shift operation (x <<= 1) until x - 1 >= n. Since x doubles in each iteration, the number of iterations required is proportional to the number of times we need to double 1 to exceed n. This requires approximately log₂(n) iterations, resulting in O(log n) time complexity.

Space Complexity Analysis: The algorithm only uses a single variable x to store the intermediate values, regardless of the input size n. No additional data structures or recursive calls are used, making the space complexity O(1) (constant space).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Overflow for Large Values of n

When dealing with very large values of n, the left shift operation power_of_two <<= 1 can cause integer overflow if not handled properly. In some programming languages with fixed integer sizes, this could lead to unexpected behavior or wrap-around.

Example of the issue:

• If n is close to the maximum integer value, continuously doubling power_of_two might exceed the integer limit.
• In languages like C++ or Java with fixed integer sizes, this could cause undefined behavior or negative values.

Solution:

class Solution:
    def smallestNumber(self, n: int) -> int:
        # Check if n is already a number with all bits set
        # This avoids unnecessary iterations for edge cases
        if n & (n + 1) == 0:
            return n
          
        # Find the position of the most significant bit in n
        bit_position = n.bit_length()
      
        # Calculate 2^(bit_position) - 1
        result = (1 << bit_position) - 1
      
        # If result is less than n, we need the next power of 2
        if result < n:
            result = (1 << (bit_position + 1)) - 1
          
        return result

2. Misunderstanding the Problem Requirements

A common mistake is thinking that we need to find a number whose binary representation is the same length as n's binary representation, rather than finding the smallest number with all set bits that is >= n.

Incorrect approach:

# WRONG: This tries to set all bits in n's binary representation
def incorrectSolution(n: int) -> int:
    result = n
    while result & (result + 1) != 0:  # Check if all bits are set
        result |= (result + 1)  # This doesn't give the correct answer
    return result

Correct understanding: We need the smallest number from the sequence 1, 3, 7, 15, 31, 63, 127... that is >= n.

3. Off-by-One Error in Loop Condition

Using power_of_two < n instead of power_of_two - 1 < n in the while loop condition would cause incorrect results.

Incorrect version:

# WRONG: This will return a larger number than necessary
def incorrectLoop(n: int) -> int:
    power_of_two = 1
    while power_of_two < n:  # Wrong condition!
        power_of_two <<= 1
    return power_of_two - 1
    # For n=4, this returns 7 instead of 7 (happens to be correct)
    # For n=3, this returns 3 instead of 3 (correct by chance)
    # For n=2, this returns 3 instead of 3 (correct)
    # For n=1, this returns 0 instead of 1 (WRONG!)

The correct condition power_of_two - 1 < n ensures we find the smallest power of 2 where subtracting 1 gives us a value >= n.