Problem Description

You are given a binary string s that represents a positive integer. Your task is to count how many steps it takes to reduce this number to 1 by repeatedly applying these operations:

• If the number is even (ends with '0' in binary), divide it by 2
• If the number is odd (ends with '1' in binary), add 1 to it

The solution processes the binary string from right to left (least significant bit to most significant bit), simulating the operations while tracking carry-overs from additions.

When examining each bit from right to left:

• If there's a carry from a previous addition and the current bit is '0', it becomes '1' and the carry is cleared
• If there's a carry and the current bit is '1', it remains '0' and the carry propagates
• When we encounter a '1' (either original or from carry), we need an addition operation (incrementing the step count and setting carry to true)
• Each bit position requires one step (either division for '0' or addition for '1')

The algorithm counts operations by:

1. Processing each bit from the second-to-last position to the first (skipping the leftmost bit initially)
2. Handling carry propagation from addition operations
3. Adding steps for each bit processed
4. Adding an extra step if there's a final carry beyond the original string length

For example, with binary string "1101":

• Start from rightmost '1': odd, add 1 → becomes "1110" (1 step, carry set)
• Next '0': even, divide by 2 → becomes "111" (1 step)
• Next '1': odd, add 1 → becomes "1000" (1 step, carry set)
• Continue until reaching 1

Intuition

The key insight is that we don't need to actually convert the binary string to a number and perform the operations. Instead, we can simulate the process directly on the binary representation.

When working with binary numbers:

• Dividing by 2 is simply removing the rightmost bit (right shift)
• Adding 1 to an odd number changes the rightmost '1' to '0' and creates a carry that propagates leftward until it finds a '0' to flip to '1'

Starting from the rightmost bit and moving left mimics how we would actually perform these operations. Each bit position we examine represents one step in our reduction process - either a division (for '0') or an addition followed by a division (for '1').

The clever part is recognizing that when we add 1 to an odd binary number, it creates a carry effect. For instance, adding 1 to 1011 gives 1100. The rightmost '1' becomes '0', and we carry over, flipping bits until we hit a '0'. This carry mechanism is crucial because it affects how we interpret subsequent bits as we move leftward.

By tracking this carry as we traverse from right to left, we can determine the exact number of operations needed without performing actual arithmetic. Each bit contributes at least one step (the division step), and '1' bits contribute an additional step (the addition before division). The carry modifies which bits we see as '0' or '1' as we progress through the string.

This approach is efficient because it processes each bit exactly once, avoiding the need for repeated string manipulations or number conversions that would be required in a direct simulation.

Solution Approach

The solution implements a simulation approach that processes the binary string from right to left, tracking carry-over from addition operations.

Algorithm Steps:

1. Initialize variables:

   • carry = False: Tracks whether there's a carry from the previous addition operation
   • ans = 0: Counts the total number of steps

2. Process bits from right to left using s[:0:-1] (reverse string excluding the first character):

   • This skips the leftmost bit initially since we process from the second-to-last position

3. Handle carry propagation:

   if carry:
       if c == '0':
           c = '1'
           carry = False
       else:
           c = '0'

   • If there's a carry and current bit is '0', it becomes '1' and carry is cleared
   • If there's a carry and current bit is '1', it becomes '0' and carry continues

4. Count operations based on current bit:

   if c == '1':
       ans += 1
       carry = True
   ans += 1

   • If the bit (after carry handling) is '1', we need an addition operation (increment ans and set carry = True)
   • Every bit position requires one step (division), so we always increment ans by 1

5. Handle final carry:

   if carry:
       ans += 1

   • If there's still a carry after processing all bits, it means we've extended beyond the original number's length, requiring one more division

Why this works:

• Each '0' bit requires 1 step (division by 2)
• Each '1' bit requires 2 steps (add 1, then divide by 2)
• The carry mechanism correctly simulates how addition affects subsequent bits
• By processing right-to-left, we naturally follow the order of operations needed to reduce the number to 1

Time Complexity: O(n) where n is the length of the binary string Space Complexity: O(1) as we only use a constant amount of extra space

Example Walkthrough

Let's trace through the algorithm with the binary string s = "1011" (which represents 11 in decimal).

Initial state:

• carry = False
• ans = 0
• We'll process the string from right to left, excluding the leftmost '1'

Step 1: Process rightmost bit '1'

• Current bit: '1'
• No carry to handle yet
• Since c == '1':
  • Add 1 to make it even: ans = 1
  • Set carry = True (adding 1 to '1011' gives '1100')
• Division step: ans = 2

Step 2: Process middle bit '1'

• Current bit: '1'
• Handle carry: Since we have carry and bit is '1', it becomes '0'
  • c = '0', carry = True (remains)
• Since c == '0' (after carry):
  • No addition needed
• Division step: ans = 3

Step 3: Process second bit '0'

• Current bit: '0'
• Handle carry: Since we have carry and bit is '0', it becomes '1'
  • c = '1', carry = False
• Since c == '1' (after carry):
  • Add 1 to make it even: ans = 4
  • Set carry = True
• Division step: ans = 5

Step 4: Check final carry

• We still have carry = True after processing all bits
• This means the number extended beyond original length
• Add one more step: ans = 6

Result: 6 steps to reduce "1011" to "1"

Verification:

• 1011 (11) → add 1 → 1100 (12)
• 1100 (12) → divide by 2 → 110 (6)
• 110 (6) → divide by 2 → 11 (3)
• 11 (3) → add 1 → 100 (4)
• 100 (4) → divide by 2 → 10 (2)
• 10 (2) → divide by 2 → 1 (1)

Total: 6 steps ✓

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the string s. The algorithm iterates through the string once from right to left (excluding the first character) using the slice s[:0:-1], which takes O(n-1) iterations. Each iteration performs constant time operations: checking the carry flag, updating the character value, checking if the character is '1', and incrementing the answer counter. Therefore, the overall time complexity is linear with respect to the input string length.

The space complexity is O(1). The algorithm uses only a constant amount of extra space regardless of the input size. The variables used are: carry (a boolean flag), ans (an integer counter), and c (which holds a single character at a time during iteration). The string slice s[:0:-1] creates a reversed view but in Python string slicing for iteration doesn't create a new string in memory when used directly in a for loop. Thus, only constant additional space is required.

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

Common Pitfalls

1. Incorrectly handling the iteration range

A common mistake is iterating through the entire string instead of excluding the leftmost bit. Since we want to reduce the number to 1 (not 0), we should stop processing when we reach the most significant bit.

Incorrect approach:

for bit in s[::-1]:  # Processes ALL bits including the leftmost
    # ... process bit

Correct approach:

for bit in s[:0:-1]:  # Excludes the first character (leftmost bit)
    # ... process bit

2. Modifying the loop variable directly

Another pitfall is trying to modify the loop variable bit and expecting it to affect subsequent logic. In Python, reassigning the loop variable doesn't change the original value being processed.

Incorrect approach:

for bit in s[:0:-1]:
    if carry:
        if bit == '0':
            bit = '1'  # This reassignment creates confusion
            carry = False
        else:
            bit = '0'
  
    if bit == '1':  # Using modified bit here
        steps += 1
        carry = True

Better approach - Use a separate variable:

for original_bit in s[:0:-1]:
    current_bit = original_bit
    if carry:
        if current_bit == '0':
            current_bit = '1'
            carry = False
        else:
            current_bit = '0'
  
    if current_bit == '1':
        steps += 1
        carry = True

3. Forgetting to handle the final carry

When the binary string is all 1's (like "111"), the final addition creates a carry that extends beyond the original number's length. Forgetting to account for this final carry leads to an incorrect step count.

Incorrect approach:

def numSteps(self, s: str) -> int:
    carry = False
    steps = 0
  
    for bit in s[:0:-1]:
        # ... process bits
      
    # Missing: if carry: steps += 1
    return steps

Correct approach:

def numSteps(self, s: str) -> int:
    carry = False
    steps = 0
  
    for bit in s[:0:-1]:
        # ... process bits
  
    if carry:  # Don't forget this!
        steps += 1
      
    return steps

4. Misunderstanding when to increment the step counter

Some might think we only need to increment steps when performing an addition (odd number), but we need to count both additions AND divisions.

Incorrect approach:

for bit in s[:0:-1]:
    if carry:
        # handle carry
  
    if bit == '1':
        steps += 1  # Only counting additions
        carry = True
    # Missing the division step count!

Correct approach:

for bit in s[:0:-1]:
    if carry:
        # handle carry
  
    if bit == '1':
        steps += 1  # Count addition
        carry = True
  
    steps += 1  # Count division (always happens)