Summary: This guide explains how to solve LeetCode problem 1404, where you must count the number of steps to reduce a binary string to 1 using only two operations: divide by 2 when the number is even and add 1 when the number is odd. The solution avoids converting the full binary string to a numeric type and instead simulates operations on the string efficiently.

Problem Summary

Input: A string s representing a positive integer in binary format.
Goal: Return the total number of steps to reduce the number to 1.
Operations: If even, divide by 2. If odd, add 1.

Intuition and Key Observations

Divide by 2 in binary is equivalent to removing the trailing ‘0’. That is one step when the last digit is ‘0’.
Adding 1 to an odd binary number (ending in ‘1’) causes a carry. That carry can ripple left through consecutive ‘1’ bits, turning them into ‘0’ until a ‘0’ is found or a new leading ‘1’ is created.
Carry trick: Instead of performing repeated string modifications, track a single carry flag. Walk the string from right to left, excluding the most significant bit, and compute steps by combining the current bit value and the carry.

Algorithm Outline

Initialize steps to 0 and carry to 0.
Iterate from the last character to the second character (index length-1 down to 1).
For each bit:

If the bit is ‘0’ and there is no carry, it takes 1 step to divide by 2.
If the bit is ‘0’ and carry is 1, the bit effectively becomes ‘1’, so it will take 2 steps: add 1 then divide.
If the bit is ‘1’ and carry is 0, adding 1 will create a carry and takes 2 steps total (add then divide), and set carry to 1.
If the bit is ‘1’ and carry is 1, then ‘1’+carry becomes ‘0’ with carry still 1, so this state costs 1 step (divide) and carry remains 1.

After the loop, if carry is 1, add a final step for the leftmost carry resolution.

Walkthrough Example

Example: s = “1101”

Start from the rightmost bit and use the carry method.
Processing yields a total of 6 steps. The operations sequence is: add, divide, add, divide, divide, divide leading to 6 steps.
This matches step by step simulation but uses O(n) time and O(1) extra space.

Compact Implementations

C++ implementation:

int numSteps(string s) { int n = s.size(), steps = 0, carry = 0; for (int i = n – 1; i >= 1; –i) { if (s[i] == ‘0’) steps += (carry == 0) ? 1 : 2; else { steps += (carry == 0) ? 2 : 1; carry = 1; } } return steps + carry; }

Python implementation:

def numSteps(s):
steps = 0
carry = 0
for ch in reversed(s[:-1]):
if ch == ‘0’:
steps += 1 if carry == 0 else 2
else:
steps += 2 if carry == 0 else 1
carry = 1
return steps + carry

JavaScript implementation:

Time and Space Complexity

Time complexity: O(n), where n is the length of the binary string. Each bit is processed once.
Space complexity: O(1) additional space. Only a few integers for counters and the carry are used.

Tips and Common Pitfalls

Do not convert the full binary string to an integer type for large inputs, as it may overflow typical numeric types.
Remember to exclude the most significant bit from the main loop and handle any leftover carry after the loop.
Test edge cases: s = “1” should return 0, s = “10” should return 1, and strings with long runs of trailing ones will exercise carries.

Conclusion: Tracking a carry and processing the binary string from right to left yields a simple, efficient solution to LeetCode 1404. The approach is language agnostic and easy to implement in C++, Python, or JavaScript while remaining optimal in time and space.

Beginner-Friendly Guide: LeetCode 1404 Binary String Reduction (C++, Python, JavaScript)