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Number of Steps to Reduce a Number in Binary Representation to One

Number: 1520

Difficulty: Medium

Paid? No

Companies: Microsoft, Google, Amazon

Problem Description

Given a binary representation of an integer as a string s, you must return the number of steps to reduce it to 1. In each step, if the number is even, you divide it by 2; if it is odd, you add 1. It is guaranteed that these operations will eventually reduce any valid input to 1.

Key Insights

  • The problem simulates a process on the binary representation without converting it directly to an integer.
  • When the number is even (last bit is '0'), division by 2 is equivalent to removing the least significant bit.
  • When the number is odd (last bit is '1'), an addition of 1 might propagate a carry through consecutive ones.
  • A reverse iteration (from least significant bit to most significant) with a carry flag can handle the chain reaction efficiently.
  • Special handling is needed for the most significant bit; once you reach it, the process terminates (considering any remaining carry).

Space and Time Complexity

Time Complexity: O(n) where n is the length of the binary string.
Space Complexity: O(n) if converting the string to an array for in-place operations, but O(1) additional space if processing the string directly with indices and a carry flag.

Solution

We can solve the problem by iterating from the end of the binary string (least significant bit) towards the beginning. We maintain a carry variable to simulate the effect of addition when the current bit is '1'. For each bit (except the most significant bit), we:

  • If the bit (considering the carry) is 0 (i.e., even), we perform one division operation.
  • If it is 1 (i.e., odd), then we simulate adding one (which may turn a series of 1’s into 0’s with a carry propagated) and then the following division, thus counting two operations and setting the carry to 1. Finally, we add any leftover carry to the step count for the most significant bit.

This approach allows us to handle the ripple effect of carries in a single pass from right to left, ensuring that we count the operations accurately without converting the entire binary string into a large integer.

Code Solutions

def numSteps(s: str) -> int:
    # Initialize the steps counter and carry flag.
    steps = 0
    carry = 0
    n = len(s)
    # Traverse from the least significant bit (rightmost) to the second bit.
    for i in range(n - 1, 0, -1):
        # Convert the current character to an integer and add the carry.
        bit = int(s[i])
        if bit + carry == 0:
            # Even result: one division step.
            steps += 1
        else:
            # Odd result: perform an addition (possibly causing a carry) and then a division.
            steps += 2
            carry = 1  # Mark the carry for the next iteration.
    # The final most significant bit: add any leftover carry.
    return steps + carry

# Example usage:
print(numSteps("1101"))
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