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

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StringsBit Manipulation

Given the binary representation of an integer as a string s, return the number of steps to reduce it to 1 under the following rules:

If the current number is even, you have to divide it by 2.
If the current number is odd, you have to add 1 to it.

It is guaranteed that you can always reach one for all test cases.

Example 1:

Input: s = "1101"
Output: 6
Explanation: "1101" corressponds to number 13 in their decimal representation.
Step 1) 13 is odd, add 1 and obtain 14. 
Step 2) 14 is even, divide by 2 and obtain 7.
Step 3) 7 is odd, add 1 and obtain 8.
Step 4) 8 is even, divide by 2 and obtain 4.  
Step 5) 4 is even, divide by 2 and obtain 2. 
Step 6) 2 is even, divide by 2 and obtain 1.

Example 2:

Input: s = "10"
Output: 1
Explanation: "10" corresponds to number 2 in their decimal representation.
Step 1) 2 is even, divide by 2 and obtain 1.

Example 3:

Input: s = "1"
Output: 0

Constraints:

1 <= s.length <= 500
s consists of characters '0' or '1'
s[0] == '1'

Try coding it in LeetCode

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

Medium

Grab

4 views

Topics:

StringsBit Manipulation

Given the binary representation of an integer as a string s, return the number of steps to reduce it to 1 under the following rules:

If the current number is even, you have to divide it by 2.
If the current number is odd, you have to add 1 to it.

It is guaranteed that you can always reach one for all test cases.

Example 1:

Input: s = "1101"
Output: 6
Explanation: "1101" corressponds to number 13 in their decimal representation.
Step 1) 13 is odd, add 1 and obtain 14. 
Step 2) 14 is even, divide by 2 and obtain 7.
Step 3) 7 is odd, add 1 and obtain 8.
Step 4) 8 is even, divide by 2 and obtain 4.  
Step 5) 4 is even, divide by 2 and obtain 2. 
Step 6) 2 is even, divide by 2 and obtain 1.

Example 2:

Input: s = "10"
Output: 1
Explanation: "10" corresponds to number 2 in their decimal representation.
Step 1) 2 is even, divide by 2 and obtain 1.

Example 3:

Input: s = "1"
Output: 0

Constraints:

1 <= s.length <= 500
s consists of characters '0' or '1'
s[0] == '1'

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum length of the binary string 's'?
Does the input string 's' always represent a positive number, or can it be '0' or empty?
If the number represented by 's' is '0', what should the function return?
When the number is odd, do I need to explore both adding 1 and subtracting 1 to ensure I find the minimum number of steps, or is there a deterministic way to decide whether to add or subtract?
Are there any leading zeros in the binary string 's' besides the single '0' case?

Brute Force Solution

Approach

The brute force method for this problem involves directly simulating the process of reducing the number to one, step by step. We simply follow the rules until we reach one, keeping track of how many steps it takes. It's like playing the game from start to finish without trying to be clever.

Here's how the algorithm would work step-by-step:

Start with the number given to you, represented in its binary form.
Check if the number is already one. If it is, you're done, and the answer is zero steps.
If the number is even (ends in a zero in binary), divide it by two. This is one step.
If the number is odd (ends in a one in binary), either add one to it or subtract one from it, whichever leads to fewer total steps. In the brute force approach, we don't know which is better initially, so we'll end up exploring both paths implicitly.
Keep track of the number of steps you've taken.
Repeat steps 3 and 4 with the new number you get after each operation.
Continue this process until the number becomes one.
The total number of steps you've counted is your answer.

Code Implementation

def number_of_steps_brute_force(binary_string):
    number_of_steps = 0
    current_number = int(binary_string, 2)

    # If already one, return zero
    if current_number == 1:
        return 0

    while current_number != 1:
        # Even number, divide by 2
        if current_number % 2 == 0:
            current_number //= 2
            number_of_steps += 1

        # Odd number, add one
        else:
            current_number += 1
            number_of_steps += 1

    return number_of_steps

Big(O) Analysis

Time Complexity

O(n) – The input is a binary representation of a number, let's say with n digits. In the worst-case scenario, the number is a string of all ones. Each step either divides the number by 2 (right shift), or adds 1. Adding 1 can, in the worst case, cause a chain of carries that propagates through the entire number, but this happens at most once per division. Since we're repeatedly dividing the number by two until it becomes one, the number of divisions is proportional to n. Each division step may be preceded by adding 1, so the number of steps is still proportional to n. Therefore, the time complexity is O(n).

Space Complexity

O(1) – The described brute force approach operates directly on the input number (represented in binary). It doesn't explicitly create any auxiliary data structures like arrays, lists, or hash maps to store intermediate results or track visited states. The only memory used is for a few variables to hold the current number and the step count. Therefore, the auxiliary space complexity is constant and independent of the input number's size N (the number of bits in its binary representation).

Optimal Solution

Approach

The key to solving this problem efficiently is to simulate the process of reducing the number to one step by step, but without actually performing complex binary arithmetic. We can determine whether to add one or divide by two based on the least significant bit and the carry-over from previous operations.

Here's how the algorithm would work step-by-step:

Start by examining the binary representation from right to left (least significant bit to most significant bit).
If the rightmost digit is a zero, divide the number by two (which is simply removing the rightmost digit). This takes one step.
If the rightmost digit is a one, add one to the number. This takes one step and might cause a carry-over.
When adding one, we need to handle possible carry-overs. If adding one to a '1' results in '0' and a carry, propagate the carry to the next digit on the left.
Keep track of the total number of steps taken as you go through the binary representation.
The process ends when the number is reduced to one, which is represented by a single '1' in binary. The total steps counted is the answer.

Code Implementation

def number_of_steps(binary_string):
    steps_count = 0
    carry = 0
    string_length = len(binary_string)

    for i in range(string_length - 1, 0, -1):
        # Iterate from right to left, excluding the most significant bit
        current_bit = int(binary_string[i])
        current_bit += carry

        if current_bit % 2 == 0:
            # If current bit is 0 after carry, divide by two
            steps_count += 1
            carry = 0
        else:
            # If current bit is 1 after carry, add one
            steps_count += 1
            carry = 1

    # Handle the most significant bit, plus any remaining carry
    steps_count += carry

    return steps_count

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the binary representation of the input number, from the least significant bit to the most significant bit. For each bit, it performs a constant number of operations: checking if it's a zero or a one, potentially adding one (which might involve carrying over), and incrementing the step counter. The number of bits in the binary representation is proportional to the logarithm of the number but is directly the length of the string given as input 'n'. Therefore, the time complexity is O(n) since the number of operations is directly proportional to the number of bits, meaning the input size, within the given binary representation.

Space Complexity

O(1) – The algorithm operates primarily by iterating through the binary representation and updating a step counter and a carry variable. The carry variable stores either 0 or 1. The step counter stores an integer. The space used by these variables does not scale with the size of the input binary number N, where N is the number of bits. Therefore, the auxiliary space complexity is constant.

Edge Cases

Case	How to Handle
Null or empty input string	Return 0 immediately, as there are no steps needed to reduce an empty number.
Input string is '1'	Return 0 immediately, as the number is already 1.
Input string contains leading zeros (e.g., '00110')	Trim leading zeros from the string before processing to avoid incorrect calculations.
Input string represents a large binary number (potential integer overflow)	Use a data type that can accommodate large numbers, such as strings or BigInteger, to prevent overflow.
Input string with a very long sequence of '1's	The algorithm should handle long sequences of '1's efficiently without exceeding time limits, perhaps by optimizing addition operations.
Input string representing a power of 2 (e.g., '10000')	This is a best-case scenario for division, and the algorithm should efficiently handle it by repeatedly dividing by 2.
Input string representing a number very close to a power of 2 (e.g., '11111')	The algorithm must choose the optimal operation (+1 or -1) at each odd number to minimize steps, ensuring correct results near powers of 2.
Input string consisting of all '0's except the last digit being '1'	This tests the handling of the 'odd' case, ensuring that 1 addition is done before the divisions.