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

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

Transform Binary Number to One

Given a binary string s representing a positive integer, you need to count the minimum number of operations to reduce it to "1".

The transformation rules are simple:
• If the number is even (ends with '0'), divide by 2 (remove last '0')
• If the number is odd (ends with '1'), add 1 to make it even

Example: "1101" (13 in decimal) → "1110" (+1) → "111" (÷2) → "1000" (+1) → "100" (÷2) → "10" (÷2) → "1" (÷2) = 6 steps

Your goal is to find the minimum number of steps needed for any valid binary string input.

Input & Output

example_1.py — Basic Case

$ Input: s = "1101"

› Output: 6

💡 Note: "1101" (13) → "1110" (+1) → "111" (÷2) → "1000" (+1) → "100" (÷2) → "10" (÷2) → "1" (÷2). Total: 6 steps

example_2.py — Single Digit

$ Input: s = "1"

› Output: 0

💡 Note: Already at "1", no operations needed

example_3.py — Power of Two

$ Input: s = "10000"

› Output: 4

💡 Note: "10000" (16) → "1000" (÷2) → "100" (÷2) → "10" (÷2) → "1" (÷2). Only divisions needed: 4 steps

Constraints

1 ≤ s.length ≤ 500
s consists only of characters '0' and '1'
s[0] == '1' (no leading zeros)
s represents a valid binary number

Visualization

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Understanding the Visualization

Identify Pattern

Scan the binary string to identify operation patterns

Count Operations

Each '0' needs division, each '1' needs addition then division

Handle Carries

Propagate carries efficiently without actual computation

Sum Results

Total all operations needed across the entire string

Key Takeaway

🎯 Key Insight: We don't need to perform actual arithmetic - analyzing binary patterns with carry propagation gives us the answer in O(n) time!

Asked in

f Meta 25 G Google 20 a Amazon 15 ⊞ Microsoft 12

The optimal solution uses string pattern analysis with carry propagation. Instead of performing actual arithmetic operations, we analyze the binary string directly: each '0' contributes 1 step (division), each '1' contributes 2 steps (addition + division) with carry handling. This approach runs in O(n) time and handles very large numbers without overflow issues.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Simulation	O(n²)	O(1)	Simulate the exact process: convert to integer, apply operations, count steps
String Pattern Analysis (Optimal)	O(n)	O(1)	Analyze binary string patterns with carry propagation to count operations efficiently

Brute Force Simulation — Algorithm Steps

Convert binary string to integer
While number > 1:
If even: divide by 2, increment counter
If odd: add 1, increment counter
Return counter

Visualization

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Step-by-Step Walkthrough

Start with binary

Convert binary string to decimal number

Check even/odd

If even divide by 2, if odd add 1

Repeat until 1

Continue until we reach 1

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int numStepsToReduceToOne(char* s) {
    long long num = 0;
    int len = strlen(s);
    
    for (int i = 0; i < len; i++) {
        num = num * 2 + (s[i] - '0');
    }
    
    int steps = 0;
    while (num != 1) {
        if (num % 2 == 0) {
            num /= 2;
        } else {
            num += 1;
        }
        steps++;
    }
    
    return steps;
}

int main() {
    char s[1001];
    scanf("%s", s);
    printf("%d\n", numStepsToReduceToOne(s));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Each operation can take O(n) time for large numbers, and we need O(n) operations

⚠ Quadratic Growth

Space Complexity

O(1)

Only storing the current number and counter

✓ Linear Space

28.4K Views

Medium Frequency

~15 min Avg. Time

892 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Simulation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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