LeetCode #1404 — MEDIUM

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

Move from brute-force thinking to an efficient approach using bit manipulation strategy.

The Problem

Problem Statement

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'

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Bit Manipulation

Example 1

"1101"

Example 2

"10"

Example 3

"1"

Core Insight

What unlocks the optimal approach

Read the string from right to left, if the string ends in '0' then the number is even otherwise it is odd.
Simulate the steps described in the binary string.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1404: Number of Steps to Reduce a Number in Binary Representation to One
class Solution {
    public int numSteps(String s) {
        boolean carry = false;
        int ans = 0;
        for (int i = s.length() - 1; i > 0; --i) {
            char c = s.charAt(i);
            if (carry) {
                if (c == '0') {
                    c = '1';
                    carry = false;
                } else {
                    c = '0';
                }
            }
            if (c == '1') {
                ++ans;
                carry = true;
            }
            ++ans;
        }
        if (carry) {
            ++ans;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1404: Number of Steps to Reduce a Number in Binary Representation to One
func numSteps(s string) int {
	ans := 0
	carry := false
	for i := len(s) - 1; i > 0; i-- {
		c := s[i]
		if carry {
			if c == '0' {
				c = '1'
				carry = false
			} else {
				c = '0'
			}
		}
		if c == '1' {
			ans++
			carry = true
		}
		ans++
	}
	if carry {
		ans++
	}
	return ans
}

# Accepted solution for LeetCode #1404: Number of Steps to Reduce a Number in Binary Representation to One
class Solution:
    def numSteps(self, s: str) -> int:
        carry = False
        ans = 0
        for c in s[:0:-1]:
            if carry:
                if c == '0':
                    c = '1'
                    carry = False
                else:
                    c = '0'
            if c == '1':
                ans += 1
                carry = True
            ans += 1
        if carry:
            ans += 1
        return ans

// Accepted solution for LeetCode #1404: Number of Steps to Reduce a Number in Binary Representation to One
struct Solution;

impl Solution {
    fn num_steps(s: String) -> i32 {
        let mut carry = 0;
        let mut res = 0;
        let n = s.len();
        let s: Vec<u8> = s.bytes().collect();
        for i in (1..n).rev() {
            res += 1;
            if s[i] - b'0' + carry == 1 {
                carry = 1;
                res += 1;
            }
        }
        res + carry as i32
    }
}

#[test]
fn test() {
    let s = "1101".to_string();
    let res = 6;
    assert_eq!(Solution::num_steps(s), res);
    let s = "10".to_string();
    let res = 1;
    assert_eq!(Solution::num_steps(s), res);
    let s = "1".to_string();
    let res = 0;
    assert_eq!(Solution::num_steps(s), res);
}

// Accepted solution for LeetCode #1404: Number of Steps to Reduce a Number in Binary Representation to One
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1404: Number of Steps to Reduce a Number in Binary Representation to One
// class Solution {
//     public int numSteps(String s) {
//         boolean carry = false;
//         int ans = 0;
//         for (int i = s.length() - 1; i > 0; --i) {
//             char c = s.charAt(i);
//             if (carry) {
//                 if (c == '0') {
//                     c = '1';
//                     carry = false;
//                 } else {
//                     c = '0';
//                 }
//             }
//             if (c == '1') {
//                 ++ans;
//                 carry = true;
//             }
//             ++ans;
//         }
//         if (carry) {
//             ++ans;
//         }
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.