Formatted question description: https://leetcode.ca/all/1404.html

1404. Number of Steps to Reduce a Number in Binary Representation to One (Medium)

Given a number s in their binary representation. 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's guaranteed that you can always reach to one for all testcases.

 

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" corressponds 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'

Related Topics:
String, Bit Manipulation

Solution 1.

// OJ: https://leetcode.com/problems/number-of-steps-to-reduce-a-number-in-binary-representation-to-one/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int numSteps(string s) {
        int ans = 0;
        while (s != "1") {
            if (s.back() == '1') {
                int i = s.size() - 1;
                while (i >= 0 && s[i] == '1') {
                    s[i] = '0';
                    --i;
                }
                if (i == -1) s.insert(s.begin(), '1');
                else s[i] = '1';
            } else s.pop_back();
            ++ans;
        }
        return ans;
    }
};

Java

  • class Solution {
        public int numSteps(String s) {
            int steps = 0;
            StringBuffer sb = new StringBuffer(s);
            while (sb.length() > 1) {
                char lastChar = sb.charAt(sb.length() - 1);
                if (lastChar == '0')
                    sb.deleteCharAt(sb.length() - 1);
                else
                    sb = addOne(sb);
                steps++;
            }
            return steps;
        }
    
        public StringBuffer addOne(StringBuffer sb) {
            int length = sb.length();
            int lastZeroIndex = -1;
            for (int i = length - 1; i >= 0; i--) {
                if (sb.charAt(i) == '0') {
                    lastZeroIndex = i;
                    break;
                }
            }
            if (lastZeroIndex < 0) {
                for (int i = length - 1; i >= 0; i--)
                    sb.setCharAt(i, '0');
                sb.insert(0, '1');
            } else {
                sb.setCharAt(lastZeroIndex, '1');
                for (int i = lastZeroIndex + 1; i < length; i++)
                    sb.setCharAt(i, '0');
            }
            return sb;
        }
    }
    
  • // OJ: https://leetcode.com/problems/number-of-steps-to-reduce-a-number-in-binary-representation-to-one/
    // Time: O(N)
    // Space: O(1)
    class Solution {
    public:
        int numSteps(string s) {
            int ans = 0;
            while (s != "1") {
                if (s.back() == '1') {
                    int i = s.size() - 1;
                    while (i >= 0 && s[i] == '1') {
                        s[i] = '0';
                        --i;
                    }
                    if (i == -1) s.insert(s.begin(), '1');
                    else s[i] = '1';
                } else s.pop_back();
                ++ans;
            }
            return ans;
        }
    };
    
  • class Solution:
        def numSteps(self, s: str) -> int:
            step = 0
            s_int = int(s, 2)
            while s_int != 1:
                if s_int % 2 == 0:
                    s_int //= 2
                else:
                    s_int += 1
                step += 1
            return step
    

All Problems

All Solutions