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

Description

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" 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'

Solutions

Solution 1: Simulation

We simulate operations $1$ and $2$, while using carry to record the carry-over.

The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.

  • 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;
    }
}

  • class Solution {
public:
    int numSteps(string s) {
        int ans = 0;
        bool carry = false;
        for (int i = s.size() - 1; i; --i) {
            char 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;
    }
};

  • 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


  • 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
}

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