Question

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

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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:
  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

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;
  }
};

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
}