• 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'

Solution: Simulation

Time complexity: O(n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int numSteps(string s) {
    int ans = 0;
    int carry = 0;
    // The highest bit must be 1, 
    // process to the 2nd highest bit
    for (int i = s.length() - 1; i > 0; --i) {
      if (s[i] - '0' + carry == 1) {
        ans += 2; // odd: +1, even: / 2
        carry = 1; // always has a carry
      } else {
        ans += 1; // even: / 2
        // carray remains e.g. 1 + 1 = 10, or 0 + 0 = 00
      }
    }
    // If there is a carry, then it's even, one more step.
    return ans + carry;
  }
};

# Author: Huahua
class Solution:
  def numSteps(self, s: str) -> int:
    ans, carry = 0, 0
    for i in range(1, len(s)):
      if ord(s[-i]) - ord('0') + carry == 1:
        ans += 2
        carry = 1
      else:
        ans += 1
    return ans + carry

The post 花花酱 LeetCode 1404. Number of Steps to Reduce a Number in Binary Representation to One appeared first on Huahua's Tech Road.

Problem:

https://leetcode.com/problems/count-binary-substrings/description/

Give a string s, count the number of non-empty (contiguous) substrings that have the same number of 0’s and 1’s, and all the 0’s and all the 1’s in these substrings are grouped consecutively.

Substrings that occur multiple times are counted the number of times they occur.

Example 1:

Input: "00110011"
Output: 6
Explanation: There are 6 substrings that have equal number of consecutive 1's and 0's: "0011", "01", "1100", "10", "0011", and "01".

Notice that some of these substrings repeat and are counted the number of times they occur.

Also, "00110011" is not a valid substring because all the 0's (and 1's) are not grouped together.

Example 2:

Input: "10101"
Output: 4
Explanation: There are 4 substrings: "10", "01", "10", "01" that have equal number of consecutive 1's and 0's.

Note:

• s.length will be between 1 and 50,000.
• s will only consist of “0” or “1” characters.

Solution 0: Search

Try all possible substrings and check whether it’s a valid one or not.

Time complexity O(2^n * n) TLE

Space complexity: O(n)

Solution 1: Running Length

For S = “000110”, there are two virtual blocks “[00011]0” and “000[110]” which contains consecutive 0s and 1s or (1s and 0s)

Keep tracking of the running length of 0, 1 for each block

“00011” => ‘0’: 3, ‘1’:2

ans += min(3, 2) => ans = 2 (“[0011]”, “0[01]1”)

“110” => ‘0’: 1, ‘1’: 2

ans += min(1, 2) => ans = 3 (“[0011]”, “0[01]1”, “1[10]”)

We can reuse part of the running length from last block.

Time complexity: O(n)

Space complexity: O(1)

C++

// Author: Huahua
// Running time: 43 ms
class Solution {
public:
  int countBinarySubstrings(string s) {        
    vector<int> lens(128, 0);
    int i = 0;
    int l = 0;
    int ans = 0;
    while (true) {
      while (i < s.length() && s[i] == s[l]) ++i;      
      lens[s[l]] = i - l;
      ans += min(lens['0'], lens['1']);
      if (i == s.length()) break;
      lens[s[i]] = 0;
      l = i;
    }
    return ans;
  }
};

The post 花花酱 LeetCode 696. Count Binary Substrings appeared first on Huahua's Tech Road.