696. Count Binary Substrings

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

Python

Two Pointer

列表如下

String	Group Count	Groups	Details
0011	2	01, 0011	1個group，min(0,1) = 2
00011	2	01, 0011	1個group，min(0,1) = 2
10101	4	10, 01, 10, 01	4個group，min(0,1)各別為1
000111	3	000111, 0011, 01	1個group，min(0,1) = 3
00011100111	7	000111, 0011, 01, 1100, 10, 0011, 01	3個group，min(0,1)各為3, 2, 2

歸納後會發現，以連續的0跟1組成的group，取其中各group小的數字加總為答案

class Solution:
    def countBinarySubstrings(self, s: str) -> int:
        left, right = 0, 1
        count = 0

        for i in range(1, len(s)):
            if s[i] != s[i-1]:
                count += min(left, right)
                left = right
                right = 1
            else:
                right += 1

        count += min(left, right)
        return count

JS

 var countBinarySubstrings = function(s) {
   let curr = 1;
   let prev = 0;
   let sum = 0;

   for (let i = 1; i < s.length; i++) {
     if (s[i] === s[i - 1]) curr++;
     else {       
       sum += Math.min(curr, prev);
       prev = curr;
       curr = 1;
     }
   }
   return sum + Math.min(curr, prev)
};