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;
  }
};

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