Another Casual Coder

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.

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

• With 2 variables you'll keep track of the number of consecutive ones, and the number of consecutive zeros, one variable for each
• Make sure to reset the count exactly when you transition from one to the other
• Every time that you increment either the consecutive ones, or the consecutive zeros, you check whether that count is still less than or equal than the other one. The idea is that if the current count (say for the '1's) is still less than the other count (say for the '0's), it means that you have found a new substring matching the criteria. Then, increment the return value

public class Solution
{
 public int CountBinarySubstrings(string s)
 {
 if (String.IsNullOrEmpty(s)) return 0;
 int numberConsecutiveOnes = 0;
 int numberConsecutiveZeroes = 0;
 int result = 0;
 for (int i = 0; i < s.Length; i++)
 {
 if (s[i] == '0')
 {
 if (i - 1 >= 0 && s[i - 1] == '1') numberConsecutiveZeroes = 0;
 numberConsecutiveZeroes++;
 if (numberConsecutiveZeroes <= numberConsecutiveOnes) result++;
 }
 else
 {
 if (i - 1 >= 0 && s[i - 1] == '0') numberConsecutiveOnes = 0;
 numberConsecutiveOnes++;
 if (numberConsecutiveOnes <= numberConsecutiveZeroes) result++;
 }
 }
 return result;
 }
}