LeetCode //C - 696. Count Binary Substrings

696. Count Binary Substrings

Given a binary string s, return the number of non-empty 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: s = “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: s = “10101”
Output: 4
Explanation: There are 4 substrings: “10”, “01”, “10”, “01” that have equal number of consecutive 1’s and 0’s.

Constraints:

• $1<=s.length<=10^5$
• s[i] is either ‘0’ or ‘1’.

From: LeetCode
Link: 696. Count Binary Substrings

Solution:

Ideas:

• prev tracks the length of the previous group (e.g., “00” → prev = 2).

• curr tracks the length of the current group (e.g., “11” → curr = 2).

• Whenever the character changes (s[i] != s[i-1]), it means a new group starts:

  • Add min(prev, curr) to the result because only that many valid substrings can be formed.

  • Then update prev = curr and reset curr = 1.

• At the end, add the last min(prev, curr) again.

Code:

int countBinarySubstrings(char* s) {int prev = 0, curr = 1;int count = 0;for (int i = 1; s[i]; i++) {if (s[i] == s[i - 1]) {curr++;} else {count += prev < curr ? prev : curr;prev = curr;curr = 1;}}count += prev < curr ? prev : curr;return count;
}