# 3084. Count Substrings Starting and Ending with Given Character

## Description

You are given a string s and a character c. Return the total number of substrings of s that start and end with c.

Example 1:

Input: s = "abada", c = "a"

Output: 6

Explanation: Substrings starting and ending with "a" are: "abada", "abada", "abada", "abada", "abada", "abada".

Example 2:

Input: s = "zzz", c = "z"

Output: 6

Explanation: There are a total of 6 substrings in s and all start and end with "z".

Constraints:

• 1 <= s.length <= 105
• s and c consist only of lowercase English letters.

## Solutions

### Solution 1: Mathematics

First, we can count the number of character $c$ in string $s$, denoted as $cnt$.

Each character $c$ can form a substring on its own, so there are $cnt$ substrings that meet the condition. Each character $c$ can form a substring with other $c$ characters, so there are $\frac{cnt \times (cnt - 1)}{2}$ substrings that meet the condition.

Therefore, the answer is $cnt + \frac{cnt \times (cnt - 1)}{2}$.

The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.

• class Solution {
public long countSubstrings(String s, char c) {
long cnt = s.chars().filter(ch -> ch == c).count();
return cnt + cnt * (cnt - 1) / 2;
}
}

• class Solution {
public:
long long countSubstrings(string s, char c) {
long long cnt = ranges::count(s, c);
return cnt + cnt * (cnt - 1) / 2;
}
};

• class Solution:
def countSubstrings(self, s: str, c: str) -> int:
cnt = s.count(c)
return cnt + cnt * (cnt - 1) // 2


• func countSubstrings(s string, c byte) int64 {
cnt := int64(strings.Count(s, string(c)))
return cnt + cnt*(cnt-1)/2
}

• function countSubstrings(s: string, c: string): number {
const cnt = s.split('').filter(ch => ch === c).length;
return cnt + Math.floor((cnt * (cnt - 1)) / 2);
}