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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 \(O(1)\).

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

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

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

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

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

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