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2083. Substrings That Begin and End With the Same Letter πŸ”’

Description

You are given a 0-indexed string s consisting of only lowercase English letters. Return the number of substrings in s that begin and end with the same character.

A substring is a contiguous non-empty sequence of characters within a string.

 

Example 1:

Input: s = "abcba"
Output: 7
Explanation:
The substrings of length 1 that start and end with the same letter are: "a", "b", "c", "b", and "a".
The substring of length 3 that starts and ends with the same letter is: "bcb".
The substring of length 5 that starts and ends with the same letter is: "abcba".

Example 2:

Input: s = "abacad"
Output: 9
Explanation:
The substrings of length 1 that start and end with the same letter are: "a", "b", "a", "c", "a", and "d".
The substrings of length 3 that start and end with the same letter are: "aba" and "aca".
The substring of length 5 that starts and ends with the same letter is: "abaca".

Example 3:

Input: s = "a"
Output: 1
Explanation:
The substring of length 1 that starts and ends with the same letter is: "a".

 

Constraints:

  • 1 <= s.length <= 105
  • s consists only of lowercase English letters.

Solutions

Solution 1

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class Solution:
    def numberOfSubstrings(self, s: str) -> int:
        cnt = Counter()
        ans = 0
        for c in s:
            cnt[c] += 1
            ans += cnt[c]
        return ans

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class Solution {
    public long numberOfSubstrings(String s) {
        int[] cnt = new int[26];
        long ans = 0;
        for (int i = 0; i < s.length(); ++i) {
            int j = s.charAt(i) - 'a';
            ++cnt[j];
            ans += cnt[j];
        }
        return ans;
    }
}

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class Solution {
public:
    long long numberOfSubstrings(string s) {
        int cnt[26]{};
        long long ans = 0;
        for (char& c : s) {
            ans += ++cnt[c - 'a'];
        }
        return ans;
    }
};

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func numberOfSubstrings(s string) (ans int64) {
    cnt := [26]int{}
    for _, c := range s {
        c -= 'a'
        cnt[c]++
        ans += int64(cnt[c])
    }
    return ans
}

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