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Formatted question description: https://leetcode.ca/all/2083.html
2083. Substrings That Begin and End With the Same Letter (Medium)
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.
Companies:
Google
Related Topics:
Hash Table, Math, String, Counting, Prefix Sum
Similar Questions:
- Number of Good Pairs (Easy)
- Sum of Beauty of All Substrings (Medium)
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Solution 1.
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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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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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func numberOfSubstrings(s string) (ans int64) { cnt := [26]int{} for _, c := range s { c -= 'a' cnt[c]++ ans += int64(cnt[c]) } return ans }