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class Solution {
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 public:
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  long long appealSum(string s) {
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    int n = s.size();
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    long long ans = 0;
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    for (char ch = 'a'; ch <= 'z'; ch++) // we are finding the number of substrings containing at
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                                         // least 1 occurence of ch
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    {
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      int prev = 0; // prev will store the previous index of the charcter ch
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      for (int i = 0; i < n; i++) {
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        if (s[i] == ch)
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          prev = i + 1; // if the current character is equal to ch , then the
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                        // no. of substring ending at i and having at least one
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                        // occurence of ch will be i+1 .
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        ans += prev; // else the no. of substrings ending at i and having at
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                     // least one occurence of ch will be the equal to, the
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                     // previous index of ch.
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      }
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    }
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    return ans; // TC - O(n*26) , SC - O(1)
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  }
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};