Total Appeal of A String LeetCode Solution

Problem – Total Appeal of A String LeetCode Solution

The appeal of a string is the number of distinct characters found in the string.

  • For example, the appeal of "abbca" is 3 because it has 3 distinct characters: 'a''b', and 'c'.

Given a string s, return the total appeal of all of its substrings.

substring is a contiguous sequence of characters within a string.

Example 1:

Input: s = "abbca"
Output: 28
Explanation: The following are the substrings of "abbca":
- Substrings of length 1: "a", "b", "b", "c", "a" have an appeal of 1, 1, 1, 1, and 1 respectively. The sum is 5.
- Substrings of length 2: "ab", "bb", "bc", "ca" have an appeal of 2, 1, 2, and 2 respectively. The sum is 7.
- Substrings of length 3: "abb", "bbc", "bca" have an appeal of 2, 2, and 3 respectively. The sum is 7.
- Substrings of length 4: "abbc", "bbca" have an appeal of 3 and 3 respectively. The sum is 6.
- Substrings of length 5: "abbca" has an appeal of 3. The sum is 3.
The total sum is 5 + 7 + 7 + 6 + 3 = 28.

Example 2:

Input: s = "code"
Output: 20
Explanation: The following are the substrings of "code":
- Substrings of length 1: "c", "o", "d", "e" have an appeal of 1, 1, 1, and 1 respectively. The sum is 4.
- Substrings of length 2: "co", "od", "de" have an appeal of 2, 2, and 2 respectively. The sum is 6.
- Substrings of length 3: "cod", "ode" have an appeal of 3 and 3 respectively. The sum is 6.
- Substrings of length 4: "code" has an appeal of 4. The sum is 4.
The total sum is 4 + 6 + 6 + 4 = 20.

Constraints:

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

Total Appeal of A String LeetCode Solution in Java

    public long appealSum(String s) {
        int last[] = new int[26];
        long res = 0;
        for (int i = 0; i < s.length(); ++i) {
            last[s.charAt(i) - 'a'] = i + 1;
            for (int j: last)
                res += j;
        }
        return res;
    }

Total Appeal of A String LeetCode Solution in C++

    long long appealSum(string s) {
        vector<int> last(26);
        long res = 0, n = s.size();
        for (int i = 0; i < n; ++i) {
            last[s[i] - 'a'] = i + 1;
            for (int j: last)
                res += j;
        }
        return res;
    }

Total Appeal of A String LeetCode Solution in Python

    def appealSum(self, s):
        last = {}
        res = 0
        for i,c in enumerate(s):
            last[c] = i + 1
            res += sum(last.values())
        return res
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