2262. Total Appeal of A String

Difficulty:
Related Topics:
Similar Questions:

Problem

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

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

A 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:

Solution

class Solution {
    public long appealSum(String s) {
        int len = s.length();
        int[] lastPos = new int[26];
        Arrays.fill(lastPos, -1);
        long res = 0;
        for (int i = 0; i < len; i++) {
            int idx = s.charAt(i) - 'a';
            res += (i - lastPos[idx]) * (len - i);
            lastPos[idx] = i;
        }
        return res;
    }
}

Explain:

nope.

Complexity: