Count Subarray by Element

2262. Total Appeal of A String

Hard

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.

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:

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

left表示能为左侧作出贡献的个数
right 表示能为右侧作出贡献的个数
5      01234
-1     abbca
left:  12144
right: 54321

        5+8+3+8+4

 

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