Hard
The appeal of a string is the number of distinct characters found in the string.
"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.import java.util.Arrays;
public 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 += (long) (i - lastPos[idx]) * (len - i);
lastPos[idx] = i;
}
return res;
}
}