2262. Total Appeal of A String

Description

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.

Solutions

Solution 1: Enumeration

We can enumerate all the substrings that end with each character \(s[i]\) and calculate their gravitational value sum \(t\). Finally, we add up all the \(t\) to get the total gravitational value sum.

When we reach \(s[i]\), which is added to the end of the substring that ends with \(s[i-1]\), we consider the change of the gravitational value sum \(t\):

If \(s[i]\) has not appeared before, then the gravitational value of all substrings that end with \(s[i-1]\) will increase by \(1\), and there are a total of \(i\) such substrings. Therefore, \(t\) increases by \(i\), plus the gravitational value of \(s[i]\) itself, which is \(1\). Therefore, \(t\) increases by a total of \(i+1\).

If \(s[i]\) has appeared before, let the last appearance position be \(j\). Then we add \(s[i]\) to the end of the substrings \(s[0..i-1]\), \([1..i-1]\), \(s[2..i-1]\), \(\cdots\), \(s[j..i-1]\). The gravitational value of these substrings will not change because \(s[i]\) has already appeared in these substrings. The gravitational value of the substrings \(s[j+1..i-1]\), \(s[j+2..i-1]\), \(\cdots\), \(s[i-1]\) will increase by \(1\), and there are a total of \(i-j-1\) such substrings. Therefore, \(t\) increases by \(i-j-1\), plus the gravitational value of \(s[i]\) itself, which is \(1\). Therefore, \(t\) increases by a total of \(i-j\). Therefore, we can use an array \(pos\) to record the last appearance position of each character. Initially, all positions are set to \(-1\).

Next, we traverse the string, and each time we update the gravitational value sum \(t\) of the substring that ends with the current character to \(t = t + i - pos[c]\), where \(c\) is the current character. We add \(t\) to the answer. Then we update \(pos[c]\) to the current position \(i\). We continue to traverse until the end of the string.

Python3JavaC++GoTypeScript

class Solution:
    def appealSum(self, s: str) -> int:
        ans = t = 0
        pos = [-1] * 26
        for i, c in enumerate(s):
            c = ord(c) - ord('a')
            t += i - pos[c]
            ans += t
            pos[c] = i
        return ans

class Solution {
    public long appealSum(String s) {
        long ans = 0;
        long t = 0;
        int[] pos = new int[26];
        Arrays.fill(pos, -1);
        for (int i = 0; i < s.length(); ++i) {
            int c = s.charAt(i) - 'a';
            t += i - pos[c];
            ans += t;
            pos[c] = i;
        }
        return ans;
    }
}

class Solution {
public:
    long long appealSum(string s) {
        long long ans = 0, t = 0;
        vector<int> pos(26, -1);
        for (int i = 0; i < s.size(); ++i) {
            int c = s[i] - 'a';
            t += i - pos[c];
            ans += t;
            pos[c] = i;
        }
        return ans;
    }
};

func appealSum(s string) int64 {
    var ans, t int64
    pos := make([]int, 26)
    for i := range pos {
        pos[i] = -1
    }
    for i, c := range s {
        c -= 'a'
        t += int64(i - pos[c])
        ans += t
        pos[c] = i
    }
    return ans
}

function appealSum(s: string): number {
    const pos: number[] = Array(26).fill(-1);
    const n = s.length;
    let ans = 0;
    let t = 0;
    for (let i = 0; i < n; ++i) {
        const c = s.charCodeAt(i) - 97;
        t += i - pos[c];
        ans += t;
        pos[c] = i;
    }
    return ans;
}

2262. Total Appeal of A String

Description

Solutions

Solution 1: Enumeration

Comments