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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.

The time complexity is $O(n)$, and the space complexity is $O(|\Sigma|)$, where $n$ is the length of the string $s$, and $|\Sigma|$ is the size of the character set. In this problem, $|\Sigma| = 26$.

  • 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;
    }
};

  • 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


  • 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;
}

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