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809. Expressive Words

Description

Sometimes people repeat letters to represent extra feeling. For example:

  • "hello" -> "heeellooo"
  • "hi" -> "hiiii"

In these strings like "heeellooo", we have groups of adjacent letters that are all the same: "h", "eee", "ll", "ooo".

You are given a string s and an array of query strings words. A query word is stretchy if it can be made to be equal to s by any number of applications of the following extension operation: choose a group consisting of characters c, and add some number of characters c to the group so that the size of the group is three or more.

  • For example, starting with "hello", we could do an extension on the group "o" to get "hellooo", but we cannot get "helloo" since the group "oo" has a size less than three. Also, we could do another extension like "ll" -> "lllll" to get "helllllooo". If s = "helllllooo", then the query word "hello" would be stretchy because of these two extension operations: query = "hello" -> "hellooo" -> "helllllooo" = s.

Return the number of query strings that are stretchy.

 

Example 1:

Input: s = "heeellooo", words = ["hello", "hi", "helo"]
Output: 1
Explanation: 
We can extend "e" and "o" in the word "hello" to get "heeellooo".
We can't extend "helo" to get "heeellooo" because the group "ll" is not size 3 or more.

Example 2:

Input: s = "zzzzzyyyyy", words = ["zzyy","zy","zyy"]
Output: 3

 

Constraints:

  • 1 <= s.length, words.length <= 100
  • 1 <= words[i].length <= 100
  • s and words[i] consist of lowercase letters.

Solutions

Solution 1: Traversal Counting + Two Pointers

We can traverse the array $\textit{words}$, and for each word $t$ in the array, check if $t$ can be expanded to obtain $s$. If it can, increment the answer by one.

Therefore, the key problem is to determine if the word $t$ can be expanded to obtain $s$. We use a function $\textit{check}(s, t)$ to determine this. The implementation logic of the function is as follows:

First, check the length relationship between $s$ and $t$. If the length of $t$ is greater than $s$, return $\textit{false}$ directly. Otherwise, use two pointers $i$ and $j$ to point to $s$ and $t$, respectively, both initially set to $0$.

If the characters pointed to by $i$ and $j$ are different, then $t$ cannot be expanded to obtain $s$, and we return $\textit{false}$. Otherwise, we need to check the relationship between the consecutive occurrence counts of the characters pointed to by $i$ and $j$, denoted as $c_1$ and $c_2$. If $c_1 \lt c_2$ or $c_1 \lt 3$ and $c_1 \neq c_2$, then $t$ cannot be expanded to obtain $s$, and we return $\textit{false}$. Otherwise, move $i$ and $j$ to the right by $c_1$ and $c_2$ times, respectively, and continue checking.

If both $i$ and $j$ reach the end of the strings, then $t$ can be expanded to obtain $s$, and we return $\textit{true}$. Otherwise, we return $\textit{false}$.

The time complexity is $O(n \times m + \sum_{i=0}^{m-1} w_i)$, where $n$ and $m$ are the lengths of the string $s$ and the array $\textit{words}$, respectively, and $w_i$ is the length of the $i$-th word in the array $\textit{words}$.

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class Solution:
    def expressiveWords(self, s: str, words: List[str]) -> int:
        def check(s, t):
            m, n = len(s), len(t)
            if n > m:
                return False
            i = j = 0
            while i < m and j < n:
                if s[i] != t[j]:
                    return False
                k = i
                while k < m and s[k] == s[i]:
                    k += 1
                c1 = k - i
                i, k = k, j
                while k < n and t[k] == t[j]:
                    k += 1
                c2 = k - j
                j = k
                if c1 < c2 or (c1 < 3 and c1 != c2):
                    return False
            return i == m and j == n

        return sum(check(s, t) for t in words)

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class Solution {
    public int expressiveWords(String s, String[] words) {
        int ans = 0;
        for (String t : words) {
            if (check(s, t)) {
                ++ans;
            }
        }
        return ans;
    }

    private boolean check(String s, String t) {
        int m = s.length(), n = t.length();
        if (n > m) {
            return false;
        }
        int i = 0, j = 0;
        while (i < m && j < n) {
            if (s.charAt(i) != t.charAt(j)) {
                return false;
            }
            int k = i;
            while (k < m && s.charAt(k) == s.charAt(i)) {
                ++k;
            }
            int c1 = k - i;
            i = k;
            k = j;
            while (k < n && t.charAt(k) == t.charAt(j)) {
                ++k;
            }
            int c2 = k - j;
            j = k;
            if (c1 < c2 || (c1 < 3 && c1 != c2)) {
                return false;
            }
        }
        return i == m && j == n;
    }
}

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class Solution {
public:
    int expressiveWords(string s, vector<string>& words) {
        auto check = [](string& s, string& t) -> int {
            int m = s.size(), n = t.size();
            if (n > m) return 0;
            int i = 0, j = 0;
            while (i < m && j < n) {
                if (s[i] != t[j]) return 0;
                int k = i;
                while (k < m && s[k] == s[i]) ++k;
                int c1 = k - i;
                i = k, k = j;
                while (k < n && t[k] == t[j]) ++k;
                int c2 = k - j;
                j = k;
                if (c1 < c2 || (c1 < 3 && c1 != c2)) return 0;
            }
            return i == m && j == n;
        };

        int ans = 0;
        for (string& t : words) ans += check(s, t);
        return ans;
    }
};

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func expressiveWords(s string, words []string) (ans int) {
    check := func(s, t string) bool {
        m, n := len(s), len(t)
        if n > m {
            return false
        }
        i, j := 0, 0
        for i < m && j < n {
            if s[i] != t[j] {
                return false
            }
            k := i
            for k < m && s[k] == s[i] {
                k++
            }
            c1 := k - i
            i, k = k, j
            for k < n && t[k] == t[j] {
                k++
            }
            c2 := k - j
            j = k
            if c1 < c2 || (c1 != c2 && c1 < 3) {
                return false
            }
        }
        return i == m && j == n
    }
    for _, t := range words {
        if check(s, t) {
            ans++
        }
    }
    return ans
}

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