README_EN.md

809. Expressive Words

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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

Python3

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)

Java

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;
    }
}

C++

class Solution {
public:
    int expressiveWords(string s, vector<string>& words) {
        int ans = 0;
        for (string& t : words) ans += check(s, t);
        return ans;
    }

    int check(string& s, string& t) {
        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;
    }
};

Go

func expressiveWords(s string, words []string) 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
	}
	ans := 0
	for _, t := range words {
		if check(s, t) {
			ans++
		}
	}
	return ans
}

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