README_EN.md

1160. Find Words That Can Be Formed by Characters

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Description

You are given an array of strings words and a string chars.

A string is good if it can be formed by characters from chars (each character can only be used once).

Return the sum of lengths of all good strings in words.

 

Example 1:

Input: words = ["cat","bt","hat","tree"], chars = "atach"

Output: 6

Explanation: 

The strings that can be formed are "cat" and "hat" so the answer is 3 + 3 = 6.

Example 2:

Input: words = ["hello","world","leetcode"], chars = "welldonehoneyr"

Output: 10

Explanation: 

The strings that can be formed are "hello" and "world" so the answer is 5 + 5 = 10.

 

Note:

1. 1 <= words.length <= 1000
2. 1 <= words[i].length, chars.length <= 100
3. All strings contain lowercase English letters only.

Solutions

Python3

class Solution:
    def countCharacters(self, words: List[str], chars: str) -> int:
        counter = Counter(chars)
        ans = 0
        for word in words:
            cnt = Counter(word)
            if all([counter[c] >= v for c, v in cnt.items()]):
                ans += len(word)
        return ans

Java

class Solution {
    public int countCharacters(String[] words, String chars) {
        int[] counter = count(chars);
        int ans = 0;
        for (String word : words) {
            int[] cnt = count(word);
            if (check(counter, cnt)) {
                ans += word.length();
            }
        }
        return ans;
    }

    private int[] count(String s) {
        int[] counter = new int[26];
        for (char c : s.toCharArray()) {
            ++counter[c - 'a'];
        }
        return counter;
    }

    private boolean check(int[] cnt1, int[] cnt2) {
        for (int i = 0; i < 26; ++i) {
            if (cnt1[i] < cnt2[i]) {
                return false;
            }
        }
        return true;
    }
}

C++

class Solution {
public:
    int countCharacters(vector<string>& words, string chars) {
        vector<int> counter = count(chars);
        int ans = 0;
        for (auto& word : words)
        {
            vector<int> cnt = count(word);
            if (check(counter, cnt)) ans += word.size();
        }
        return ans;
    }

    vector<int> count(string s) {
        vector<int> counter(26);
        for (char c : s) ++counter[c - 'a'];
        return counter;
    }

    bool check(vector<int>& cnt1, vector<int>& cnt2) {
        for (int i = 0; i < 26; ++i)
            if (cnt1[i] < cnt2[i]) return false;
        return true;
    }
};

Go

func countCharacters(words []string, chars string) int {
	counter := count(chars)
	ans := 0
	for _, word := range words {
		cnt := count(word)
		if check(counter, cnt) {
			ans += len(word)
		}
	}
	return ans
}

func count(s string) []int {
	counter := make([]int, 26)
	for _, c := range s {
		counter[c-'a']++
	}
	return counter
}

func check(cnt1, cnt2 []int) bool {
	for i := 0; i < 26; i++ {
		if cnt1[i] < cnt2[i] {
			return false
		}
	}
	return true
}

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