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.

 

Constraints:

• 1 <= words.length <= 1000
• 1 <= words[i].length, chars.length <= 100
• words[i] and chars consist of lowercase English letters.

Solutions

Python3

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

Java

class Solution {
    public int countCharacters(String[] words, String chars) {
        int[] cnt = new int[26];
        for (int i = 0; i < chars.length(); ++i) {
            ++cnt[chars.charAt(i) - 'a'];
        }
        int ans = 0;
        for (String w : words) {
            int[] wc = new int[26];
            boolean ok = true;
            for (int i = 0; i < w.length(); ++i) {
                int j = w.charAt(i) - 'a';
                if (++wc[j] > cnt[j]) {
                    ok = false;
                    break;
                }
            }
            if (ok) {
                ans += w.length();
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int countCharacters(vector<string>& words, string chars) {
        int cnt[26]{};
        for (char& c : chars) {
            ++cnt[c - 'a'];
        }
        int ans = 0;
        for (auto& w : words) {
            int wc[26]{};
            bool ok = true;
            for (auto& c : w) {
                int i = c - 'a';
                if (++wc[i] > cnt[i]) {
                    ok = false;
                    break;
                }
            }
            if (ok) {
                ans += w.size();
            }
        }
        return ans;
    }
};

Go

func countCharacters(words []string, chars string) (ans int) {
	cnt := [26]int{}
	for _, c := range chars {
		cnt[c-'a']++
	}
	for _, w := range words {
		wc := [26]int{}
		ok := true
		for _, c := range w {
			c -= 'a'
			wc[c]++
			if wc[c] > cnt[c] {
				ok = false
				break
			}
		}
		if ok {
			ans += len(w)
		}
	}
	return
}

TypeScript

function countCharacters(words: string[], chars: string): number {
    const idx = (c: string) => c.charCodeAt(0) - 'a'.charCodeAt(0);
    const cnt = new Array(26).fill(0);
    for (const c of chars) {
        cnt[idx(c)]++;
    }
    let ans = 0;
    for (const w of words) {
        const wc = new Array(26).fill(0);
        let ok = true;
        for (const c of w) {
            if (++wc[idx(c)] > cnt[idx(c)]) {
                ok = false;
                break;
            }
        }
        if (ok) {
            ans += w.length;
        }
    }
    return ans;
}

PHP

class Solution {
    /**
     * @param String[] $words
     * @param String $chars
     * @return Integer
     */
    function countCharacters($words, $chars) {
        $sum = 0;
        for ($i = 0; $i < strlen($chars); $i++) {
            $hashtable[$chars[$i]] += 1;
        }
        for ($j = 0; $j < count($words); $j++) {
            $tmp = $hashtable;
            $sum += strlen($words[$j]);
            for ($k = 0; $k < strlen($words[$j]); $k++) {
                if (!isset($tmp[$words[$j][$k]]) || $tmp[$words[$j][$k]] === 0) {
                    $sum -= strlen($words[$j]);
                    break;
                }
                $tmp[$words[$j][$k]] -= 1;
            }
        }
        return $sum;
    }
}

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