/*给你一份『词汇表』（字符串数组） words 和一张『字母表』（字符串） chars。
    假如你可以用 chars 中的『字母』（字符）拼写出 words 中的某个『单词』（字符串），那么我们就认为你掌握了这个单词。
    注意：每次拼写时，chars 中的每个字母都只能用一次。
    返回词汇表 words 中你掌握的所有单词的 长度之和。
    示例 1：
    输入：words = ["cat","bt","hat","tree"], chars = "atach"
    输出：6
    解释：
    可以形成字符串 "cat" 和 "hat"，所以答案是 3 + 3 = 6。
    示例 2：
    输入：words = ["hello","world","leetcode"], chars = "welldonehoneyr"
    输出：10
    解释：
    可以形成字符串 "hello" 和 "world"，所以答案是 5 + 5 = 10。
    链接：https://leetcode-cn.com/problems/find-words-that-can-be-formed-by-characters*/

    public static int countCharacters(String[] words, String chars) {
      int result = 0;
      int[] a = new int[26];
        char[] chars1 = chars.toCharArray();
        for (int i = 0; i < chars1.length; i++) {
            a[chars1[i]-'a'] += 1;
        }
        for(int j = 0 ; j<words.length ;j++){
            char[] word = words[j].toCharArray();
            int[] b = new int[26];
            for (int i = 0; i < word.length; i++) {
                b[word[i]-'a'] += 1;
            }
            boolean flag = true;
            for (int i = 0; i <26 ; i++) {
                if(b[i] > a[i]){
                    flag = false;
                  break;
                }
            }
            if(flag) result += word.length;
        }

      return result;
    }

    public static void main(String[] args) {
        String[] words = {"cat","bt","hat","tree"};
        String chars = "atach";

        String[] b = {"dyiclysmffuhibgfvapygkorkqllqlvokosagyelotobicwcmebnpznjbirzrzsrtzjxhsfpiwyfhzyonmuabtlwin","ndqeyhhcquplmznwslewjzuyfgklssvkqxmqjpwhrshycmvrb","ulrrbpspyudncdlbkxkrqpivfftrggemkpyjl","boygirdlggnh","xmqohbyqwagkjzpyawsydmdaattthmuvjbzwpyopyafphx","nulvimegcsiwvhwuiyednoxpugfeimnnyeoczuzxgxbqjvegcxeqnjbwnbvowastqhojepisusvsidhqmszbrnynkyop","hiefuovybkpgzygprmndrkyspoiyapdwkxebgsmodhzpx","juldqdzeskpffaoqcyyxiqqowsalqumddcufhouhrskozhlmobiwzxnhdkidr","lnnvsdcrvzfmrvurucrzlfyigcycffpiuoo","oxgaskztzroxuntiwlfyufddl","tfspedteabxatkaypitjfkhkkigdwdkctqbczcugripkgcyfezpuklfqfcsccboarbfbjfrkxp","qnagrpfzlyrouolqquytwnwnsqnmuzphne","eeilfdaookieawrrbvtnqfzcricvhpiv","sisvsjzyrbdsjcwwygdnxcjhzhsxhpceqz","yhouqhjevqxtecomahbwoptzlkyvjexhzcbccusbjjdgcfzlkoqwiwue","hwxxighzvceaplsycajkhynkhzkwkouszwaiuzqcleyflqrxgjsvlegvupzqijbornbfwpefhxekgpuvgiyeudhncv","cpwcjwgbcquirnsazumgjjcltitmeyfaudbnbqhflvecjsupjmgwfbjo","teyygdmmyadppuopvqdodaczob","qaeowuwqsqffvibrtxnjnzvzuuonrkwpysyxvkijemmpdmtnqxwekbpfzs","qqxpxpmemkldghbmbyxpkwgkaykaerhmwwjonrhcsubchs"};
        String a = "usdruypficfbpfbivlrhutcgvyjenlxzeovdyjtgvvfdjzcmikjraspdfp";
        int i = countCharacters(b, a);
        System.out.println("args = " + i);
    }