(Easy) Find Words That Can Be Formed by Characters LeetCode

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

 

Solution:

class Solution {
    public int countCharacters(String[] words, String chars) {
        
        if(words==null|| chars==null||words.length==0||chars.length()==0){
            return 0;
        }
        
        int result = 0;
        
        for(int i = 0; i<words.length; i++){
            
            boolean flag = true;
            
            for(int j = 0; j<words[i].length();j++){
             
                if(!check(chars,words[i].charAt(j),Count(words[i],words[i].charAt(j)))){
                    flag = false;
                }
            }
            
            if(flag){
           
                result = result + words[i].length();
            }
            
        }
        
        return result;
        
    }
    
    boolean check(String str, Character ch, int n){
        int tmp =0;
        for(int i = 0; i<str.length();i++){
            if(str.charAt(i)==ch){
                
                tmp++;
            }
            
        }
        
        return tmp>=n? true:false;
        
    }
    
    int Count(String str, Character ch){
        
        int res = 0; 
        
        for(int i = 0; i<str.length(); i++){
            
            if(str.charAt(i)==ch ){
                res++;
            }
        }
        
        return res;
    }
    
    
}