Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the ransom note can be constructed from the magazines ; otherwise, it will return false.

Each letter in the magazine string can only be used once in your ransom note.

Note:
You may assume that both strings contain only lowercase letters.

canConstruct("a", "b") -> false
canConstruct("aa", "ab") -> false
canConstruct("aa", "aab") -> true

一刷
題解：給了一個magazine字符串，判斷這個字符串里面所有的字符能不能組成ransom note string
方法：有一個長度為26的統計magazine中 character frequency的array, 然后對ransom中的字符遍歷，如果存在，frequency--， 如果有frequency<0, 不能構成ransom， return false

public class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        int[] arr = new int[26];
        //count the character frequency
        for (int i = 0; i < magazine.length(); i++) {
            arr[magazine.charAt(i) - 'a']++;
        }
        for (int i = 0; i < ransomNote.length(); i++) {
            if(--arr[ransomNote.charAt(i)-'a'] < 0) {
                return false;
            }
        }
        return true;
    }
}

二刷
同上，頻率數組

class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        int[] mag = new int[26];
        for(int i=0; i<magazine.length(); i++){
            mag[magazine.charAt(i) - 'a']++;
        }
        
        for(int i=0; i<ransomNote.length(); i++){
            if(--mag[ransomNote.charAt(i) - 'a']<0) return false;
        }
        return true;
    }
}