383. Ransom Note

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

1. Analyse

It is a easy problem, we just need to count the quantity of each letter in magazine and compare with quantity of each letter in ransom not. We can use a map to store the quantity.

2. AC Code

(12 Lines)

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        map<char,int> mp; 
        for(int i=0; i<magazine.size(); i++)
            mp[magazine[i]]++;
        for(int i=0; i<ransomNote.size(); i++)
            if(--mp[ransomNote[i]] < 0)
                return false;
        return true;
    }
};

Here we can learn one thing: if a map does not have a key, its value will initial to be 0.