345 leetcode ransom letter

If ransomNote Is longer than magazine longer , That must be wrong

First traversal magazine, Count the number of occurrences of each character , Re traversal ransomNote, Reduce the number of corresponding characters in the array by one at a time , If the value after subtracting one is less than 0 了 , This character is no longer in the array , So it's wrong

class Solution 
{
    
public:
	bool canConstruct(string ransomNote, string magazine)
	{
    
		int len1 = ransomNote.size();
		int len2 = magazine.size();
		if (len1 > len2)  return false;
		vector<int>ve(26,0);
		for (auto& ch : magazine) 
		{
    
			++ve[ch - 'a'];
		}
		for (auto& ch : ransomNote) 
		{
    
			if (--ve[ch - 'a'] < 0) 
			{
    
				return false;
			}
		}
		return true;
	}
};
int main()
{
    
	Solution A;
	cout << A.canConstruct("bg","bbg") << endl;
	return 0;
}

Time complexity ：O(m+n), among m Is string ransomNote The length of ,n Is string magazine The length of , We only need to traverse two characters once
Spatial complexity ：O(∣S∣),S It's a character set. , In this question S All lowercase English letters , therefore |S| = 26