Understanding the Problem

There are two strings give as parameters, ransomNote and magazine. I need to return True if ransomNote can be constructed with the characters of magazine. As an example,

# sample 1
ransomNote = "ba"
magazine = "aab"

this sample above is True because ransomNote characters can be made with “ab” from magazine, but,

# sample 2
ransomNote = "baa"
magazine = "ab"

this case comes with the output of False as magazine characters lack to form ransomNote.

Approach

I was in between the decisions. One was to go with pop() method, which I do not know why I like this name of the method, and the other one was to go with remove(), which I recently got familiar with. But after a while trying to solve the problem, pop() was not friendly with me so I went with remove().

I would chage first the ransomNote into a list in the form of which will be easier for me to handle with remove(), the for looping magazine string to check characters.

Solution

class Solution:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        l = list(ransomNote)
        for c in magazine:
            if c in l:
                l.remove(c)

        if len(l) == 0:
            return True
        return False

Characters from magazine will be checked if the same character exists in list(ransomNote). If it does exist, the character will be remove from the list so the miss match won’t happen.

Finally, if the list has length 0, it means every character of ransomNone was used in magazine, thus it fulfills the instruction, which finishes by returning True

Reflection

Other people were solving the problem with count() method, and some other were returning the function quicker by comparing the characters so the function will not have to check every character until the very end, which I had never thought of while solving this problem. I know I could do better though, hence will try it with shorter time and better efficiency.