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Playing Hacks and Stuffs!

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Project maintained by h4ckyou Hosted on GitHub Pages — Theme by mattgraham

Ransom Note

Given two strings ransomNote and magazine, return True if ransomNote can be constructed by using the letters from magazine and False otherwise.

Note: Each letter in magazine can only be used once in ransomNote.

We will be given two strings and our goal is to return True is the first string can be formed from the second one

Here’s my solution approach:

First I’ll gather each character occurrence of the two string and store in a hashtable (one for each)

The reason for doing that is because I’ll need it in the future also we need to make sure that each character in string a has the same occurrence in string b

So assuming we’re trying a brute force approach that particular test case will make the program return the wrong answer let’s say:

ransomNote = "aa"
magazine = "ab"

Because we’ll be likely checking if ransomNote[i] in magazine and that would presumely return True that ransomNote can be formed by using letters from magazine

That’s because the occurrence of a in ransomNote is twice while in magazine it’s just once and since we’re using the in command in python it would return that yes there are two a’s

So inorder to not meet that I’m saving each occurrence of letters of both strings in a hashtable

Then I will iterate through the key and value pairs of the ransomNote hashtable and check if the key exists in the magazine hashtable AND the value of magazine[key] is equal to or greater than the current iterate value

If that returns True I’ll append it to a list tt else I’ll append False to the tt list

At the end of the iteration I’ll check if False exists in tt and if it does then I return False or True

Here’s my solve script: link

I spent some amount of minutes coming up with various approach since the test cases where messing with my program 😭🥲

Leetcode Submission Script

class Solution:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        magazine_table = {}
        ransom_table = {}
        tt = []

        for i in ransomNote:
            if i in ransom_table:
                ransom_table[i] += 1
            else:
                ransom_table[i] = 1
        
        for i in magazine:
            if i in magazine_table:
                magazine_table[i] += 1
            else:
                magazine_table[i] = 1

        for keyR, valueR in ransom_table.items():
            if keyR in magazine_table and magazine_table[keyR] >= valueR:
                tt.append(True)
            else:
                tt.append(False)
        
        if False in tt:
            return False
        else:
            return True