Grind 75 - 15 - Ransom Note

Problem Statement:

LeetCode Number: 383

Difficulty Level: Easy

Explanation:

• You are given two strings, ransomNote and magazine.
• A ransom note can be constructed from a magazine if the magazine contains all the letters in the ransom note, including multiplicity.
• Every letter in the magazine string can only be used once in your ransom note.
• Write a function that will return True if the ransom note can be constructed from the magazine and False if it cannot.

Sample Data:

• Input: ransomNote = "aa", magazine = "aab"
• Output: True

Questions to be asked to the interviewer:

• Should I consider case sensitivity? (i.e., are "a" and "A" considered different characters?)
• What should be returned if the ransomNote or magazine string is empty?
• Are there any constraints on the lengths of the ransomNote and magazine strings?

Explain any edge cases for this problem:

• When both the ransomNote and magazine are empty.
• When ransomNote is longer than the magazine, it's impossible to construct the note.

Available Approaches to Solve this Problem:

• Using Counter: Utilize a counter to count the occurrences of characters in the magazine, and then check if the ransom note can be constructed.
• Brute Force: Iterate through the ransom note and for each character, search and replace it in the magazine string.

My Approach to Solve this Problem:

I will use the Counter approach as it is more efficient. By counting the occurrences of characters in the magazine, I can efficiently check if each character in the ransom note can be constructed from the magazine.

from collections import Counter


def canConstruct(ransomNote, magazine):
? ? magazine_counter = Counter(magazine)
? ? for char in ransomNote:
? ? ? ? if magazine_counter[char] == 0:
? ? ? ? ? ? return False
? ? ? ? magazine_counter[char] -= 1
? ? return True

        

Explain the Solution to a Non-Programmer:

• Imagine you want to create a message by cutting out letters from a magazine.
• You have a specific message in mind (the ransom note) and a page from a magazine with various letters.
• You start by counting how many times each letter appears on the magazine page.
• Then, you go through your desired message letter by letter, checking if that letter is available in the magazine.
• If a letter is available, you "cut it out" by reducing the count of that letter in the magazine.
• If you find a letter in your message that's not available in the magazine, you know you can't create the message.
• If you can find every letter of your message in the magazine, then you can create the message.

Code in Detail :

• Line 1: Import the Counter class, which helps us count occurrences of elements in a collection.
• Line 3: Define the function canConstruct, taking the ransom note and magazine as parameters.
• Line 4: Use the Counter to count the occurrences of each character in the magazine.
• Line 5-8: Iterate through the ransom note. For each character, if its count in the magazine is zero, return False. If it's available, decrease its count in the magazine.
• Line 9: If we successfully iterate through the ransom note without returning False, it means the note can be constructed, so return True.

Pattern of this Problem:

This problem follows the pattern of utilizing a frequency counter to keep track of the occurrences of characters, enabling us to quickly check if the ransom note can be formed.

Big O Notation:

• Time Complexity: O(m+n)
• O(m+n), where m is the length of the magazine and n is the length of the ransom note. This accounts for the time it takes to count the characters in the magazine and iterate through the ransom note.
• Space Complexity: O(m)
• O(m), m is the length of the magazine
• m is the length of the magazine. We use extra space to store the counts of the characters from the magazine.

Points to Remember to Solve this Problem in the Future:

• Utilize the Counter class for efficient counting of character occurrences.
• Iterate through the ransom note, decrementing the count for each character found, and handle the case when the count becomes zero.

Code Line by Line with Some Sample Data:

• Input: ransomNote = "aa", magazine = "aab"
• Line 4: magazine_counter = {'a': 2, 'b': 1}
• Line 5: For the first 'a' in ransomNote, count is 2 in magazine, so decrement it by 1.
• Line 5: For the second 'a' in ransomNote, count is 1 in magazine, so decrement it by 1.
• Line 9: No character in ransomNote has a count of 0 in magazine, so return True.

Key Syntax Used in this Solution:

• Counter Class for Frequency Counting: Counter(magazine)
• Looping Through Characters in the Ransom Note: for char in ransomNote
• Checking and Decrementing the Count of a Character: if magazine_counter[char] == 0 and magazine_counter[char] -= 1

Understanding these key syntax points, the pattern of frequency counting, and handling the zero count case will help in solving similar problems in the future.

The next problem is

Climbing Stairs