Can You Solve This Facebook Interview Problem?

@that_swe_intern

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Why it worked

The video leverages a common pain point for aspiring tech professionals: coding interview preparation. By presenting a challenging problem (Ransom Note) and a successful solution, it offers value and positions the creator as knowledgeable. The "Facebook interview" hook and the "Follow for more!" call to action encourage engagement and audience growth.

Summary

The video explains how to solve the 'Ransom Note' coding problem, which involves determining if one string can be constructed from the letters of another. It demonstrates a Python solution using a hash map to count character frequencies and then checks if the ransom note can be formed.

Structure

1Introduce the problem and its context (Facebook interview)
2Explain the problem statement and constraints
3Walk through examples to illustrate the logic
4Present the Python code solution using a hash map
5Show the accepted submission and performance metrics
6Call to action for engagement

Original caption

Would you pass Facebook's interview? Follow to level up in tech! #leetcode #techinterview #compscimajor #facebook #breakintotech

More from @that_swe_intern

Product placement

The video features a screen recording of the LeetCode platform, displaying the 'Ransom Note' problem description and the accepted Python solution. The LeetCode platform is the primary visual subject related to the product.

Call to action

Comment down below if you would have passed this interview and follow for more!

On-screen text

383. Ransom Note Easy Companies Given two strings ransomNote and magazine, return true if ransomNote can be constructed by using the letters from magazine and false otherwise. Each letter in magazine can only be used once in ransomNote. Example 1: Input: ransomNote = "a", magazine = "b" Output: false Example 2: Input: ransomNote = "aa", magazine = "ab" Output: false Example 3: Input: ransomNote = "aa", magazine = "aab" Output: true Constraints: Python Auto from collections import defaultdict class Solution(object): def canConstruct(self, ransomNote: str, magazine: str) -> bool: # type ransomNote: str # type magazine: str # rtype: bool need = defaultdict(int) # (letter, count) for char in ransomNote: need[char] += 1 for char in magazine: need[char] -= 1 return not any([need_count > 0 for need_count in need.values()]) Accepted Runtime 29 ms Beats 91.47% of users with Python Memory 13.40 MB Beats 85.30% of users with Python More challenges 691. Stickers to Spell Word Status Accepted a few seconds ago Python 29 ms 13.4 MB need[char] -= 1 return not any([need_count > 0 for need_count in in need.values()]) Follow for more! Comment down below if you would have passed this interview and follow for more!

Transcript

Can You Solve This Facebook Interview Problem? All right today, we're gonna cover Ransom Note. Given two strings, Ransom Note and magazine, return true if Ransom Note can be constructed using the letters from magazine and false otherwise. Each letter in magazine can only be used once in Ransom Note. Example 1: Input: ransomNote = "a", magazine = "b" Output: false. Example 2: Input: ransomNote = "aa", magazine = "ab" Output: false. Example 3: Input: ransomNote = "aa", magazine = "aab" Output: true. Ransom note is 'aa', magazine is 'aab'. Output should be true because there's 2 A's inside our magazine, 2 A's inside our ransom note. All right, let's solve this question. We already know the counts of each letter. So let's make a hash map. Need = defaultdict(int). This hash map stores our letter and its count. Then for each letter we find inside the ransom note, let's increment the count. for char in ransomNote: need[char] += 1. Then for each letter we find inside the magazine, we can subtract that amount from what we need. for char in magazine: need[char] -= 1. All right, now we know the values of our hash map are the counts of the letters. If any of those values are above 0, then we have to return false. return not any([need_count > 0 for need_count in need.values()]). All right, let's see if this works. All right, there we go. We beat 91% of Python submissions. Comment down below if you would have passed this interview and follow for more!