Ransom Note 15 XP Easy

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

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

ransomNote = 'aa'
magazine  = 'aab'
# True — magazine has 2 a's, we need 2 a's

The hash map approach: count all letters in magazine, then check if ransomNote's letters are available.

📋 Instructions

Implement `can_construct(ransomNote, magazine)` that returns True/False. Test with: - `can_construct('a', 'b')` → False - `can_construct('aa', 'aab')` → True Print both results.

Use a dictionary to count characters in magazine. Then iterate through ransomNote — for each char, check if count > 0 in the dict, and decrement.

🧪 Test Cases

Input

can_construct('a', 'b')

Expected

False

Boolean check

Input

can_construct('aa', 'aab')

Expected

True

Boolean check

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main.py

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Output

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