How do you solve Ransom Note on LeetCode?

Ransom Note (LeetCode #383) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a frequency count allows us to efficiently track character availability.

What is the brute force solution for Ransom Note?

The brute force approach for Ransom Note is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks if each character in the ransom note can be found in the magazine. This is a straightforward method but can be inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Ransom Note?

Ransom Note uses the following concepts: Hash Table, String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Ransom Note?

Always clarify the problem constraints and edge cases with the interviewer. Explain your thought process clearly while coding; it shows your understanding. Consider both time and space complexity when discussing your solution.

What are common mistakes when solving Ransom Note?

Not accounting for character frequency, leading to false negatives. Overlooking edge cases where characters are present but insufficient in number.

#383

Ransom Note

Easy

Hash TableStringCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a frequency count of characters in the magazine. This allows us to efficiently check if we have enough letters to construct the ransom note.

⚙️

Algorithm

4 steps

1Step 1: Create a frequency map (dictionary) to count occurrences of each character in magazine.
2Step 2: For each character in ransomNote, check if it exists in the frequency map and if the count is greater than zero.
3Step 3: Decrease the count in the frequency map for each character used.
4Step 4: If all characters are accounted for, return true; otherwise, return false.

solution.py10 lines

1# Full working Python code
2
3def canConstruct(ransomNote, magazine):
4    from collections import Counter
5    magazine_count = Counter(magazine)
6    for char in ransomNote:
7        if magazine_count[char] <= 0:
8            return False
9        magazine_count[char] -= 1
10    return True

ℹ

Complexity note: The time complexity is O(n) because we traverse the magazine and ransomNote strings once each. The space complexity is O(n) for storing the frequency counts.

1Using a frequency count allows us to efficiently track character availability.
2The optimal solution scales better with larger inputs due to linear time complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.