#383

Easy

LeetCode Problem Workspace

Ransom Note

Determine if a ransom note can be constructed from a magazine's letters using hash tables and string counting techniques.

hash table string counting

Problem Statement

You are given two strings, ransomNote and magazine. Return true if ransomNote can be constructed from the letters of magazine, and false otherwise. Each letter in the magazine can only be used once, and the ransomNote can contain any characters, as long as each character is present in the magazine with sufficient frequency.

For example, if ransomNote is "aa" and magazine is "aab", the function should return true, since the magazine has enough occurrences of the letter 'a' to construct the ransom note. However, if the ransomNote is "aa" and the magazine is "ab", it should return false, as the magazine lacks a second 'a'.

Examples

Example 1

Input: ransomNote = "a", magazine = "b"

Output: false

Example details omitted.

Example 2

Input: ransomNote = "aa", magazine = "ab"

Output: false

Example details omitted.

Example 3

Input: ransomNote = "aa", magazine = "aab"

Output: true

Example details omitted.

Constraints

1 <= ransomNote.length, magazine.length <= 105
ransomNote and magazine consist of lowercase English letters.

Solution Approach

Using a Hash Table for Letter Frequency

One efficient solution is to create a hash table (or frequency map) to track how many times each letter appears in the magazine. Then, for each character in the ransomNote, check if it appears enough times in the magazine, adjusting the hash table as characters are used. This method ensures that all letters are counted and used properly.

String Counting Approach

Alternatively, you can use an array or hash map to count occurrences of characters in both the ransomNote and the magazine. Once the counts are collected, compare them to verify that the magazine has enough occurrences of each character. This solution has an O(n) time complexity, where n is the length of the longer string.

Optimizing with Early Termination

To optimize performance, consider early termination in the solution. Once you find that a letter in the ransomNote cannot be fully matched by letters in the magazine (e.g., if there are insufficient occurrences), you can immediately return false without further checks. This can save time for larger inputs.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of this problem depends on the approach. Using a hash table or array for counting results in an O(n) time complexity, where n is the length of the longer string (either the ransomNote or the magazine). The space complexity is O(1), as the hash table or array size is constant (26 for lowercase English letters).

What Interviewers Usually Probe

Checks for efficient handling of string operations and hash table manipulation.
Evaluates understanding of counting characters and early termination optimization.
Tests ability to correctly implement hash table usage and manage frequency comparisons.

Common Pitfalls or Variants

Common pitfalls

Incorrectly assuming that characters can be reused from the magazine when they can only be used once.
Failing to account for the scenario where a character in the ransomNote does not appear enough times in the magazine.
Overcomplicating the solution by using unnecessary data structures or algorithms, instead of leveraging a simple hash table or frequency map.

Follow-up variants

What if the magazine and ransomNote contain uppercase letters? Modify the hash table to account for both cases.
Extend the problem to handle more complex scenarios like punctuations or numbers. How would the approach change?
Consider the impact on performance if the input strings were significantly larger (e.g., with lengths up to 10^6).

FAQ

What is the primary pattern for solving the Ransom Note problem?

The primary pattern involves using a hash table or frequency map to count occurrences of characters in both the ransomNote and the magazine, then comparing these counts to ensure the ransomNote can be constructed.

What is the time complexity of the Ransom Note problem?

The time complexity is O(n), where n is the length of the longer string (either ransomNote or magazine). This is because each string is processed once to count letter frequencies.

What if the ransomNote has more characters than the magazine?

If the ransomNote contains more characters than the magazine can supply, the answer should be false. The hash table comparison will show insufficient occurrences of required characters.

How does early termination improve the solution?

Early termination helps by stopping further processing once it's clear that the ransomNote cannot be fully constructed from the magazine, thus saving time and reducing unnecessary computation.

Can the Ransom Note problem be solved using brute force?

While a brute-force solution could work, it would be inefficient. Using hash tables or frequency maps to count characters is a more optimal approach with O(n) time complexity.

terminal

Solution

Solution 1: Hash Table or Array

We can use a hash table or an array $cnt$ of length $26$ to record the number of times each character appears in the string `magazine`. Then traverse the string `ransomNote`, for each character $c$ in it, we decrease the number of $c$ by $1$ in $cnt$. If the number of $c$ is less than $0$ after the decrease, it means that the number of $c$ in `magazine` is not enough, so it cannot be composed of `ransomNote`, just return $false$.

class Solution:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        cnt = Counter(magazine)
        for c in ransomNote:
            cnt[c] -= 1
            if cnt[c] < 0:
                return False
        return True

Continue Practicing

#387 first unique character in a string #451 sort characters by frequency #299 bulls and cows