383. Ransom Note 🔍

Difficulty: Easy - Tags: Hash Table, String

LeetCode Problem Link

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Problem Statement 📜

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.

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Examples 🌟

🔹 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

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Constraints ⚙️

• 1 <= ransomNote.length, magazine.length <= 10^5

• ransomNote and magazine consist of lowercase English letters.

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Solution 💡

To solve this problem, we can use a hash map to count the occurrences of each character in magazine. Then, we check if each character in ransomNote can be satisfied using the character counts.

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Java Solution

import java.util.HashMap;
import java.util.Map;

class Solution1 {
    public boolean canConstruct(String ransomNote, String magazine) {
        Map<Character, Integer> charCountMap = new HashMap<>();

        // Count characters in magazine
        for (char c : magazine.toCharArray()) {
            if (charCountMap.containsKey(c)) {
                charCountMap.put(c, charCountMap.get(c) + 1);
            } else {
                charCountMap.put(c, 1);
            }
        }

        // Check if ransomNote can be formed
        for (char c : ransomNote.toCharArray()) {
            if (charCountMap.containsKey(c) && charCountMap.get(c) > 0) {
                charCountMap.put(c, charCountMap.get(c) - 1);
            } else {
                return false;
            }
        }

        return true;
    }
}

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Explanation of the Solution

1. Count Characters in Magazine:

   • Use a HashMap to store the frequency of each character in magazine.

2. Validate Ransom Note:

   • Iterate through ransomNote and check if the required character is available in the HashMap with a count greater than 0.

   • If available, decrement the count. If not, return false.

3. Result:

   • If all characters in ransomNote are satisfied, return true.

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Time Complexity ⏳

• O(m + n):

  • m is the length of ransomNote.

  • n is the length of magazine.

  • Both strings are traversed once.

Space Complexity 💾

• O(k):

  • k is the number of unique characters in magazine.

  • Space used by the HashMap.

You can find the full solution here.