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Ransom Note⚓︎

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Description⚓︎

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:

  • 1 <= ransomNote.length, magazine.length <= 10^5
  • ransomNote and magazine consist of lowercase English letters.

Solution⚓︎

Hash Map⚓︎

We can check if the count of each character in the ransomNote is less than or equal to the count of that character in the magazine. If it's greater, it means there aren't enough characters in the magazine to form the ransomNote.

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class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        unordered_map<int, int> mapping;
        for (auto c : magazine) {
            //Record the number of occurrences of each character in the "magazine" through "mapping"
            mapping[c - 'a']++;
        }
        // Traverse "randomNote" and do the -- operation on the corresponding number of characters in the "mapping"
        for (auto c : ransomNote) {
            // If it is less than zero, it means that the characters appearing in "randomNote" are not in the "magazine"
            if (--mapping[c - 'a'] < 0)
                return false;
        }
        return true;
    }
};

Another way of writing:

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class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        unordered_map<char, int> hash;
        for (auto c : magazine) hash[c]++;
        for (auto c : ransomNote) {
            if (!hash[c]) return false;
            else hash[c]--;
        }
        return true;
    }
};

Time complexity: \(O(n)\)