Ransom Note

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.

class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        if (ransomNote.length() > magazine.length()) {
            return false;
        }
        
        int[] magazineChars = new int[26];
        for (char c : magazine.toCharArray()) {
            magazineChars[c - 'a']++;
        }
        
        for (char c : ransomNote.toCharArray()) {
            if (magazineChars[c - 'a'] == 0) {
                return false;
            }
            magazineChars[c - 'a']--;
        }
        
        return true;
    }
}

Brute Force Solution

A naive approach would involve iterating through the ransomNote and for each character, searching for it in magazine. If found, remove it from magazine and continue. If not found, return false. This approach has a time complexity of O(m*n), where n is the length of ransomNote and m is the length of magazine.

Optimal Solution

The optimal solution involves using a frequency array (or HashMap) to count the occurrences of each character in the magazine. Then, iterate through the ransomNote and decrement the count for each character. If the count becomes negative, it means the magazine does not have enough characters to construct the ransomNote, so return false. This approach has a time complexity of O(m + n), where n is the length of ransomNote and m is the length of magazine.

Big(O) Run-time Analysis

The time complexity of the provided solution is O(m + n), where n is the length of ransomNote and m is the length of magazine.

1. Frequency Counting in magazine: The code iterates through the magazine string once to count the frequency of each character. This loop takes O(m) time, where m is the length of the magazine.
2. Checking ransomNote: The code iterates through the ransomNote string once to check if each character is available in the required quantity in the magazine. This loop takes O(n) time, where n is the length of the ransomNote.

Since these two loops are performed sequentially, the overall time complexity is O(m + n).

Big(O) Space Usage Analysis

The space complexity of the provided solution is O(1), which means it uses a constant amount of extra space regardless of the input size.

1. Frequency Array: The code uses an integer array magazineChars of size 26 to store the frequency of each lowercase English letter in the magazine. Since the size of the array is fixed (26), it occupies a constant amount of space.

Therefore, the overall space complexity is O(1).

Edge Cases

1. ransomNote longer than magazine: If the ransomNote is longer than the magazine, it's impossible to construct the ransomNote, so return false immediately.
2. Empty strings: If either ransomNote or magazine is an empty string, handle appropriately. If ransomNote is empty, return true. If magazine is empty but ransomNote is not, return false.
3. Non-lowercase characters: The problem statement specifies lowercase English letters. If there are other characters, the character counting array would need to be adjusted (e.g., using a HashMap instead of an array).
4. Very large strings: For extremely large strings, memory usage could become a concern. The current solution, using a fixed-size array, avoids memory issues for the constraints specified. If handling Unicode characters, consider using a HashMap.