LeetCode 383: Ransom Note Solution

Problem Explanation

Explanation

To solve this problem, we can iterate over the characters in the magazine string and count the frequency of each character. Then, we iterate over the characters in the ransomNote string and decrement the corresponding character count from the magazine map. If at any point the count goes negative or the character is not present in the map, we return false. If we are able to successfully construct the ransomNote using the characters from magazine, we return true.

Time Complexity

The time complexity of this approach is O(m + n), where m is the length of the magazine string and n is the length of the ransomNote string.

Space Complexity

The space complexity of this approach is O(1) since the frequency map will at most contain 26 lowercase English letters.

Solution Code

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

Frequently Asked Questions

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What is the time complexity of LeetCode 383 (Ransom Note)?

The time complexity for LeetCode 383 (Ransom Note) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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