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This approach involves using hash maps to count the frequency of each letter in both ransomNote and magazine. The idea is to determine if magazine has enough of each letter to build ransomNote.
magazine.ransomNote and check if the magazine frequency is sufficient.ransomNote has a higher frequency than in magazine, return false.true.Time Complexity: O(m + n), where m and n are the lengths of magazine and ransomNote, respectively.
Space Complexity: O(1), as the space is allocated for a fixed-size frequency array.
1from collections import Counter
2
3def canConstruct(ransomNote: str, magazine: str) -> bool:
4 magazine_count = Counter(magazine)
5Python's Counter from the collections module provides an elegant way to count frequencies. The solution compares the counters of ransomNote and magazine to determine constructibility.
Instead of counting character frequencies, we can sort both ransomNote and magazine and use a two-pointer technique to check if magazine contains all letters needed for ransomNote in a sorted order.
ransomNote and magazine.ransomNote are matched, return true; otherwise, return false.Time Complexity: O(n log n + m log m), dominated by sorting
Space Complexity: O(1)
This C solution sorts both strings using qsort and iterates using two indices, attempting to match each character from ransomNote with a character from magazine.