Understanding the Problem
We are given two strings and asked to determine if they are anagrams. That means we need to check whether the two strings contain the same characters with the same frequency, but possibly in a different order. Think of it like this: Can we rearrange the letters of the first string to form the second one?
We also need to ignore the case of letters (uppercase vs lowercase), and carefully handle cases like different lengths or empty strings.
Step-by-Step Solution with Example
Step 1: Convert both strings to lowercase
This ensures that 'A' and 'a' are treated as the same character. For example, if we are comparing "Listen" and "Silent", we first convert both to "listen" and "silent".
Step 2: Check the lengths
If the strings have different lengths, they cannot be anagrams. For example, "hello" and "helloo" are not anagrams because one has an extra character.
Step 3: Count the frequency of each character
We use a dictionary (also called a map) to count how many times each character appears in each string.
Example: Compare "listen" and "silent".
- 'l': appears once in both
- 'i': once in both
- 's': once in both
- 't': once in both
- 'e': once in both
- 'n': once in both
Since every character and its count matches, they are anagrams.
Step 4: Compare both frequency maps
If the dictionaries for both strings are equal, the strings are anagrams. If even one character count is different, they are not.
Example: "aabbcc" vs "abc" → Not anagrams, because "aabbcc" has two of each letter, while "abc" has only one.
Edge Cases
- Empty strings: Two empty strings are considered anagrams because they contain the same (no) characters.
- One empty, one non-empty: Not anagrams.
- Case sensitivity: Convert to lowercase before comparison to ensure fair evaluation.
- Completely different letters: For example, "abc" and "xyz" → Not anagrams.
Finally
This problem tests your understanding of strings, maps, and case handling. By walking through the example step-by-step and handling edge cases carefully, you ensure a solid and beginner-friendly solution.
The overall time complexity of the solution is O(n), where n is the length of the strings, because we loop through each string a limited number of times.
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