HashMap is useful for counting character frequencies efficiently. Instead of sorting both strings, counting occurrences allows us to compare them in a single traversal. Always check the string lengths first because strings of different lengths can never be anagrams. This frequency-counting technique is widely used in hashing and string interview problems.
Problem stats
Before
If the lengths of both strings are different, they cannot be anagrams.
Otherwise:
1. Convert the second string into a character array.
2. For every character in the first string, search for the same character in the second array.
3. If found, mark it as used.
4. If any character cannot be found, return false.
5. If all characters are matched, return true.
This solution repeatedly searches through the string, making it inefficient.
After
If both strings have different lengths, return false immediately.
Use a HashMap to count the frequency of each character in the first string.
Then traverse the second string and decrease the count.
If any character is missing or its count becomes negative, return false.
If every frequency becomes zero after processing both strings, the strings are valid anagrams.
The optimized implementation, ready for interview preparation and quick revision.
Solution.java
1import java.util.HashMap;2import java.util.Map;3 4public class Solution {5 6 public boolean isAnagram(String s, String t) {7 8 if (s.length() != t.length()) {9 return false;10 }11 12 Map<Character, Integer> frequency = new HashMap<>();13 14 for (char ch : s.toCharArray()) {15 frequency.put(ch, frequency.getOrDefault(ch, 0) + 1);16 }17 18 for (char ch : t.toCharArray()) {19 20 if (!frequency.containsKey(ch)) {21 return false;22 }23 24 frequency.put(ch, frequency.get(ch) - 1);25 26 if (frequency.get(ch) == 0) {27 frequency.remove(ch);28 }29 }30 31 return frequency.isEmpty();32 }33 34 public static void main(String[] args) {35 36 Solution solution = new Solution();37 38 System.out.println(solution.isAnagram("anagram", "nagaram"));39 System.out.println(solution.isAnagram("rat", "car"));40 }41}Analyze the efficiency of the algorithm by understanding its time and space complexity across different execution scenarios.
Time complexity
Space complexity
By execution case
Best case
O(n)
Average case
O(n)
Worst case
O(n)
This solution achieves a O(n) time complexity while using O(n) extra memory. It is considered the optimal approach for this problem and is suitable for coding interviews as well as competitive programming.
Every coding problem teaches a pattern. Focus on the concepts, avoid common mistakes, and remember the interview-worthy takeaways instead of memorizing code.
The biggest takeaway from this problem.
HashMap is useful for counting character frequencies efficiently. Instead of sorting both strings, counting occurrences allows us to compare them in a single traversal. Always check the string lengths first because strings of different lengths can never be anagrams. This frequency-counting technique is widely used in hashing and string interview problems.
Explain why the optimized solution works before writing the final code. Interviewers care about your thinking process as much as your implementation.
Avoid jumping directly to coding. Always analyze edge cases, constraints, and the optimal approach before implementation.
Focus on understanding the algorithm's pattern instead of memorizing the code. Once the logic becomes clear, implementing the solution in any programming language becomes much easier.
Every DSA problem introduces a reusable pattern. Instead of remembering the exact solution, remember the thought process that led to it. Over time, these patterns will help you solve new problems much faster and perform better in coding interviews.
Practice consistently and move through the roadmap one problem at a time.
Practice similar problems to strengthen your understanding of the underlying algorithm and improve pattern recognition.