Valid Anagram
Detailed guide and Python implementation for the 'Valid Anagram' problem.
1. Concept Overview
The 'Valid Anagram' problem is a key challenge in the Arrays & Hashing section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Valid Anagram.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Valid Anagram carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
Given two strings s and t, return True if t is an anagram of s, and False otherwise.
An anagram is a word or phrase formed by rearranging the letters of a different word or phrase, using all the original letters exactly once.
Write a function isAnagram(s: str, t: str) -> bool.
- •1 <= len(s), len(t) <= 5 * 10^4
- •s and t consist of lowercase English letters
Examples
s = "anagram", t = "nagaram"
True
Both strings contain the same characters with the same frequencies: a(3), n(1), g(1), r(1), m(1).
s = "rat", t = "car"
False
'rat' contains 't' but 'car' does not. They have different character compositions.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
Open the problem in PyRun's browser-based Python editor. Your code runs fully offline — no server required.
Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def is_anagram_opt(s, t):
if len(s) != len(t):
return False
count = {}
for char in s:
count[char] = count.get(char, 0) + 1
for char in t:
if char not in count or count[char] == 0:
return False
count[char] -= 1
return TrueBrute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def is_anagram_brute(s, t):
if len(s) != len(t):
return False
return sorted(s) == sorted(t)Algorithm Pattern Checklist
When dealing with Arrays & Hashing data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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