Create 0242-valid-anagram.md

Author: mahek0620

dsa-solutions/lc-solutions/0200-0299/0242-valid-anagram.md

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+---
+id: valid-anagram
+title: Valid Anagram
+sidebar_label: 0242 Valid Anagram
+tags:
+  - String
+  - Hash Table
+  - Sorting
+  - LeetCode
+  - Python
+  - Java
+  - C++
+description: "This is a solution to the Valid Anagram problem on LeetCode."
+---
+
+## Problem Description
+
+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, typically using all the original letters exactly once.
+
+### Examples
+
+**Example 1:**
+
+```
+Input: s = "anagram", t = "nagaram"
+Output: true
+```
+
+**Example 2:**
+
+```
+Input: s = "rat", t = "car"
+Output: false
+```
+
+### Constraints
+
+- $1 <= s.length, t.length <= 5 * 10^4$
+- s and t consist of lowercase English letters.
+
+## Solution for Valid Anagram Problem
+
+### Approach
+
+To determine if two strings `s` and `t` are anagrams, we can use the following approach:
+
+1. **Character Counting**: Count the occurrences of each character in both strings using arrays of size 26 (since there are 26 lowercase letters).
+2. **Comparison**: Compare the two arrays. If they are identical, then `t` is an anagram of `s`.
+
+### Code in Different Languages
+
+<Tabs>
+  <TabItem value="Python" label="Python" default>
+  <SolutionAuthor name="@mahek0620"/>
+   ```python
+
+    class Solution:
+        def isAnagram(self, s: str, t: str) -> bool:
+            if len(s) != len(t):
+                return False
+
+            # Initialize arrays to count occurrences of each character
+            count_s = [0] * 26
+            count_t = [0] * 26
+
+            # Count occurrences of each character in s
+            for char in s:
+                count_s[ord(char) - ord('a')] += 1
+
+            # Count occurrences of each character in t
+            for char in t:
+                count_t[ord(char) - ord('a')] += 1
+
+            # Compare the two arrays
+            return count_s == count_t
+    ```
+  </TabItem>
+  <TabItem value="Java" label="Java">
+  <SolutionAuthor name="@mahek0620"/>
+   ```java
+    
+    class Solution {
+        public boolean isAnagram(String s, String t) {
+            if (s.length() != t.length()) {
+                return false;
+            }
+
+            // Initialize arrays to count occurrences of each character
+            int[] count_s = new int[26];
+            int[] count_t = new int[26];
+
+            // Count occurrences of each character in s
+            for (char ch : s.toCharArray()) {
+                count_s[ch - 'a']++;
+            }
+
+            // Count occurrences of each character in t
+            for (char ch : t.toCharArray()) {
+                count_t[ch - 'a']++;
+            }
+
+            // Compare the two arrays
+            return Arrays.equals(count_s, count_t);
+        }
+    }
+    ```
+  </TabItem>
+  <TabItem value="C++" label="C++">
+  <SolutionAuthor name="@mahek0620"/>
+   ```cpp
+   
+    #include <vector>
+    using namespace std;
+
+    class Solution {
+    public:
+        bool isAnagram(string s, string t) {
+            if (s.length() != t.length()) {
+                return false;
+            }
+
+            // Initialize arrays to count occurrences of each character
+            vector<int> count_s(26, 0);
+            vector<int> count_t(26, 0);
+
+            // Count occurrences of each character in s
+            for (char ch : s) {
+                count_s[ch - 'a']++;
+            }
+
+            // Count occurrences of each character in t
+            for (char ch : t) {
+                count_t[ch - 'a']++;
+            }
+
+            // Compare the two arrays
+            return count_s == count_t;
+        }
+    };
+    ```
+
+  </TabItem>
+</Tabs>
+
+### Complexity Analysis
+
+- **Time complexity**: O(n), where n is the length of the strings `s` and `t`. We iterate through both strings once to count characters and compare counts, which are both O(n).
+- **Space complexity**: O(1) for the fixed-size arrays in Python and O(26) = O(1) for arrays in Java and C++, since there are only 26 lowercase letters.
+
+## References
+
+- **LeetCode Problem:** [Valid Anagram](https://leetcode.com/problems/valid-anagram/)
+- **Solution Link:** [Valid Anagram Solution on LeetCode](https://leetcode.com/problems/valid-anagram/solution/)
+- **Author's GeeksforGeeks Profile:** [Mahek Patel](https://leetcode.com/u/mahekrpatel611/)