From ee51ac57054d1afc8e60dc5dadbffa76903a0f01 Mon Sep 17 00:00:00 2001
From: T-Akshay-Raj <161284052+T-Akshay-Raj@users.noreply.github.com>
Date: Wed, 27 Mar 2024 13:28:25 +0530
Subject: [PATCH] Update AnagramDemo.java

Included Complexity Analysis and Optimized Solution
---
 src/Anagram/AnagramDemo.java | 37 ++++++++++++++++++++++++++++++++++++
 1 file changed, 37 insertions(+)

diff --git a/src/Anagram/AnagramDemo.java b/src/Anagram/AnagramDemo.java
index bedd07e..797a942 100644
--- a/src/Anagram/AnagramDemo.java
+++ b/src/Anagram/AnagramDemo.java
@@ -18,6 +18,7 @@ public static void main(String[] args) {
                     (str1 + " and " + str2 + " are not anagrams.");
         }
     }
+    //Approach 1: Using char array and sorting : Time Complexity: O(n log n) and Space Complexity: O(n)
     public static boolean checkAnagram(String str1, String str2) {
         if(str1.length() != str2.length()) return false;
         // Remove spaces and convert to lowercase for comparison
@@ -32,5 +33,41 @@ public static boolean checkAnagram(String str1, String str2) {
                 str2.chars().sorted().toArray()
         );
     }
+    /*
+    Complexity Analysis: 
+      The time complexity of this function is O(n log n), where n is the length of the input strings. This is because the function involves sorting the characters in the strings, which has a time complexity of O(n log n) for each string.
+      The space complexity of this function is O(n), where n is the length of the input strings. This is because the function creates character arrays of the input strings, which have a space complexity of O(n) each.
+    */
 
+    //Approach 2: Using Arrays : Time Complexity: O(n) and Space Complexity: O(1)
+    private static boolean isAnagram2(String str1, String str2) {
+    //replace all white spaces and convert to lower case
+    str1 = str1.replace(" ", "").toLowerCase();
+    str2 = str2.replace(" ", "").toLowerCase();
+    //check if length of both strings are equal
+    if (str1.length() != str2.length()) {
+      return false;
+    }
+    //create a frequency array for each string
+    int[] frequencyArray1 = new int[26];
+    int[] frequencyArray2 = new int[26];
+    for (char c : str1.toCharArray()) {
+      frequencyArray1[c - 'a']++;
+    }
+    for (char c : str2.toCharArray()) {
+      frequencyArray2[c - 'a']++;
+    }
+    //compare the frequency arrays
+    for (int i = 0; i < 26; i++) {
+      if (frequencyArray1[i] != frequencyArray2[i]) {
+        return false;
+      }
+    }
+    return true;
+  }
+    /*
+    Complexity Analysis:
+        The time complexity of this function is O(n), where n is the length of the input strings. This is because the function iterates through the characters of the strings to build the frequency arrays.
+        The space complexity of this function is O(1). This is because the function uses two integer arrays of fixed size (26) to store the frequency maps of the characters in the input strings, regardless of the length of the input strings.
+    */
 }