Check whether two Strings are anagram of each other

Given two strings. The task is to check whether the given strings are anagrams of each other or not. An anagram of a string is another string that contains the same characters, only the order of characters can be different. For example, “abcd” and “dabc” are an anagram of each other.

Examples:

Input: str1 = “listen”  str2 = “silent”
Output: “Anagram”
Explanation: All characters of “listen” and “silent” are the same.

Input: str1 = “gram”  str2 = “arm”
Output: “Not Anagram”

We strongly recommend that you click here and practice it, before moving on to the solution.

Check whether two strings are anagrams of each other using sorting:

Sort the two given strings and compare, if they are equal then they are anagram of each other.

Follow the steps to implement the idea:-

• Sort both strings.
• Compare the sorted strings:
• If they are equal return True.
• Else return False.

Below is the implementation of the above idea:

C++

 `// C++ program to check whether two strings are anagrams` `// of each other` `#include ` `using` `namespace` `std;`   `/* function to check whether two strings are anagram of` `each other */` `bool` `areAnagram(string str1, string str2)` `{` `    ``// Get lengths of both strings` `    ``int` `n1 = str1.length();` `    ``int` `n2 = str2.length();`   `    ``// If length of both strings is not same, then they` `    ``// cannot be anagram` `    ``if` `(n1 != n2)` `        ``return` `false``;`   `    ``// Sort both the strings` `    ``sort(str1.begin(), str1.end());` `    ``sort(str2.begin(), str2.end());`   `    ``// Compare sorted strings` `    ``for` `(``int` `i = 0; i < n1; i++)` `        ``if` `(str1[i] != str2[i])` `            ``return` `false``;`   `    ``return` `true``;` `}`   `// Driver code` `int` `main()` `{` `    ``string str1 = ``"gram"``;` `    ``string str2 = ``"arm"``;`   `    ``// Function Call` `    ``if` `(areAnagram(str1, str2))` `        ``cout << ``"The two strings are anagram of each other"``;` `    ``else` `        ``cout << ``"The two strings are not anagram of each "` `                ``"other"``;`   `    ``return` `0;` `}`

Java

 `// JAVA program to check whether two strings` `// are anagrams of each other` `import` `java.io.*;` `import` `java.util.Arrays;` `import` `java.util.Collections;`   `class` `GFG {`   `    ``/* function to check whether two strings are` `    ``anagram of each other */` `    ``static` `boolean` `areAnagram(``char``[] str1, ``char``[] str2)` `    ``{` `        ``// Get lengths of both strings` `        ``int` `n1 = str1.length;` `        ``int` `n2 = str2.length;`   `        ``// If length of both strings is not same,` `        ``// then they cannot be anagram` `        ``if` `(n1 != n2)` `            ``return` `false``;`   `        ``// Sort both strings` `        ``Arrays.sort(str1);` `        ``Arrays.sort(str2);`   `        ``// Compare sorted strings` `        ``for` `(``int` `i = ``0``; i < n1; i++)` `            ``if` `(str1[i] != str2[i])` `                ``return` `false``;`   `        ``return` `true``;` `    ``}`   `    ``/* Driver Code*/` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``char` `str1[] = { ``'g'``, ``'r'``, ``'a'``, ``'m'` `};` `        ``char` `str2[] = { ``'a'``, ``'r'``, ``'m'` `};` `      `  `        ``// Function Call` `        ``if` `(areAnagram(str1, str2))` `            ``System.out.println(``"The two strings are"` `                               ``+ ``" anagram of each other"``);` `        ``else` `            ``System.out.println(``"The two strings are not"` `                               ``+ ``" anagram of each other"``);` `    ``}` `}`   `// This code is contributed by Nikita Tiwari.`

Python

 `class` `Solution:`   `    ``# Function is to check whether two strings are anagram of each other or not.` `    ``def` `isAnagram(``self``, a, b):`   `        ``if` `sorted``(a) ``=``=` `sorted``(b):` `            ``return` `True` `        ``else``:` `            ``return` `False`   `# {` `#  Driver Code Starts`   `if` `__name__ ``=``=` `'__main__'``:` `    ``a ``=` `"gram"` `    ``b ``=` `"arm"` `    ``if``(Solution().isAnagram(a, b)):` `      ``print``(``"The two strings are anagram of each other"``)` `    ``else``:` `      ``print``(``"The two strings are not anagram of each other"``)` `# } Driver Code Ends`

C#

 `// C# program to check whether two` `// strings are anagrams of each other` `using` `System;` `using` `System.Collections;` `class` `GFG {`   `    ``/* function to check whether two` `strings are anagram of each other */` `    ``public` `static` `bool` `areAnagram(ArrayList str1,` `                                  ``ArrayList str2)` `    ``{` `        ``// Get lengths of both strings` `        ``int` `n1 = str1.Count;` `        ``int` `n2 = str2.Count;`   `        ``// If length of both strings is not` `        ``// same, then they cannot be anagram` `        ``if` `(n1 != n2) {` `            ``return` `false``;` `        ``}`   `        ``// Sort both strings` `        ``str1.Sort();` `        ``str2.Sort();`   `        ``// Compare sorted strings` `        ``for` `(``int` `i = 0; i < n1; i++) {` `            ``if` `(str1[i] != str2[i]) {` `                ``return` `false``;` `            ``}` `        ``}`   `        ``return` `true``;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `Main(``string``[] args)` `    ``{` `        ``// create and initialize new ArrayList` `        ``ArrayList str1 = ``new` `ArrayList();` `        ``str1.Add(``'g'``);` `        ``str1.Add(``'r'``);` `        ``str1.Add(``'a'``);` `        ``str1.Add(``'m'``);` `        ``// create and initialize new ArrayList` `        ``ArrayList str2 = ``new` `ArrayList();` `        ``str2.Add(``'a'``);` `        ``str2.Add(``'r'``);` `        ``str2.Add(``'m'``);`   `        ``// Function call ` `        ``if` `(areAnagram(str1, str2)) {` `            ``Console.WriteLine(``"The two strings are"` `                              ``+ ``" anagram of each other"``);` `        ``}` `        ``else` `{` `            ``Console.WriteLine(``"The two strings are not"` `                              ``+ ``" anagram of each other"``);` `        ``}` `    ``}` `}`   `// This code is contributed by Shrikant13`

Javascript

 ``

Output

```The two strings are not anagram of each other
```

Time Complexity: O(N * logN), For sorting.
Auxiliary Space: O(1) as it is using constant extra space

Check whether two strings are anagrams of each other by counting frequency:

The idea is based in an assumption that the set of possible characters in both strings is small. that the characters are stored using 8 bit and there can be 256 possible characters.

So count the frequency of the characters and if the frequency of characters in both strings are the same, they are anagram of each other.

Follow the below steps to Implement the idea:

• Create count arrays of size 256 for both strings. Initialize all values in count arrays as 0.
• Iterate through every character of both strings and increment the count of characters in the corresponding count arrays.
• Compare count arrays. If both count arrays are the same, then return true else return false.

Below is the implementation of the above approach:

C++

 `// C++ program to check if two strings` `// are anagrams of each other` `#include ` `using` `namespace` `std;` `#define NO_OF_CHARS 256`   `/* function to check whether two strings are anagram of` `each other */` `bool` `areAnagram(``char``* str1, ``char``* str2)` `{` `    ``// Create 2 count arrays and initialize all values as 0` `    ``int` `count1[NO_OF_CHARS] = { 0 };` `    ``int` `count2[NO_OF_CHARS] = { 0 };` `    ``int` `i;`   `    ``// For each character in input strings, increment count` `    ``// in the corresponding count array` `    ``for` `(i = 0; str1[i] && str2[i]; i++) {` `        ``count1[str1[i]]++;` `        ``count2[str2[i]]++;` `    ``}`   `    ``// If both strings are of different length. Removing` `    ``// this condition will make the program fail for strings` `    ``// like "aaca" and "aca"` `    ``if` `(str1[i] || str2[i])` `        ``return` `false``;`   `    ``// Compare count arrays` `    ``for` `(i = 0; i < NO_OF_CHARS; i++)` `        ``if` `(count1[i] != count2[i])` `            ``return` `false``;`   `    ``return` `true``;` `}`   `/* Driver code*/` `int` `main()` `{` `    ``char` `str1[] = ``"gram"``;` `    ``char` `str2[] = ``"arm"``;`   `    ``// Function Call` `    ``if` `(areAnagram(str1, str2))` `        ``cout << ``"The two strings are anagram of each other"``;` `    ``else` `        ``cout << ``"The two strings are not anagram of each "` `                ``"other"``;`   `    ``return` `0;` `}`   `// This is code is contributed by rathbhupendra`

C

 `// C program to check if two strings` `// are anagrams of each other` `#include ` `#define NO_OF_CHARS 256`   `/* function to check whether two strings are anagram of` `   ``each other */` `bool` `areAnagram(``char``* str1, ``char``* str2)` `{` `    ``// Create 2 count arrays and initialize all values as 0` `    ``int` `count1[NO_OF_CHARS] = { 0 };` `    ``int` `count2[NO_OF_CHARS] = { 0 };` `    ``int` `i;`   `    ``// For each character in input strings, increment count` `    ``// in the corresponding count array` `    ``for` `(i = 0; str1[i] && str2[i]; i++) {` `        ``count1[str1[i]]++;` `        ``count2[str2[i]]++;` `    ``}`   `    ``// If both strings are of different length. Removing` `    ``// this condition will make the program fail for strings` `    ``// like "aaca" and "aca"` `    ``if` `(str1[i] || str2[i])` `        ``return` `false``;`   `    ``// Compare count arrays` `    ``for` `(i = 0; i < NO_OF_CHARS; i++)` `        ``if` `(count1[i] != count2[i])` `            ``return` `false``;`   `    ``return` `true``;` `}`   `/* Driver code*/` `int` `main()` `{` `    ``char` `str1[] = ``"gram"``;` `    ``char` `str2[] = ``"arm"``;`   `    ``// Function Call` `    ``if` `(areAnagram(str1, str2))` `        ``printf``(``"The two strings are anagram of each other"``);` `    ``else` `        ``printf``(``"The two strings are not anagram of each "` `               ``"other"``);`   `    ``return` `0;` `}`

Java

 `// JAVA program to check if two strings` `// are anagrams of each other` `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG {`   `    ``static` `int` `NO_OF_CHARS = ``256``;`   `    ``/* function to check whether two strings` `    ``are anagram of each other */` `    ``static` `boolean` `areAnagram(``char` `str1[], ``char` `str2[])` `    ``{` `        ``// Create 2 count arrays and initialize` `        ``// all values as 0` `        ``int` `count1[] = ``new` `int``[NO_OF_CHARS];` `        ``Arrays.fill(count1, ``0``);` `        ``int` `count2[] = ``new` `int``[NO_OF_CHARS];` `        ``Arrays.fill(count2, ``0``);` `        ``int` `i;`   `        ``// For each character in input strings,` `        ``// increment count in the corresponding` `        ``// count array` `        ``for` `(i = ``0``; i < str1.length && i < str2.length;` `             ``i++) {` `            ``count1[str1[i]]++;` `            ``count2[str2[i]]++;` `        ``}`   `        ``// If both strings are of different length.` `        ``// Removing this condition will make the program` `        ``// fail for strings like "aaca" and "aca"` `        ``if` `(str1.length != str2.length)` `            ``return` `false``;`   `        ``// Compare count arrays` `        ``for` `(i = ``0``; i < NO_OF_CHARS; i++)` `            ``if` `(count1[i] != count2[i])` `                ``return` `false``;`   `        ``return` `true``;` `    ``}`   `    ``/* Driver code*/` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``char` `str1[] = (``"gram"``).toCharArray();` `        ``char` `str2[] = (``"arm"``).toCharArray();`   `        ``// Function call` `        ``if` `(areAnagram(str1, str2))` `            ``System.out.println(``"The two strings are"` `                               ``+ ``" anagram of each other"``);` `        ``else` `            ``System.out.println(``"The two strings are not"` `                               ``+ ``" anagram of each other"``);` `    ``}` `}`   `// This code is contributed by Nikita Tiwari.`

Python

 `# Python program to check if two strings are anagrams of` `# each other` `NO_OF_CHARS ``=` `256`   `# Function to check whether two strings are anagram of` `# each other`     `def` `areAnagram(str1, str2):`   `    ``# Create two count arrays and initialize all values as 0` `    ``count1 ``=` `[``0``] ``*` `NO_OF_CHARS` `    ``count2 ``=` `[``0``] ``*` `NO_OF_CHARS`   `    ``# For each character in input strings, increment count` `    ``# in the corresponding count array` `    ``for` `i ``in` `str1:` `        ``count1[``ord``(i)] ``+``=` `1`   `    ``for` `i ``in` `str2:` `        ``count2[``ord``(i)] ``+``=` `1`   `    ``# If both strings are of different length. Removing this` `    ``# condition will make the program fail for strings like` `    ``# "aaca" and "aca"` `    ``if` `len``(str1) !``=` `len``(str2):` `        ``return` `0`   `    ``# Compare count arrays` `    ``for` `i ``in` `xrange``(NO_OF_CHARS):` `        ``if` `count1[i] !``=` `count2[i]:` `            ``return` `0`   `    ``return` `1`     `# Driver code` `str1 ``=` `"gram"` `str2 ``=` `"arm"`   `# Function call` `if` `areAnagram(str1, str2):` `    ``print` `"The two strings are anagram of each other"` `else``:` `    ``print` `"The two strings are not anagram of each other"`   `# This code is contributed by Bhavya Jain`

C#

 `// C# program to check if two strings` `// are anagrams of each other`   `using` `System;`   `public` `class` `GFG {`   `    ``static` `int` `NO_OF_CHARS = 256;`   `    ``/* function to check whether two strings` `    ``are anagram of each other */` `    ``static` `bool` `areAnagram(``char``[] str1, ``char``[] str2)` `    ``{` `        ``// Create 2 count arrays and initialize` `        ``// all values as 0` `        ``int``[] count1 = ``new` `int``[NO_OF_CHARS];` `        ``int``[] count2 = ``new` `int``[NO_OF_CHARS];` `        ``int` `i;`   `        ``// For each character in input strings,` `        ``// increment count in the corresponding` `        ``// count array` `        ``for` `(i = 0; i < str1.Length && i < str2.Length;` `             ``i++) {` `            ``count1[str1[i]]++;` `            ``count2[str2[i]]++;` `        ``}`   `        ``// If both strings are of different length.` `        ``// Removing this condition will make the program` `        ``// fail for strings like "aaca" and "aca"` `        ``if` `(str1.Length != str2.Length)` `            ``return` `false``;`   `        ``// Compare count arrays` `        ``for` `(i = 0; i < NO_OF_CHARS; i++)` `            ``if` `(count1[i] != count2[i])` `                ``return` `false``;`   `        ``return` `true``;` `    ``}`   `    ``/* Driver code*/` `    ``public` `static` `void` `Main()` `    ``{` `        ``char``[] str1 = (``"gram"``).ToCharArray();` `        ``char``[] str2 = (``"arm"``).ToCharArray();`   `        ``// Function Call` `        ``if` `(areAnagram(str1, str2))` `            ``Console.WriteLine(``"The two strings are"` `                              ``+ ``" anagram of each other"``);` `        ``else` `            ``Console.WriteLine(``"The two strings are not"` `                              ``+ ``" anagram of each other"``);` `    ``}` `}`   `// This code contributed by 29AjayKumar`

Javascript

 ``

Output

```The two strings are not anagram of each other
```

Time Complexity: O(n)
Auxiliary Space: O(256) i.e. O(1) for constant space.

Check whether two strings are anagrams of each other by storing all characters in HashMap:

The idea is a modification of the above approach where instead of creating an array of 256 characters HashMap is used to store characters and count of characters in HashMap. Idea is to put all characters of one String in HashMap and reduce them as they are encountered while looping over other the string.

Below is the implementation of the above approach:

C++

 `// C++ implementation of the approach` `// Function that returns true if a and b` `// are anagarams of each other` `#include ` `using` `namespace` `std;`   `bool` `isAnagram(string a, string b)` `{`   `    ``// Check if length of both strings is same or not` `    ``if` `(a.length() != b.length()) {` `        ``return` `false``;` `    ``}` `    ``// Create a HashMap containing Character as Key and` `    ``// Integer as Value. We will be storing character as` `    ``// Key and count of character as Value.` `    ``unordered_map<``char``, ``int``> Map;` `    ``// Loop over all character of String a and put in` `    ``// HashMap.` `    ``for` `(``int` `i = 0; i < a.length(); i++) {` `        ``Map[a[i]]++;` `    ``}` `    ``// Now loop over String b` `    ``for` `(``int` `i = 0; i < b.length(); i++) {` `        ``// Check if current character already exists in` `        ``// HashMap/map` `        ``if` `(Map.find(b[i]) != Map.end()) {` `            ``// If contains reduce count of that` `            ``// character by 1 to indicate that current` `            ``// character has been already counted as` `            ``// idea here is to check if in last count of` `            ``// all characters in last is zero which` `            ``// means all characters in String a are` `            ``// present in String b.` `            ``Map[b[i]] -= 1;` `        ``}` `        ``else` `{` `            ``return` `false``;` `        ``}` `    ``}`   `    ``// Loop over all keys and check if all keys are 0.` `    ``// If so it means it is anagram.` `    ``for` `(``auto` `items : Map) {` `        ``if` `(items.second != 0) {` `            ``return` `false``;` `        ``}` `    ``}` `    ``// Returning True as all keys are zero` `    ``return` `true``;` `}`   `// Driver code` `int` `main()` `{` `    ``string str1 = ``"gram"``;` `    ``string str2 = ``"arm"``;`   `    ``// Function call` `    ``if` `(isAnagram(str1, str2))` `        ``cout << ``"The two strings are anagram of each other"` `             ``<< endl;` `    ``else` `        ``cout << ``"The two strings are not anagram of each "` `                ``"other"` `             ``<< endl;` `}`   `// This code is contributed by shinjanpatra`

Java

 `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG {` `    ``public` `static` `boolean` `isAnagram(String a, String b)` `    ``{` `        ``// Check if length of both strings is same or not` `        ``if` `(a.length() != b.length()) {` `            ``return` `false``;` `        ``}` `        ``// Create a HashMap containing Character as Key and` `        ``// Integer as Value. We will be storing character as` `        ``// Key and count of character as Value.` `        ``HashMap map = ``new` `HashMap<>();` `        ``// Loop over all character of String a and put in` `        ``// HashMap.` `        ``for` `(``int` `i = ``0``; i < a.length(); i++) {` `            ``// Check if HashMap already contain current` `            ``// character or not` `            ``if` `(map.containsKey(a.charAt(i))) {` `                ``// If contains increase count by 1 for that` `                ``// character` `                ``map.put(a.charAt(i),` `                        ``map.get(a.charAt(i)) + ``1``);` `            ``}` `            ``else` `{` `                ``// else put that character in map and set` `                ``// count to 1 as character is encountered` `                ``// first time` `                ``map.put(a.charAt(i), ``1``);` `            ``}` `        ``}` `        ``// Now loop over String b` `        ``for` `(``int` `i = ``0``; i < b.length(); i++) {` `            ``// Check if current character already exists in` `            ``// HashMap/map` `            ``if` `(map.containsKey(b.charAt(i))) {` `                ``// If contains reduce count of that` `                ``// character by 1 to indicate that current` `                ``// character has been already counted as` `                ``// idea here is to check if in last count of` `                ``// all characters in last is zero which` `                ``// means all characters in String a are` `                ``// present in String b.` `                ``map.put(b.charAt(i),` `                        ``map.get(b.charAt(i)) - ``1``);` `            ``}` `            ``else` `{` `                ``return` `false``;` `            ``}` `        ``}` `        ``// Extract all keys of HashMap/map` `        ``Set keys = map.keySet();` `        ``// Loop over all keys and check if all keys are 0.` `        ``// If so it means it is anagram.` `        ``for` `(Character key : keys) {` `            ``if` `(map.get(key) != ``0``) {` `                ``return` `false``;` `            ``}` `        ``}` `        ``// Returning True as all keys are zero` `        ``return` `true``;` `    ``}` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``String str1 = ``"gram"``;` `        ``String str2 = ``"arm"``;`   `        ``// Function call` `        ``if` `(isAnagram(str1, str2))` `            ``System.out.print(``"The two strings are "` `                             ``+ ``"anagram of each other"``);` `        ``else` `            ``System.out.print(``"The two strings are "` `                             ``+ ``"not anagram of each other"``);` `    ``}` `}`

Python3

 `# Python3 implementation of the approach` `# Function that returns True if a and b` `# are anagarams of each other`     `def` `isAnagram(a, b):`   `    ``# Check if length of both strings is same or not` `    ``if` `(``len``(a) !``=` `len``(b)):` `        ``return` `False`   `    ``# Create a HashMap containing Character as Key and` `    ``# Integer as Value. We will be storing character as` `    ``# Key and count of character as Value.` `    ``map` `=` `{}` `    ``# Loop over all character of String a and put in` `    ``# HashMap.` `    ``for` `i ``in` `range``(``len``(a)):` `        ``# Check if HashMap already contain current` `        ``# character or not` `        ``if` `(a[i] ``in` `map``):` `            ``# If contains increase count by 1 for that` `            ``# character` `            ``map``[a[i]] ``+``=` `1`   `        ``else``:` `            ``# else set that character in map and set` `            ``# count to 1 as character is encountered` `            ``# first time` `            ``map``[a[i]] ``=` `1`   `    ``# Now loop over String b` `    ``for` `i ``in` `range``(``len``(b)):` `        ``# Check if current character already exists in` `        ``# HashMap/map` `        ``if` `(b[i] ``in` `map``):` `            ``# If contains reduce count of that` `            ``# character by 1 to indicate that current` `            ``# character has been already counted as` `            ``# idea here is to check if in last count of` `            ``# all characters in last is zero which` `            ``# means all characters in String a are` `            ``# present in String b.` `            ``map``[b[i]] ``-``=` `1` `        ``else``:` `            ``return` `False`   `    ``# Extract all keys of HashMap/map` `    ``keys ``=` `map``.keys()` `    ``# Loop over all keys and check if all keys are 0.` `    ``# If so it means it is anagram.` `    ``for` `key ``in` `keys:` `        ``if` `(``map``[key] !``=` `0``):` `            ``return` `False`   `    ``# Returning True as all keys are zero` `    ``return` `True`     `# Driver code` `str1 ``=` `"gram"` `str2 ``=` `"arm"`   `# Function call` `if` `(isAnagram(str1, str2)):` `    ``print``(``"The two strings are anagram of each other"``)` `else``:` `    ``print``(``"The two strings are not anagram of each other"``)`     `# This code is contributed by shinjanpatra`

C#

 `using` `System;` `using` `System.Collections.Generic;`   `public` `class` `GFG {` `    ``public` `static` `bool` `isAnagram(String a, String b)` `    ``{`   `        ``// Check if length of both strings is same or not` `        ``if` `(a.Length != b.Length) {` `            ``return` `false``;` `        ``}`   `        ``// Create a Dictionary containing char as Key and` `        ``// int as Value. We will be storing character as` `        ``// Key and count of character as Value.` `        ``Dictionary<``char``, ``int``> map` `            ``= ``new` `Dictionary<``char``, ``int``>();`   `        ``// Loop over all character of String a and put in` `        ``// Dictionary.` `        ``for` `(``int` `i = 0; i < a.Length; i++) {`   `            ``// Check if Dictionary already contain current` `            ``// character or not` `            ``if` `(map.ContainsKey(a[i])) {`   `                ``// If contains increase count by 1 for that` `                ``// character` `                ``map[a[i]] = map[a[i]] + 1;` `            ``}` `            ``else` `{`   `                ``// else put that character in map and set` `                ``// count to 1 as character is encountered` `                ``// first time` `                ``map.Add(a[i], 1);` `            ``}` `        ``}`   `        ``// Now loop over String b` `        ``for` `(``int` `i = 0; i < b.Length; i++) {`   `            ``// Check if current character already exists in` `            ``// Dictionary/map` `            ``if` `(map.ContainsKey(b[i])) {`   `                ``// If contains reduce count of that` `                ``// character by 1 to indicate that current` `                ``// character has been already counted as` `                ``// idea here is to check if in last count of` `                ``// all characters in last is zero which` `                ``// means all characters in String a are` `                ``// present in String b.` `                ``map[b[i]] = map[b[i]] - 1;` `            ``}` `            ``else` `{` `                ``return` `false``;` `            ``}` `        ``}`   `        ``// Extract all keys of Dictionary/map` `        ``var` `keys = map.Keys;`   `        ``// Loop over all keys and check if all keys are 0.` `        ``// If so it means it is anagram.` `        ``foreach``(``char` `key ``in` `keys)` `        ``{` `            ``if` `(map[key] != 0) {` `                ``return` `false``;` `            ``}` `        ``}`   `        ``// Returning True as all keys are zero` `        ``return` `true``;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main(String[] args)` `    ``{` `        ``String str1 = ``"gram"``;` `        ``String str2 = ``"arm"``;`   `        ``// Function call` `        ``if` `(isAnagram(str1, str2))` `            ``Console.Write(``"The two strings are "` `                          ``+ ``"anagram of each other"``);` `        ``else` `            ``Console.Write(``"The two strings are "` `                          ``+ ``"not anagram of each other"``);` `    ``}` `}`   `// This code is contributed by shikhasingrajput`

Javascript

 ``

Output

```The two strings are not anagram of each other

```

Time Complexity: O(N)
Auxiliary Space: O(N), Hashmap uses an extra space