# Check if two strings are k-anagrams or not

• Difficulty Level : Easy
• Last Updated : 06 Jul, 2022

Given two strings of lowercase alphabets and a value k, the task is to find if two strings are K-anagrams of each other or not.
Two strings are called k-anagrams if following two conditions are true.

1. Both have same number of characters.
2. Two strings can become anagram by changing at most k characters in a string.

Examples :

```Input:  str1 = "anagram" , str2 = "grammar" , k = 3
Output:  Yes
Explanation: We can update maximum 3 values and
it can be done in changing only 'r' to 'n'
and 'm' to 'a' in str2.

Input:  str1 = "geeks", str2 = "eggkf", k = 1
Output:  No
Explanation: We can update or modify only 1
value but there is a need of modifying 2 characters.
i.e. g and f in str 2.```

Method 1:

Below is a solution to check if two strings are k-anagrams of each other or not.

1. Stores occurrence of all characters of both strings in separate count arrays.
2. Count number of different characters in both strings (in this if a string has 4 a and second has 3 ‘a’ then it will be also counted.
3. If count of different characters is less than or equal to k, then return true else false.

Implementation:

## C++

 `// C++ program to check if two strings are k anagram` `// or not.` `#include` `using` `namespace` `std;` `const` `int` `MAX_CHAR = 26;`   `// Function to check that string is k-anagram or not` `bool` `arekAnagrams(string str1, string str2, ``int` `k)` `{` `    ``// If both strings are not of equal` `    ``// length then return false` `    ``int` `n = str1.length();` `    ``if` `(str2.length() != n)` `        ``return` `false``;`   `    ``int` `count1[MAX_CHAR] = {0};` `    ``int` `count2[MAX_CHAR] = {0};`   `    ``// Store the occurrence of all characters` `    ``// in a hash_array` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``count1[str1[i]-``'a'``]++;` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``count2[str2[i]-``'a'``]++;` `     `  `    ``int` `count = 0;`   `    ``// Count number of characters that are` `    ``// different in both strings` `    ``for` `(``int` `i = 0; i < MAX_CHAR; i++)` `        ``if` `(count1[i] > count2[i])` `            ``count = count + ``abs``(count1[i]-count2[i]);`   `    ``// Return true if count is less than or` `    ``// equal to k` `    ``return` `(count <= k);` `}`   `// Driver code` `int` `main()` `{` `    ``string str1 = ``"anagram"``;` `    ``string str2 = ``"grammar"``;` `    ``int` `k = 2;` `    ``if` `(arekAnagrams(str1, str2, k))` `        ``cout << ``"Yes"``;` `    ``else` `        ``cout<< ``"No"``;` `    ``return` `0;` `}`

## Java

 `// Java program to check if two strings are k anagram` `// or not.` `public` `class` `GFG {` `     `  `    ``static` `final` `int` `MAX_CHAR = ``26``;`   `    ``// Function to check that string is k-anagram or not` `    ``static` `boolean` `arekAnagrams(String str1, String str2, ` `                                                 ``int` `k)` `    ``{` `        ``// If both strings are not of equal` `        ``// length then return false` `        ``int` `n = str1.length();` `        ``if` `(str2.length() != n)` `            ``return` `false``;`   `        ``int``[] count1 = ``new` `int``[MAX_CHAR];` `        ``int``[] count2 = ``new` `int``[MAX_CHAR];` `        ``int` `count = ``0``;` `       `  `        ``// Store the occurrence of all characters` `        ``// in a hash_array` `        ``for` `(``int` `i = ``0``; i < n; i++)` `            ``count1[str1.charAt(i) - ``'a'``]++;` `        ``for` `(``int` `i = ``0``; i < n; i++)` `            ``count2[str2.charAt(i) - ``'a'``]++;`   `        ``// Count number of characters that are` `        ``// different in both strings` `        ``for` `(``int` `i = ``0``; i < MAX_CHAR; i++)` `            ``if` `(count1[i] > count2[i])` `                ``count = count + Math.abs(count1[i] - ` `                                          ``count2[i]);`   `        ``// Return true if count is less than or` `        ``// equal to k` `        ``return` `(count <= k);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``String str1 = ``"anagram"``;` `        ``String str2 = ``"grammar"``;` `        ``int` `k = ``2``;` `        ``if` `(arekAnagrams(str1, str2, k))` `            ``System.out.println(``"Yes"``);` `        ``else` `            ``System.out.println(``"No"``);` `    ``}` `}` `// This code is contributed by Sumit Ghosh`

## Python3

 `# Python3 program to check if two ` `# strings are k anagram or not.` `MAX_CHAR ``=` `26`   `# Function to check that is ` `# k-anagram or not ` `def` `arekAnagrams(str1, str2, k) :`   `    ``# If both strings are not of equal ` `    ``# length then return false ` `    ``n ``=` `len``(str1)` `    ``if` `(``len``(str2)!``=` `n) :` `        ``return` `False`   `    ``count1 ``=` `[``0``] ``*` `MAX_CHAR ` `    ``count2 ``=` `[``0``] ``*` `MAX_CHAR`   `    ``# Store the occurrence of all ` `    ``# characters in a hash_array ` `    ``for` `i ``in` `range``(n): ` `        ``count1[``ord``(str1[i]) ``-` `               ``ord``(``'a'``)] ``+``=` `1` `    ``for` `i ``in` `range``(n): ` `        ``count2[``ord``(str2[i]) ``-` `               ``ord``(``'a'``)] ``+``=` `1` `        `  `    ``count ``=` `0`   `    ``# Count number of characters that` `    ``# are different in both strings ` `    ``for` `i ``in` `range``(MAX_CHAR):` `        ``if` `(count1[i] > count2[i]) :` `            ``count ``=` `count ``+` `abs``(count1[i] ``-` `                                ``count2[i]) `   `    ``# Return true if count is less` `    ``# than or equal to k ` `    ``return` `(count <``=` `k) `   `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``:` `    ``str1 ``=` `"anagram"` `    ``str2 ``=` `"grammar"` `    ``k ``=` `2` `    ``if` `(arekAnagrams(str1, str2, k)): ` `        ``print``(``"Yes"``) ` `    ``else``:` `        ``print``(``"No"``)`   `# This code is contributed` `# by SHUBHAMSINGH10`

## C#

 `// C# program to check if two ` `// strings are k anagram or not.` `using` `System;` `class` `GFG {` `    `  `    ``static` `int` `MAX_CHAR = 26;`   `    ``// Function to check that ` `    ``// string is k-anagram or not` `    ``static` `bool` `arekAnagrams(``string` `str1, ` `                             ``string` `str2, ` `                                      ``int` `k)` `    ``{` `        `  `        ``// If both strings are not of equal` `        ``// length then return false` `        ``int` `n = str1.Length;` `        ``if` `(str2.Length != n)` `            ``return` `false``;`   `        ``int``[] count1 = ``new` `int``[MAX_CHAR];` `        ``int``[] count2 = ``new` `int``[MAX_CHAR];` `        ``int` `count = 0;` `        `  `        ``// Store the occurrence` `        ``// of all characters` `        ``// in a hash_array` `        ``for` `(``int` `i = 0; i < n; i++)` `            ``count1[str1[i] - ``'a'``]++;` `        ``for` `(``int` `i = 0; i < n; i++)` `            ``count2[str2[i] - ``'a'``]++;`   `        ``// Count number of characters that are` `        ``// different in both strings` `        ``for` `(``int` `i = 0; i < MAX_CHAR; i++)` `            ``if` `(count1[i] > count2[i])` `                ``count = count + Math.Abs(count1[i] - ` `                                         ``count2[i]);`   `        ``// Return true if count is` `        ``// less than or equal to k` `        ``return` `(count <= k);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``string` `str1 = ``"anagram"``;` `        ``string` `str2 = ``"grammar"``;` `        ``int` `k = 2;` `        ``if` `(arekAnagrams(str1, str2, k))` `            ``Console.Write(``"Yes"``);` `        ``else` `            ``Console.Write(``"No"``);` `    ``}` `}` `// This code is contributed by nitin mittal.`

## PHP

 ` ``\$count2``[``\$i``])` `            ``\$count` `= ``\$count` `+ ``abs``(``\$count1``[``\$i``] - ` `                                  ``\$count2``[``\$i``]);`   `    ``// Return true if count is ` `    ``// less than or equal to k` `    ``return` `(``\$count` `<= ``\$k``);` `}`   `// Driver Code` `\$str1` `= ``"anagram"``;` `\$str2` `= ``"grammar"``;` `\$k` `= 2;` `if` `(arekAnagrams(``\$str1``, ``\$str2``, ``\$k``))` `    ``echo` `"Yes"``;` `else` `    ``echo` `"No"``;`   `// This code is contributed by m_kit` `?>`

## Javascript

 ``

Output

`Yes`

Time complexity: O(n)
Auxiliary Space: O(1)

Method 2: We can optimize above solution. Here we use only one count array to store counts of characters in str1. We traverse str2 and decrement occurrence of every character in count array that is present in str2. If we find a character that is not there in str1, we increment count of different characters. If count of different character become more than k, we return false.

Implementation:

## C++

 `// Optimized C++ program to check if two strings` `// are k anagram or not.` `#include` `using` `namespace` `std;`   `const` `int` `MAX_CHAR = 26;`   `// Function to check if str1 and str2 are k-anagram` `// or not` `bool` `areKAnagrams(string str1, string str2, ``int` `k)` `{` `    ``// If both strings are not of equal` `    ``// length then return false` `    ``int` `n = str1.length();` `    ``if` `(str2.length() != n)` `        ``return` `false``;`   `    ``int` `hash_str1[MAX_CHAR] = {0};`   `    ``// Store the occurrence of all characters` `    ``// in a hash_array` `    ``for` `(``int` `i = 0; i < n ; i++)` `        ``hash_str1[str1[i]-``'a'``]++;`   `    ``// Store the occurrence of all characters` `    ``// in a hash_array` `    ``int` `count = 0;` `    ``for` `(``int` `i = 0; i < n ; i++)` `    ``{` `        ``if` `(hash_str1[str2[i]-``'a'``] > 0)` `            ``hash_str1[str2[i]-``'a'``]--;` `        ``else` `            ``count++;`   `        ``if` `(count > k)` `            ``return` `false``;` `    ``}`   `    ``// Return true if count is less than or` `    ``// equal to k` `    ``return` `true``;` `}`   `// Driver code` `int` `main()` `{` `    ``string str1 = ``"fodr"``;` `    ``string str2 = ``"gork"``;` `    ``int` `k = 2;` `    ``if` `(areKAnagrams(str1, str2, k) == ``true``)` `        ``cout << ``"Yes"``;` `    ``else` `        ``cout << ``"No"``;` `    ``return` `0;` `}`

## Java

 `// Optimized Java program to check if two strings` `// are k anagram or not.` `public` `class` `GFG {` `    `  `    ``static` `final` `int` `MAX_CHAR = ``26``;` `     `  `    ``// Function to check if str1 and str2 are k-anagram` `    ``// or not` `    ``static` `boolean` `areKAnagrams(String str1, String str2, ` `                                                  ``int` `k)` `    ``{` `        ``// If both strings are not of equal` `        ``// length then return false` `        ``int` `n = str1.length();` `        ``if` `(str2.length() != n)` `            ``return` `false``;` `     `  `        ``int``[] hash_str1 = ``new` `int``[MAX_CHAR];` `     `  `        ``// Store the occurrence of all characters` `        ``// in a hash_array` `        ``for` `(``int` `i = ``0``; i < n ; i++)` `            ``hash_str1[str1.charAt(i)-``'a'``]++;` `     `  `        ``// Store the occurrence of all characters` `        ``// in a hash_array` `        ``int` `count = ``0``;` `        ``for` `(``int` `i = ``0``; i < n ; i++)` `        ``{` `            ``if` `(hash_str1[str2.charAt(i)-``'a'``] > ``0``)` `                ``hash_str1[str2.charAt(i)-``'a'``]--;` `            ``else` `                ``count++;` `     `  `            ``if` `(count > k)` `                ``return` `false``;` `        ``}` `     `  `        ``// Return true if count is less than or` `        ``// equal to k` `        ``return` `true``;` `    ``}` `     `  `    ``// Driver code` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``String str1 = ``"fodr"``;` `        ``String str2 = ``"gork"``;` `        ``int` `k = ``2``;` `        ``if` `(areKAnagrams(str1, str2, k) == ``true``)` `            ``System.out.println(``"Yes"``);` `        ``else` `            ``System.out.println(``"No"``);` `    ``}` `}` `// This code is contributed by Sumit Ghosh`

## Python3

 `# Optimized Python3 program ` `# to check if two strings` `# are k anagram or not.` `MAX_CHAR ``=` `26``;`   `# Function to check if str1 ` `# and str2 are k-anagram or not` `def` `areKAnagrams(str1, str2, k):` `    ``# If both strings are ` `    ``# not of equal length ` `    ``# then return false`   `    ``n ``=` `len``(str1);` `    ``if` `(``len``(str2) !``=` `n):` `        ``return` `False``;`   `    ``hash_str1 ``=` `[``0``]``*``(MAX_CHAR);`   `    ``# Store the occurrence of ` `    ``# all characters in a hash_array` `    ``for` `i ``in` `range``(n):` `        ``hash_str1[``ord``(str1[i]) ``-` `ord``(``'a'``)]``+``=``1``;`   `    ``# Store the occurrence of all ` `    ``# characters in a hash_array` `    ``count ``=` `0``;` `    ``for` `i ``in` `range``(n):` `        ``if` `(hash_str1[``ord``(str2[i]) ``-` `ord``(``'a'``)] > ``0``):` `            ``hash_str1[``ord``(str2[i]) ``-` `ord``(``'a'``)]``-``=``1``;` `        ``else``:` `            ``count``+``=``1``;`   `        ``if` `(count > k):` `            ``return` `False``;`   `    ``# Return true if count is ` `    ``# less than or equal to k` `    ``return` `True``;`   `# Driver code` `str1 ``=` `"fodr"``;` `str2 ``=` `"gork"``;` `k ``=` `2``;` `if` `(areKAnagrams(str1, str2, k) ``=``=` `True``):` `    ``print``(``"Yes"``);` `else``:` `    ``print``(``"No"``);` `        `  `# This code is contributed by mits`

## C#

 `// Optimized C# program to check if two strings` `// are k anagram or not.` `using` `System;`   ` ``class` `GFG {` `    `  `    ``static`  `int` `MAX_CHAR = 26;` `    `  `    ``// Function to check if str1 and str2 are k-anagram` `    ``// or not` `    ``static` `bool` `areKAnagrams(String str1, String str2, ` `                                                ``int` `k)` `    ``{` `        ``// If both strings are not of equal` `        ``// [i] then return false` `        ``int` `n = str1.Length;` `        ``if` `(str2.Length != n)` `            ``return` `false``;` `    `  `        ``int``[] hash_str1 = ``new` `int``[MAX_CHAR];` `    `  `        ``// Store the occurrence of all characters` `        ``// in a hash_array` `        ``for` `(``int` `i = 0; i < n ; i++)` `            ``hash_str1[str1[i]-``'a'``]++;` `    `  `        ``// Store the occurrence of all characters` `        ``// in a hash_array` `        ``int` `count = 0;` `        ``for` `(``int` `i = 0; i < n ; i++)` `        ``{` `            ``if` `(hash_str1[str2[i]-``'a'``] > 0)` `                ``hash_str1[str2[i]-``'a'``]--;` `            ``else` `                ``count++;` `    `  `            ``if` `(count > k)` `                ``return` `false``;` `        ``}` `    `  `        ``// Return true if count is less than or` `        ``// equal to k` `        ``return` `true``;` `    ``}` `    `  `    ``// Driver code` `     ``static` `void` `Main()` `    ``{` `        ``String str1 = ``"fodr"``;` `        ``String str2 = ``"gork"``;` `        ``int` `k = 2;`   `        ``if` `(areKAnagrams(str1, str2, k) == ``true``)` `            ``Console.Write(``"Yes"``);` `        ``else` `            ``Console.Write(``"No"``);` `    ``}` `}` `// This code is contributed by Anuj_67`

## PHP

 ` 0)` `            ``\$hash_str1``[``\$str2``[``\$i``] - ``'a'``]--;` `        ``else` `            ``\$count``++;`   `        ``if` `(``\$count` `> ``\$k``)` `            ``return` `false;` `    ``}`   `    ``// Return true if count is ` `    ``// less than or equal to k` `    ``return` `true;` `}`   `// Driver code` `\$str1` `= ``"fodr"``;` `\$str2` `= ``"gork"``;` `\$k` `= 2;` `if` `(areKAnagrams(``\$str1``, ``\$str2``, ``\$k``) == true)` `    ``echo` `"Yes"``;` `else` `    ``echo` `"No"``;` `        `  `// This code is contributed by ajit` `?>`

## Javascript

 ``

Output

`Yes`

Time complexity: O(n)
Auxiliary Space: O(1)

Method 3:

• In this method the idea is to initialize an array or a list and the base case would be to check the strings and if they have different lengths then they cannot form an anagram.
• Otherwise convert the strings into character array and sort them.
• Then iterate till the length of the string and if the character are not equal then add it to the list and check if list size is less than equal to K then it is possible to form K anagram else not.

Below is the implementation of the above approach:

## C++

 `// CPP program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to check k` `// anagram of two strings` `bool` `kAnagrams(string str1, string str2, ``int` `k)` `{` `    ``int` `flag = 0;`   `    ``// First Condition: If both the` `    ``// strings have different length ,` `    ``// then they cannot form anagram` `    ``if` `(str1.length() != str2.length())` `        ``return` `false``;` `    ``int` `n = str1.length();` `    ``// Converting str1 to Character Array arr1` `    ``char` `arr1[n];`   `    ``// Converting str2 to Character Array arr2` `    ``char` `arr2[n];`   `    ``strcpy``(arr1, str1.c_str());` `    ``strcpy``(arr2, str2.c_str());` `    ``// Sort arr1 in increasing order` `    ``sort(arr1, arr1 + n);`   `    ``// Sort arr2 in increasing order` `    ``sort(arr2, arr2 + n);`   `    ``vector<``char``> list;` `  `  `    ``// Iterate till str1.length()` `    ``for` `(``int` `i = 0; i < str1.length(); i++) {`   `        ``// Condition if arr1[i] is` `        ``// not equal to arr2[i]` `        ``// then add it to list` `        ``if` `(arr1[i] != arr2[i]) {` `            ``list.push_back(arr2[i]);` `        ``}` `    ``}`   `    ``// Condition to check if` `    ``// strings for K-anagram or not` `    ``if` `(list.size() <= k)` `        ``flag = 1;`   `    ``if` `(flag == 1)` `        ``return` `true``;` `    ``else` `        ``return` `false``;` `}`   `// Driver Code` `int` `main()` `{`   `    ``string str1 = ``"anagram"``, str2 = ``"grammar"``;` `    ``int` `k = 3;`   `    ``// Function Call` `    ``kAnagrams(str1, str2, k);` `    ``if` `(kAnagrams(str1, str2, k) == ``true``)` `        ``cout << ``"Yes"``;` `    ``else` `        ``cout << ``"No"``;`   `    ``// This code is contributed by bolliranadheer` `}`

## Java

 `// Java program for the above approach` `import` `java.util.*;` `class` `GFG{` `    `  `// Function to check k ` `// anagram of two strings` `public` `static` `boolean` `kAnagrams(String str1,` `                                ``String str2, ``int` `k)` `{` `    ``int` `flag = ``0``;` `    `  `    ``List list = ``new` `ArrayList<>();` `    `  `    ``// First Condition: If both the ` `    ``// strings have different length , ` `    ``// then they cannot form anagram` `    ``if` `(str1.length() != str2.length())` `        ``System.exit(``0``);` `    `  `    ``// Converting str1 to Character Array arr1` `    ``char` `arr1[] = str1.toCharArray();` `    `  `    ``// Converting str2 to Character Array arr2` `    ``char` `arr2[] = str2.toCharArray();` `    `  `    ``// Sort arr1 in increasing order` `    ``Arrays.sort(arr1);` `    `  `    ``// Sort arr2 in increasing order` `    ``Arrays.sort(arr2);` `    `  `    ``// Iterate till str1.length()` `    ``for` `(``int` `i = ``0``; i < str1.length(); i++)` `    ``{` `        `  `        ``// Condition if arr1[i] is ` `        ``// not equal to arr2[i]` `        ``// then add it to list` `        ``if` `(arr1[i] != arr2[i]) ` `        ``{` `            ``list.add(arr2[i]);` `        ``}` `    ``}` `    `  `    ``// Condition to check if ` `    ``// strings for K-anagram or not ` `    ``if` `(list.size() <= k)` `        ``flag = ``1``;`   `    ``if` `(flag == ``1``)` `        ``return` `true``;` `    ``else` `        ``return` `false``;` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{`   `    ``String str1 = ``"fodr"``;` `    ``String str2 = ``"gork"``;` `    ``int` `k = ``2``;` `    `  `    ``// Function Call` `    ``kAnagrams(str1, str2, k);` `    ``if` `(kAnagrams(str1, str2, k) == ``true``)` `        ``System.out.println(``"Yes"``);` `    ``else` `        ``System.out.println(``"No"``);` `}` `}`

## Python3

 `# Python 3 program for the above approach` `import` `sys`   `# Function to check k` `# anagram of two strings`     `def` `kAnagrams(str1, str2, k):`   `    ``flag ``=` `0`   `    ``list1 ``=` `[]`   `    ``# First Condition: If both the` `    ``# strings have different length ,` `    ``# then they cannot form anagram` `    ``if` `(``len``(str1) !``=` `len``(str2)):` `        ``sys.exit()`   `    ``# Converting str1 to Character Array arr1` `    ``arr1 ``=` `list``(str1)`   `    ``# Converting str2 to Character Array arr2` `    ``arr2 ``=` `list``(str2)`   `    ``# Sort arr1 in increasing order` `    ``arr1.sort()`   `    ``# Sort arr2 in increasing order` `    ``arr2.sort()`   `    ``# Iterate till str1.length()` `    ``for` `i ``in` `range``(``len``(str1)):`   `        ``# Condition if arr1[i] is` `        ``# not equal to arr2[i]` `        ``# then add it to list` `        ``if` `(arr1[i] !``=` `arr2[i]):` `            ``list1.append(arr2[i])`   `    ``# Condition to check if` `    ``# strings for K-anagram or not` `    ``if` `(``len``(list1) <``=` `k):` `        ``flag ``=` `1`   `    ``if` `(flag ``=``=` `1``):` `        ``return` `True` `    ``else``:` `        ``return` `False`     `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:`   `    ``str1 ``=` `"fodr"` `    ``str2 ``=` `"gork"` `    ``k ``=` `2`   `    ``# Function Call` `    ``kAnagrams(str1, str2, k)` `    ``if` `(kAnagrams(str1, str2, k) ``=``=` `True``):` `        ``print``(``"Yes"``)` `    ``else``:` `        ``print``(``"No"``)`

## C#

 `// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;`   `public` `class` `GFG` `{`   `  ``// Function to check k ` `  ``// anagram of two strings` `  ``public` `static` `bool` `kAnagrams(``string` `str1, ``string` `str2, ``int` `k)` `  ``{` `    ``int` `flag = 0;`   `    ``List<``char``> list = ``new` `List<``char``>();`   `    ``// First Condition: If both the ` `    ``// strings have different length , ` `    ``// then they cannot form anagram` `    ``if` `(str1.Length != str2.Length)` `    ``{` `      ``return` `false``;` `    ``}`   `    ``// Converting str1 to Character Array arr1` `    ``char``[] arr1 = str1.ToCharArray();`   `    ``// Converting str2 to Character Array arr2` `    ``char``[] arr2 = str2.ToCharArray();`   `    ``// Sort arr1 in increasing order` `    ``Array.Sort(arr1);`   `    ``// Sort arr2 in increasing order` `    ``Array.Sort(arr2);`   `    ``// Iterate till str1.length()` `    ``for` `(``int` `i = 0; i < str1.Length; i++)` `    ``{`   `      ``// Condition if arr1[i] is ` `      ``// not equal to arr2[i]` `      ``// then add it to list` `      ``if` `(arr1[i] != arr2[i]) ` `      ``{` `        ``list.Add(arr2[i]);` `      ``}` `    ``}`   `    ``// Condition to check if ` `    ``// strings for K-anagram or not ` `    ``if` `(list.Count <= k)` `      ``flag = 1;`   `    ``if` `(flag == 1)` `      ``return` `true``;` `    ``else` `      ``return` `false``;` `  ``}`   `  ``// Driver Code` `  ``static` `public` `void` `Main ()` `  ``{` `    ``string` `str1 = ``"fodr"``;` `    ``string` `str2 = ``"gork"``;` `    ``int` `k = 2;`   `    ``// Function Call` `    ``kAnagrams(str1, str2, k);` `    ``if` `(kAnagrams(str1, str2, k) == ``true``)` `      ``Console.WriteLine(``"Yes"``);` `    ``else` `      ``Console.WriteLine(``"No"``);` `  ``}` `}`   `// This code is contributed by avanitrachhadiya2155`

## Javascript

 ``

Output

`Yes`

Time complexity : O(n)
Auxiliary Space : O(1)

This article is contributed by Sahil Chhabra (akku). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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