# Check if two strings are k-anagrams or not

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.
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 strings has 4 a and second has 3 ‘a’ then it will be also count.
3. If count of different characters is less than or equal to k, then return true else false.

## 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 ` `?> `

Output :

```Yes
```

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.

## 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 ` `?> `

Output:

```Yes
```

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

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