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.
- Both have same number of characters.
- 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.
- Stores occurrence of all characters of both strings in separate count arrays.
- 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.
- 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<bits/stdc++.h> 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
<?php // PHP program to check // if two strings are // k anagram or not. $MAX_CHAR = 26; // Function to check that // string is k-anagram or not function arekAnagrams( $str1 , $str2 , $k ) { global $MAX_CHAR ; // If both strings are not of // equal length then return false $n = strlen ( $str1 ); if ( strlen ( $str2 ) != $n ) return false; $count1 = (0); $count2 = (0); // Store the occurrence of all // characters in a hash_array $count = 0; // Count number of characters that // are different in both strings for ( $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 $str1 = "anagram" ; $str2 = "grammar" ; $k = 2; if (arekAnagrams( $str1 , $str2 , $k )) echo "Yes" ; else echo "No" ; // This code is contributed by m_kit ?> |
Javascript
<script> // Javascript program to check if two // strings are k anagram or not. let MAX_CHAR = 26; // Function to check that string is // k-anagram or not function arekAnagrams(str1, str2, k) { // If both strings are not of equal // length then return false let n = str1.length; if (str2.length != n) return false ; let count1 = new Array(MAX_CHAR); let count2 = new Array(MAX_CHAR); let count = 0; // Store the occurrence of all characters // in a hash_array for (let i = 0; i < n; i++) count1[str1[i].charCodeAt(0) - 'a' .charCodeAt(0)]++; for (let i = 0; i < n; i++) count2[str2[i].charCodeAt(0) - 'a' .charCodeAt(0)]++; // Count number of characters that are // different in both strings for (let 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 let str1 = "anagram" ; let str2 = "grammar" ; let k = 2; if (arekAnagrams(str1, str2, k)) document.write( "Yes" ); else document.write( "No" ); // This code is contributed by rag2127 </script> |
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<bits/stdc++.h> 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
<?php // Optimized PHP 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 function areKAnagrams( $str1 , $str2 , $k ) { global $MAX_CHAR ; // If both strings are // not of equal length // then return false $n = strlen ( $str1 ); if ( strlen ( $str2 ) != $n ) return false; $hash_str1 = array (0); // Store the occurrence of // all characters in a hash_array for ( $i = 0; $i < $n ; $i ++) $hash_str1 [ $str1 [ $i ] - 'a' ]++; // Store the occurrence of all // characters in a hash_array $count = 0; for ( $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 $str1 = "fodr" ; $str2 = "gork" ; $k = 2; if (areKAnagrams( $str1 , $str2 , $k ) == true) echo "Yes" ; else echo "No" ; // This code is contributed by ajit ?> |
Javascript
<script> // Optimized Javascript program // to check if two strings // are k anagram or not. let MAX_CHAR = 26; // Function to check if str1 and // str2 are k-anagram // or not function areKAnagrams(str1, str2, k) { // If both strings are not of equal // [i] then return false let n = str1.length; if (str2.length != n) return false ; let hash_str1 = Array(MAX_CHAR); hash_str1.fill(0); // Store the occurrence of all characters // in a hash_array for (let i = 0; i < n ; i++) hash_str1[str1[i].charCodeAt()- 'a' .charCodeAt()]++; // Store the occurrence of all characters // in a hash_array let count = 0; for (let i = 0; i < n ; i++) { if (hash_str1[str2[i].charCodeAt()- 'a' .charCodeAt()] > 0) hash_str1[str2[i].charCodeAt()- 'a' .charCodeAt()]--; else count++; if (count > k) return false ; } // Return true if count is less than or // equal to k return true ; } let str1 = "fodr" ; let str2 = "gork" ; let k = 2; if (areKAnagrams(str1, str2, k) == true ) document.write( "Yes" ); else document.write( "No" ); </script> |
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 <bits/stdc++.h> 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<Character> 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
<script> // Javascript program for the above approach // Function to check k // anagram of two strings function kAnagrams(str1, str2, k) { let flag = 0; let list = []; // 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 let arr1 = str1.split( '' ); // Converting str2 to Character Array arr2 let arr2 = str2.split( '' ); // Sort arr1 in increasing order arr1.sort(); // Sort arr2 in increasing order arr2.sort(); // Iterate till str1.length() for (let 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(arr2[i]); } } // Condition to check if // strings for K-anagram or not if (list.length <= k) flag = 1; if (flag == 1) return true ; else return false ; } // Driver code let str1 = "fodr" ; let str2 = "gork" ; let k = 2; // Function Call kAnagrams(str1, str2, k); if (kAnagrams(str1, str2, k) == true ) document.write( "Yes" ); else document.write( "No" ); // This code is contributed by suresh07 </script> |
Yes
Time complexity : O(n)
Auxiliary Space : O(1)
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