Longest Common Anagram Subsequence
Given two strings str1 and str2 of length n1 and n2 respectively. The problem is to find the length of the longest subsequence which is present in both the strings in the form of anagrams.
Note: The strings contain only lowercase letters.
Examples:
Input : str1 = "abdacp", str2 = "ckamb" Output : 3 Subsequence of str1 = abc Subsequence of str2 = cab OR Subsequence of str1 = bac Subsequence of str2 = cab These are longest common anagram subsequences. Input : str1 = "abbcfke", str2 = "fbaafbly" Output : 4
Approach: Create two hash tables say freq1 and freq2. Store frequencies of each character of str1 in freq1. Likewise, store frequencies of each character of str2 in freq2. Initialize len = 0. Now, for each lowercase letter finds its lowest frequency from the two hash tables and accumulate it to len.
Implementation:
C++
// C++ implementation to find the length of the // longest common anagram subsequence #include <bits/stdc++.h> using namespace std; #define SIZE 26 // function to find the length of the // longest common anagram subsequence int longCommonAnagramSubseq( char str1[], char str2[], int n1, int n2) { // hash tables for storing frequencies of // each character int freq1[SIZE], freq2[SIZE]; memset (freq1, 0, sizeof (freq1)); memset (freq2, 0, sizeof (freq2)); int len = 0; // calculate frequency of each character // of 'str1[]' for ( int i = 0; i < n1; i++) freq1[str1[i] - 'a' ]++; // calculate frequency of each character // of 'str2[]' for ( int i = 0; i < n2; i++) freq2[str2[i] - 'a' ]++; // for each character add its minimum frequency // out of the two strings in 'len' for ( int i = 0; i < SIZE; i++) len += min(freq1[i], freq2[i]); // required length return len; } // Driver program to test above int main() { char str1[] = "abdacp" ; char str2[] = "ckamb" ; int n1 = strlen (str1); int n2 = strlen (str2); cout << "Length = " << longCommonAnagramSubseq(str1, str2, n1, n2); return 0; } |
Java
// Java implementation to find // the length() of the longest // common anagram subsequence import java.io.*; class GFG { static int SIZE = 26 ; // function to find the // length() of the longest // common anagram subsequence static int longCommonAnagramSubseq(String str1, String str2, int n1, int n2) { // hash tables for // storing frequencies // of each character int []freq1 = new int [SIZE]; int []freq2 = new int [SIZE]; for ( int i = 0 ; i < SIZE; i++) { freq1[i] = 0 ; freq2[i] = 0 ; } int len = 0 ; // calculate frequency // of each character of // 'str1[]' for ( int i = 0 ; i < n1; i++) freq1[( int )str1.charAt(i) - ( int ) 'a' ]++; // calculate frequency // of each character // of 'str2[]' for ( int i = 0 ; i < n2; i++) freq2[( int )str2.charAt(i) - ( int ) 'a' ]++; // for each character add // its minimum frequency // out of the two Strings // in 'len' for ( int i = 0 ; i < SIZE; i++) len += Math.min(freq1[i], freq2[i]); // required length() return len; } // Driver Code public static void main(String args[]) { String str1 = "abdacp" ; String str2 = "ckamb" ; int n1 = str1.length(); int n2 = str2.length(); System.out.print( "Length = " + longCommonAnagramSubseq(str1, str2, n1, n2)); } } // This code is contributed by // Manish Shaw(manishshaw1) |
Python 3
# Python 3 implementation to find # the length of the longest common # anagram subsequence SIZE = 26 # function to find the length of the # longest common anagram subsequence def longCommonAnagramSubseq(str1, str2, n1, n2): # List for storing frequencies # of each character freq1 = [ 0 ] * SIZE freq2 = [ 0 ] * SIZE l = 0 # calculate frequency of each # character of 'str1[]' for i in range (n1): freq1[ ord (str1[i]) - ord ( 'a' )] + = 1 # calculate frequency of each # character of 'str2[]' for i in range (n2) : freq2[ ord (str2[i]) - ord ( 'a' )] + = 1 # for each character add its # minimum frequency out of # the two strings in 'len' for i in range (SIZE): l + = min (freq1[i], freq2[i]) # required length return l # Driver Code if __name__ = = "__main__" : str1 = "abdacp" str2 = "ckamb" n1 = len (str1) n2 = len (str2) print ( "Length = " , longCommonAnagramSubseq(str1, str2, n1, n2)) # This code is contributed by ita_c |
C#
// C# implementation to find // the length of the longest // common anagram subsequence using System; class GFG { static int SIZE = 26; // function to find the // length of the longest // common anagram subsequence static int longCommonAnagramSubseq( string str1, string str2, int n1, int n2) { // hash tables for // storing frequencies // of each character int []freq1 = new int [SIZE]; int []freq2 = new int [SIZE]; for ( int i = 0; i < SIZE; i++) { freq1[i] = 0; freq2[i] = 0; } int len = 0; // calculate frequency // of each character of // 'str1[]' for ( int i = 0; i < n1; i++) freq1[str1[i] - 'a' ]++; // calculate frequency // of each character // of 'str2[]' for ( int i = 0; i < n2; i++) freq2[str2[i] - 'a' ]++; // for each character add // its minimum frequency // out of the two strings // in 'len' for ( int i = 0; i < SIZE; i++) len += Math.Min(freq1[i], freq2[i]); // required length return len; } // Driver Code static void Main() { string str1 = "abdacp" ; string str2 = "ckamb" ; int n1 = str1.Length; int n2 = str2.Length; Console.Write( "Length = " + longCommonAnagramSubseq(str1, str2, n1, n2)); } } // This code is contributed by // Manish Shaw(manishshaw1) |
PHP
<?php // PHP implementation to find // the length of the longest // common anagram subsequence $SIZE = 26; // function to find the // length of the longest // common anagram subsequence function longCommonAnagramSubseq( $str1 , $str2 , $n1 , $n2 ) { global $SIZE ; // hash tables for storing // frequencies of each character $freq1 = array (); $freq2 = array (); for ( $i = 0; $i < $SIZE ; $i ++) { $freq1 [ $i ] = 0; $freq2 [ $i ] = 0; } $len = 0; // calculate frequency of // each character of 'str1' for ( $i = 0; $i < $n1 ; $i ++) $freq1 [ord( $str1 [ $i ]) - ord( 'a' )]++; // calculate frequency of // each character of 'str2' for ( $i = 0; $i < $n2 ; $i ++) $freq2 [ord( $str2 [ $i ]) - ord( 'a' )]++; // for each character add // its minimum frequency // out of the two strings // in 'len' for ( $i = 0; $i < $SIZE ; $i ++) { $len += min( $freq1 [ $i ], $freq2 [ $i ]); } // required length return $len ; } // Driver Code $str1 = "abdacp" ; $str2 = "ckamb" ; $n1 = strlen ( $str1 ); $n2 = strlen ( $str2 ); echo ( "Length = " . longCommonAnagramSubseq( $str1 , $str2 , $n1 , $n2 )); // This code is contributed by // Manish Shaw(manishshaw1) ?> |
Javascript
<script> // Javascript implementation to find // the length() of the longest // common anagram subsequence let SIZE = 26; // function to find the // length() of the longest // common anagram subsequence function longCommonAnagramSubseq(str1,str2,n1,n2) { // hash tables for // storing frequencies // of each character let freq1 = new Array(SIZE); let freq2 = new Array(SIZE); for (let i = 0; i < SIZE; i++) { freq1[i] = 0; freq2[i] = 0; } let len = 0; // calculate frequency // of each character of // 'str1[]' for (let i = 0; i < n1; i++) freq1[str1[i].charCodeAt(0) - 'a' .charCodeAt(0)]++; // calculate frequency // of each character // of 'str2[]' for (let i = 0; i < n2; i++) freq2[str2[i].charCodeAt(0) - 'a' .charCodeAt(0)]++; // for each character add // its minimum frequency // out of the two Strings // in 'len' for (let i = 0; i < SIZE; i++) len += Math.min(freq1[i], freq2[i]); // required length() return len; } // Driver Code let str1 = "abdacp" ; let str2 = "ckamb" ; let n1 = str1.length; let n2 = str2.length; document.write( "Length = " + longCommonAnagramSubseq(str1, str2, n1, n2)); // This code is contributed by rag2127 </script> |
Output
Length = 3
Time Complexity: O(n+m).
Auxiliary Space: O(1).