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. Initilaize len = 0. Now, for each lowercase letter finds its lowest frequency from the two hash tables and accumulate it to len.

C++

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// 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 longCommomAnagramSubseq(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 = "
         << longCommomAnagramSubseq(str1, str2, n1, n2);
    return 0;     
}

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Java

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// 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 longCommomAnagramSubseq(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 = "
                longCommomAnagramSubseq(str1, str2, 
                                          n1, n2));
    }
}
  
// This code is contributed by 
// Manish Shaw(manishshaw1)

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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 longCommomAnagramSubseq(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 = “,
longCommomAnagramSubseq(str1, str2,
n1, n2))

# This code is contributed by ita_c

C#

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// 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 longCommomAnagramSubseq(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 = "
                longCommomAnagramSubseq(str1, str2, 
                                        n1, n2));
    }
}
  
// This code is contributed by 
// Manish Shaw(manishshaw1)

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PHP

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<?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 longCommomAnagramSubseq($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 = " .
      longCommomAnagramSubseq($str1, $str2,
                              $n1, $n2));
                                
// This code is contributed by 
// Manish Shaw(manishshaw1)
?>

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Output:

Length = 3

Time Complexity: O(n).
Auxiliary Space: O(n).



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Improved By : manishshaw1, Ita_c