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Printing Longest Common Subsequence

Here, we are going to learn to find the longest common subsequence using Dynamic programming.
Submitted by Souvik Saha, on April 04, 2020

Problem statement:

Given two strings, you have to find and print the longest common subsequence between them.

    Input:
    T Test case
    T no of input string will be given to you.

    E.g.
    3
    
    abcd abxy
    sghk rfgh
    svd vjhfd
    
    Constrain 
    1≤ length (string1) ≤100
    1≤ length (string2) ≤100
    
    Output:
    Print the length of the longest common subsequence formed from these two strings.

Example

    T=3

    Input:
    abcd abxy
    
    Output:
    2 (xy)
    
    Input:
    sghk rfgh
    
    Output:
    2 (gh)
    
    Input:
    svd vjhfd
    
    Output:
    2 (vd)

Explanation with example:

Let there are two strings str1 and str2.

    str1 = "abcd" 
    str2 = "abxy"

Using a dynamic programming algorithm to find the longest common subsequence between two given string is very efficient and fast as compared to the recursion approach.

Let f(a,b) = count the number of common subsequence from the two string starting from 0 to position a and starting from 0 to position b.

Considering the two facts:

  1. If the character of string1 at index a and character of string1 at index b are the same then we have to count how many characters are the same between the two strings before these indexes? Therefore,
        f(a,b)=f(a-1,b-1)+1
    
  2. If the character of string1 at index a and character of string1 at index b are not same then we have to calculate the maximum common character count between (0 to a-1 of string1 and 0 to b of string2) and (0 to a of string1 and 0 to b-1 of string2).
        f(a,b) = max(f(a-1,b),f(a,b-1))
    

For the two strings:

    str1 = "abcd"
    str2 = "abxy"
Printing Longest Common Subsequence

C++ Implementation:

#include <bits/stdc++.h>
using namespace std;

int count(string str1, string str2)
{
    int len1 = str1.length();
    int len2 = str2.length();
    int arr[len1 + 1][len2 + 1];
    memset(arr, 0, sizeof(arr));
    for (int i = 1; i <= len1; i++) {
        for (int j = 1; j <= len2; j++) {
            if (str1[i - 1] == str2[j - 1]) {
                arr[i][j] = arr[i - 1][j - 1] + 1;
            }
            else {
                arr[i][j] = max(arr[i - 1][j], arr[i][j - 1]);
            }
        }
    }
    return (arr[len1][len2]);
}

int main()
{
    int t;
    cout << "Test Case: ";
    cin >> t;
    while (t--) {
        string str1, str2;
        cout << "Enter the two strings: ";
        cin >> str1 >> str2;
        cout << "length of the Shorest SubString: " << count(str1, str2) << endl;
    }
    return 0;
}

Output

Test Case: 3
Enter the two strings: abcd abxy
length of the Shorest SubString: 2
Enter the two strings: sghk rfgh
length of the Shorest SubString: 2
Enter the two strings: svd svjhfd
length of the Shorest SubString: 3





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