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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:

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(a1,b1)+1

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 a1 of string1 and 0 to b of string2) and (0 to a of string1 and 0 to b1 of string2).
f(a,b) = max(f(a1,b),f(a,b1))
For the two strings:
str1 = "abcd"
str2 = "abxy"
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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