In this project, you will implement the dynamic programming-based solution to find t…

Can you do it in java or C++ for CAGATGTATCTG GAGACAGGAT ?

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In this project, you will implement the dynamic programming-based solution to find the longest common subsequence (LCS) of two sequences. Your inputs will be the two sequences (as Strings) and the outputs are the longest common subsequence (printed as a String) and the final matrix (printed as a two-dimensional array) depicting the length of the longest common subsequences (as shown in the slides) for all possible subsequences of the two input sequences The two input sequences to be used by each student are shown below. The LCS expected for the two sequences is also shown.

#include<bits/stdc++.h>

using namespace std;

int main()

{

string str1,str2,lcs;

cout<<“enter string1:”<<endl;

cin>>str1;

cout<<“enter string2:”<<endl;

cin>>str2;

int i,j;

int n = str1.length();//length of string1

int m = str2.length();//length of string2

int dp[n+1][m+1]; //matrix for computing dp values

memset(dp,0,sizeof dp);//initialising dp values with 0

//lcs dp computation

for(i=1;i<=n;i++)

{

for(j=1;j<=m;j++)

if(str1[i-1] == str2[j-1]) //if char matches

dp[i][j] = 1+dp[i-1][j-1];

else

dp[i][j] = max(dp[i-1][j],dp[i][j-1]);

}

cout<<“length of lcs:”<<dp[n][m]<<endl; //printing length of lcs

//getting the lcs of the two strings

lcs = “”;

i = n;

j = m;

while (i>0 and j>0)

{

if( str1[i-1] == str2[j-1])

{

lcs = str1[i-1]+lcs;

i-=1;

j-=1;

}

else

{

if(dp[i-1][j] < dp[i][j-1])

j-=1;

else

i-=1;

}

}

cout<<“longest common sequence:”<<lcs<<endl;

}

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