/*If you have any query do comment in the comment section else like the solution*/
#include<iostream>
using namespace std;
int LCS(string s1,string s2,int n1,int n2)
{
int dp[n1+1][n2+1],i,j;
for(i=0;i<n1+1;i++) dp[i][0]=0;
for(j=0;j<n2+1;j++) dp[0][j]=0;
for(i=1;i<n1+1;i++)
{
for(j=1;j<n2+1;j++)
{
if(s1[i-1]==s2[j-1])
dp[i][j]=1+dp[i-1][j-1];
else
dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
}
}
return dp[n1][n2];
}
int main()
{
string s1,s2;
int n1,n2;
cin>>s1>>s2;
n1 = s1.length();
n2 = s2.length();
cout<<LCS(s1,s2,n1,n2)<<endl;
}
Output:
5. Find the longest common sequence between "ababba" and "aabaa" 15 points (Dynamic Programming)
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2. (40 pts) Let A, B, and C be three strings each n characters long. We want to compute the longest subsequence that is common to all three strings. (a) Let us first consider the following greedy algorithm for this problem. Find the longest common subsequence between any pair of strings, namely, LCS(A, B) LCS(B, C), LCS(A, C). Then, find the longest common subsequence between this LCS and the 3rd string. That is, supposing that the longest common pair wise...
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