longest_common_subsequence_tabulation.cpp

/*
Given two sequences, find the length of longest subsequence present in both of them. Both the strings are of uppercase.

Input:
First line of the input contains no of test cases  T,the T test cases follow.
Each test case consist of 2 space separated integers A and B denoting the size of string str1 and str2 respectively
The next two lines contains the 2 string str1 and str2 .

Output:
For each test case print the length of longest  common subsequence of the two strings .

Constraints:
1<=T<=200
1<=size(str1),size(str2)<=100

Example:
Input:
2
6 6
ABCDGH
AEDFHR
3 2
ABC
AC

Output:
3
2

Explanation
LCS for input Sequences “ABCDGH” and “AEDFHR” is “ADH” of length 3.

LCS of "ABC" and "AC" is "AC" of length 2
*/




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

int LCS(string s1, string s2, int n, int m){
    int dp[n+1][m+1];

    for(int i=0;i<n+1;i++) dp[i][0]=0;
    for(int j=0;j<m+1;j++) dp[0][j]=0;

    for(int i=1;i<n+1;i++){
        for(int j=1;j<m+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][j-1], dp[i-1][j]);
        }
    }
    return dp[n][m];
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    int t;
    cin>>t;
    while(t--){
        int n,m;
        cin>>n>>m;
        string s1,s2;
        cin>>s1>>s2;
        cout<<LCS(s1,s2,n,m)<<endl;
    }

return 0;
}