shortest common supersequence

// shortest common supersequence
// [1 ,2 ,3, 4]
// subsequnce (order matters but maybe not continuous)
// substring or subarray .. (need to be contigous)
// subset same as subsequence ..
int solve(string s1 , string s2){
    int sa = s1.length();
    int sb = s2.length();
    vector<vector<int>> dp (sa+1 ,vector<int>(sb+1,0));
    // a represents the ath element .. b represents the knapsack present

    for(int a=1;a<=sa;a++){
        for(int b=1;b<=sb;b++){
            if(s1[a-1] == s2[b-1]){
            dp[a][b] = 1 + dp[a-1][b-1]; //here we can choose ith element again
            }else{
            dp[a][b] = max(dp[a-1][b] , dp[a][b-1]); // cant choose the ath element becuase its weight is more
            }
        }
    }

    // to print the scs
    int i = sa;
    int j = sb;
    string res = "";
    while(i>0 && j>0){
        if(s1[i-1] == s2[j-1] ){
            res+=s1[i-1];
            i--;
            j--;
        }else{
         if(dp[i-1][j] > dp[i][j-1]){
            res += s1[i-1];
             i--;
         }else{
            res+=s2[j-1];
             j--;
         }


        }
    }

    while(i>0){
        res+=s1[i-1];
        i--;
    }
    while(j>0){
        res+=s2[j-1];
        j--;
    }

    reverse(res.begin(),res.end()); /////// reversing the string is very important
    cout << res <<"\n";




    int ans =0;// gives to length of lcs
    int lcs = dp[sa][sb];
    ans = sa+sb-lcs;
    return ans;
}