[Hackerrank] The longest Common subsequence

This is the classic LCS problem. Since it requires you-to-print one longest common subsequence, just use the O (m*n)-space version here.

My accepted code is as follows.

1#include <iostream>2#include <vector>3#include <algorithm>4 5 using namespacestd;6 7vector<int> LCS (vector<int>& A, vector<int>&b) {8     intn = a.size (), M =b.size ();9vector<vector<int> > DP (n +1, vector<int> (M +1,0));Ten      for(inti =1; I <= N; i++) { One          for(intj =1; J <= M; J + +) { A             if(A[i-1] = = B[j-1]) Dp[i][j] = dp[i-1][j-1] +1; -             ElseDP[I][J] = max (Dp[i][j-1], Dp[i-1][j]); -         } the     } -vector<int>Res; -      for(inti = n, j = m; I >=1&& J >=1;) { -         if(A[i-1] = = B[j-1]) { +Res.push_back (A[i-1]); -i--; +j--; A         } at         Else if(Dp[i-1][J] >= dp[i][j-1]) i--; -         Elsej--; -     } - Reverse (Res.begin (), Res.end ()); -     returnRes; - } in  - intMain () { to     /*Enter your code here. Read input from STDIN. Print output to STDOUT*/    +     intN, M; -      while(SCANF ("%d%d", &n, &m)! =EOF) { thevector<int>a (n); *vector<int>B (m); $          for(inti =0; I < n; i++)Panax Notoginsengscanf"%d", &a[i]); -          for(inti =0; I < m; i++) thescanf"%d", &b[i]); +vector<int> res =LCS (A, b); A          for(inti =0; I < (int) Res.size (); i++) theprintf"%d", Res[i]); +printf"\ n"); -     } $     return 0; $}

Well, try this problem here and get Accepted:)

[Hackerrank] The longest Common subsequence