HackerRank - "Stock Maximize"

First I thought it should be solved using DP, and I gave a standard O(n^2) solution:

#include <iostream>
#include <vector>
#include <algorithm>
#include <numeric>
using namespace std;

#define REP(i, s, n) for(int i = s; i < n; i ++)
typedef long long LL;

LL calc(vector<LL> &in)
{
size_t len = in.size();

LL ret = 0;

/*
//    dp[i][onhand]
vector<vector<LL>> dp(len, vector<LL>(len + 1, std::numeric_limits<LL>::min()));
dp[0][0] = 0; // no action
dp[0][1] = -in[0]; // buy

REP(i, 1, len)
REP(j, 0, i + 1)
{
//    Choice 1: buy
dp[i][j + 1] = std::max(dp[i][j + 1], dp[i - 1][j] - in[i]);
//    Choice 2: no action
dp[i][j] = std::max(dp[i][j], dp[i - 1][j]);
//    Choice 3: sell all
if(j > 0)
dp[i][0] = std::max(dp[i][0], in[i] * j + dp[i-1][j]);
}
ret = *std::max_element(dp[len-1].begin(), dp[len-1].end());
*/
vector<LL> pre(len + 1, std::numeric_limits<LL>::min());
pre[0] = 0; pre[1] = -in[0];
vector<LL> now(len + 1, std::numeric_limits<LL>::min());

vector<LL> *ppre = &pre, *pnow = &now;
REP(i, 1, len)
{
REP(j, 0, i + 1)
{
//    Choice 1: buy
(*pnow)[j + 1] = std::max((*pnow)[j + 1], (*ppre)[j] - in[i]);
//    Choice 2: no action
(*pnow)[j] = std::max((*pnow)[j], (*ppre)[j]);
//    Choice 3: sell all
if(j > 0)
(*pnow)[0] = std::max((*pnow)[0], in[i] * j + (*ppre)[j]);
}
// swap
std::swap(ppre, pnow);
(*pnow).assign(len + 1, std::numeric_limits<LL>::min());
}
ret = *std::max_element((*ppre).begin(), (*ppre).end());
return ret;
}

int main()
{
int t; cin >> t;
while(t--)
{
int n; cin >> n;
vector<LL> in(n);
REP(i, 0, n) cin >> in[i];
cout << calc(in) << endl;
}
return 0;
}

But all TLE.. so there are must be a O(n) solution, and there is.. what is better than a standard DP in cerntain cases? Greedy.

#include <iostream>
#include <vector>
#include <algorithm>
#include <numeric>
using namespace std;

#define REP(i, s, n) for(int i = s; i < n; i ++)
typedef long long LL;

LL calc(vector<LL> &in)
{
size_t len = in.size();

LL ret = 0;
std::reverse(in.begin(), in.end());
LL peak = -1;
REP(i, 0, len)
{
if(in[i] > peak)
{
peak = in[i];
}
else
{
ret += peak - in[i];
}
}
return ret;
}

int main()
{
int t; cin >> t;
while(t--)
{
int n; cin >> n;
vector<LL> in(n);
REP(i, 0, n) cin >> in[i];
cout << calc(in) << endl;
}
return 0;
}