HackerRank "Jumping on the Clouds"

DP or Greedy - they are all in O(n)

In editorial, a beautiful Greedy solution is given: "To reach the last cloud in a minimum number of steps, always try make a jump from i to i + 2. If that is not possible, jump to i + 1. ". And here is my DP solution:

#include <vector>
#include <iostream>

using namespace std;

int main(){
    int n;
    cin >> n;
    vector<int> c(n);
    for(int c_i = 0;c_i < n;c_i++){
       cin >> c[c_i];
    }
    
    vector<int> dp(n, INT_MAX);
    dp[0] = 0;
    for(int i = 1; i < n; i++)
    {
        if(i > 0 && !c[i-1])
        {
            dp[i] = min(dp[i] ,dp[i - 1] + 1);
        }
        if(i>1 && !c[i - 2])
        {
            dp[i] = min(dp[i], dp[i - 2] + 1);
        }
    }
    cout << dp.back() << endl;
    return 0;
}