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