We can use straightforward dynamic programming .

Let dp[i] be the number of ways to reach index 'i' of the array .

dp[1]=1 , dp[2] = 1 , dp[3] = 1 , dp[4] = 2 will be Base cases.

If you are at index 'i' , you can reach it , either from index 'i-1' or index 'i-3' , so dp[i] = dp[i-1] + dp[i-3] .

Recurrence relation has been formed and we are done! Voial !

O(N)

Code :

#include<bits/stdc++.h>
using namespace std ;
typedef long long int ll ;
#define rep(i, l, r) for ((i) = (l); (i) <=(r); (i)++)
#define rep1(i, r, l) for ((i) = (r); (i) >=(l); (i)--)
int main()
{
ll n ;
cin>>n;
ll dp[n+3];
ll i ;
dp[1]=1;
dp[2]=1;
dp[3]=1;
dp[4]=2 ;
rep(i,5,n)
{
dp[i] = dp[i-3] + dp[i-1] ;
}
cout<<dp[n];
return 0 ;
}