The problem asks us to compute the minimum total cost it takes for a frog to travel from stone $$$1$$$ to stone $$$N (N \le 10^5)$$$ given that the frog can only jump a distance of one or two. The cost to travel between any two stones $$$i$$$ and

$$$j$$$ is given by $$$|h_i - h_j|$$$ , where $$$h_i$$$ represents the height of stone $$$i$$$ .

#include<bits/stdc++.h>
typedef long long ll;
#define mod 1000000007 
#define loop(i,a,b) for(int i= (a);i <(b) ; i++)
using namespace std ;
int solve(ll arr[], ll s, ll e, vector<vector<ll>> &dp)
{
    if(dp[s][e] != -1)  //THIS STORES COST TO GO FROM S TO E
    return dp[s][e];
    else
    {
        if(s==e)
        return dp[s][e] = 0;
        else if(s+1 == e)
        return dp[s][e] = abs(arr[e]- arr[s]);
        else if(s+2 == e)
        return dp[s][e] = min(abs(arr[s+1]-arr[e])+ abs(arr[s]-arr[s+1]), abs(arr[s]-arr[e]));
        else if(s+2<e)
        return dp[s][e] = min(solve(arr,s+1,e,dp)+abs(arr[s]-arr[s+1]), solve(arr,s+2,e,dp)+abs(arr[s]-arr[s+2]) );
    }
}
int main() 
{
    ios_base::sync_with_stdio (false);
    cin.tie(NULL); 
    ll n;
    cin>>n;
    ll arr[n];
    loop(i,0,n)
    cin>>arr[i];
    vector<vector<ll>> dp (n+1, vector<ll>(n+1, -1));
    cout<<solve(arr,0,n-1,dp);
    return 0;
}