stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;



const int N=1e5+30;

typedef long long ll;

int n,arr[N];

int dp[N];

ll solve(int i){

    if(i>=n)return 1e15;

    if(i==n-1){

        return 0;

    }



    if(dp[i]!=-1)return dp[i];



    ll path1=abs(arr[i]-arr[i+1])+solve(i+1);

    ll path2=abs(arr[i]-arr[i+2])+solve(i+2);

    return dp[i]= min(path1,path2);

}

int main()

{

    memset(dp,-1,sizeof dp);

    scanf("%d",&n);

    for(int i=0;i<n;i++)scanf("%d",arr+i);

    cout<<solve(0)<<endl;

}




// Faster C++ solution for the same problem (measured speedup: 1.365851398192245x):
#include <iostream>

#include <string>

#include<iomanip>

#include <algorithm>

#include <map>

#include <vector>

#include<queue>

#include<math.h>

#include<cmath>

#include<sstream>

#include<set>

#include <map>

#include<string.h>

#include<sstream>

#include<stack>

#include<unordered_map>

#include <iostream>

#include <stdio.h>

#include <stdlib.h>

#include <ctype.h>

#include <cstdio>

#include <cstdlib>

#include <complex>

#define ll long long

#define  point complex<double>

#define  pi acos(-1)

#define ios ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

#define arr2dof vector  vector<vector<int>>dp(N,vector<int>(2, inf));

using namespace std;

const ll N = 2e5 + 5, mod = 1e9 + 7, inff = -1e12,inf=1e12;

const double ep = 0.0000000001;

//const double pi = 3.14159265359;



double dot(point a, point b) {

	return  (conj(a)*b).real();

}

double cross(point a, point b) {

	return  (conj(a)*b).imag();

}

double angle(point a, point b)

{

	double dt = dot(a, b);

	dt /= abs(a);

	dt /= abs(b);

	return acos(dt) * 180.0 / pi;

}



bool isrightangel(int ind) {//90 degree

	vector<point>v;

	return(angle(v[ind] - v[(ind + 1) % 3], v[ind] - v[(ind + 2) % 3]) == 90);//bshof hal eldot producty3ny fe

	//dl3en 3moden 3ashn yb2a right trinagle

}



ll fact[1000006], inv[1000006];



ll fp(ll base, ll exp)

{

	if (exp == 0)

		return 1;

	ll ans = fp(base, exp / 2);

	ans = (ans * ans) % mod;

	if (exp % 2 != 0)

		ans = (ans * (base % mod)) % mod;

	return ans;

}



void calcFacAndInv(ll n)

{

	fact[0] = inv[0] = 1;

	for (ll i = 1; i <= n; i++)

	{

		fact[i] = (i * fact[i - 1]) % mod;

		inv[i] = fp(fact[i], mod - 2);

	}

}



ll ncr(ll n, ll r)

{

	return ((fact[n] * inv[r]) % mod * inv[n - r]) % mod;

}



ll npr(ll n, ll r)

{

	return (fact[n] * inv[n - r]) % mod;

}











	int dp[N];

int main()

{

	ll n,arr[100005];

	cin >> n;

	for (int i = 0; i < n; ++i) {

		cin >> arr[i];

	}

	dp[0] = 0;//land on 1

	dp[1] = abs(arr[0]-arr[1]);//to land on 2

	for (int i = 2; i < n; ++i) {

		dp[i] = min(abs(arr[i] - arr[i - 1]) + dp[i - 1], abs(arr[i] - arr[i - 2]) + dp[i - 2]);

	}

	cout << dp[n - 1];

	return 0;

}
// Slower C++ solution:
#include <iostream>

#include <cmath>

#include <cstring>

using namespace std;



int n;

int ht[100010];

int eg[100010];//eg[i] 代表到达i点消耗的最少能量 

int main(){

	cin>>n;

	for(int i=1;i<=n; ++i){

		cin>>ht[i];

		

	}//读取完毕

	memset(eg, 0, sizeof(eg));

	eg[2] = abs(ht[2]- ht[1]);

	for(int i=3; i<=n; ++i){

		int t1 = abs(ht[i] - ht[i-1]);

		int t2 = abs(ht[i] - ht[i-2]);

		eg[i] = min(eg[i-1]+t1, eg[i-2]+t2); 

	}

	cout<<eg[n]<<endl;

		

}

// Faster C++ solution for the same problem (measured speedup: 1.3712232008661223x):
#include <iostream>

#include <string>

#include<iomanip>

#include <algorithm>

#include <map>

#include <vector>

#include<queue>

#include<math.h>

#include<cmath>

#include<sstream>

#include<set>

#include <map>

#include<string.h>

#include<sstream>

#include<stack>

#include<unordered_map>

#include <iostream>

#include <stdio.h>

#include <stdlib.h>

#include <ctype.h>

#include <cstdio>

#include <cstdlib>

#include <complex>

#define ll long long

#define  point complex<double>

#define  pi acos(-1)

#define ios ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

#define arr2dof vector  vector<vector<int>>dp(N,vector<int>(2, inf));

using namespace std;

const ll N = 2e5 + 5, mod = 1e9 + 7, inff = -1e12,inf=1e12;

const double ep = 0.0000000001;

//const double pi = 3.14159265359;



double dot(point a, point b) {

	return  (conj(a)*b).real();

}

double cross(point a, point b) {

	return  (conj(a)*b).imag();

}

double angle(point a, point b)

{

	double dt = dot(a, b);

	dt /= abs(a);

	dt /= abs(b);

	return acos(dt) * 180.0 / pi;

}



bool isrightangel(int ind) {//90 degree

	vector<point>v;

	return(angle(v[ind] - v[(ind + 1) % 3], v[ind] - v[(ind + 2) % 3]) == 90);//bshof hal eldot producty3ny fe

	//dl3en 3moden 3ashn yb2a right trinagle

}



ll fact[1000006], inv[1000006];



ll fp(ll base, ll exp)

{

	if (exp == 0)

		return 1;

	ll ans = fp(base, exp / 2);

	ans = (ans * ans) % mod;

	if (exp % 2 != 0)

		ans = (ans * (base % mod)) % mod;

	return ans;

}



void calcFacAndInv(ll n)

{

	fact[0] = inv[0] = 1;

	for (ll i = 1; i <= n; i++)

	{

		fact[i] = (i * fact[i - 1]) % mod;

		inv[i] = fp(fact[i], mod - 2);

	}

}



ll ncr(ll n, ll r)

{

	return ((fact[n] * inv[r]) % mod * inv[n - r]) % mod;

}



ll npr(ll n, ll r)

{

	return (fact[n] * inv[n - r]) % mod;

}











	int dp[N];

int main()

{

	ll n,arr[100005];

	cin >> n;

	for (int i = 0; i < n; ++i) {

		cin >> arr[i];

	}

	dp[0] = 0;//land on 1

	dp[1] = abs(arr[0]-arr[1]);//to land on 2

	for (int i = 2; i < n; ++i) {

		dp[i] = min(abs(arr[i] - arr[i - 1]) + dp[i - 1], abs(arr[i] - arr[i - 2]) + dp[i - 2]);

	}

	cout << dp[n - 1];

	return 0;

}
// Slower C++ solution:
#include "bits/stdc++.h"

    

using namespace std;

vector<long long >dp(100001,-1);

vector<long long>num;

long long  erg(int n){

	if(dp[n]!=-1){

		return dp[n];

	}

	if(n<0){

		return 2000000000000;

	}

	if(n==0){

		return 0;

	}

	long long cmp1=max(num[n],num[n-1])-min(num[n],num[n-1]),cmp2=max(num[n],num[n-2])-min(num[n],num[n-2]);

	dp[n]= min(erg(n-1)+cmp1,erg(n-2)+cmp2);

	return dp[n];

}

int main() {

	int menge,numm;

	cin>>menge;

	for(int i=0;i<menge;i++){

		cin>>numm;

		num.push_back(numm);

	}

cout<<erg(menge-1);

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.7743939604875274x):
#include <iostream>

#include <string>

#include<iomanip>

#include <algorithm>

#include <map>

#include <vector>

#include<queue>

#include<math.h>

#include<cmath>

#include<sstream>

#include<set>

#include <map>

#include<string.h>

#include<sstream>

#include<stack>

#include<unordered_map>

#include <iostream>

#include <stdio.h>

#include <stdlib.h>

#include <ctype.h>

#include <cstdio>

#include <cstdlib>

#include <complex>

#define ll long long

#define  point complex<double>

#define  pi acos(-1)

#define ios ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

#define arr2dof vector  vector<vector<int>>dp(N,vector<int>(2, inf));

using namespace std;

const ll N = 2e5 + 5, mod = 1e9 + 7, inff = -1e12,inf=1e12;

const double ep = 0.0000000001;

//const double pi = 3.14159265359;



double dot(point a, point b) {

	return  (conj(a)*b).real();

}

double cross(point a, point b) {

	return  (conj(a)*b).imag();

}

double angle(point a, point b)

{

	double dt = dot(a, b);

	dt /= abs(a);

	dt /= abs(b);

	return acos(dt) * 180.0 / pi;

}



bool isrightangel(int ind) {//90 degree

	vector<point>v;

	return(angle(v[ind] - v[(ind + 1) % 3], v[ind] - v[(ind + 2) % 3]) == 90);//bshof hal eldot producty3ny fe

	//dl3en 3moden 3ashn yb2a right trinagle

}



ll fact[1000006], inv[1000006];



ll fp(ll base, ll exp)

{

	if (exp == 0)

		return 1;

	ll ans = fp(base, exp / 2);

	ans = (ans * ans) % mod;

	if (exp % 2 != 0)

		ans = (ans * (base % mod)) % mod;

	return ans;

}



void calcFacAndInv(ll n)

{

	fact[0] = inv[0] = 1;

	for (ll i = 1; i <= n; i++)

	{

		fact[i] = (i * fact[i - 1]) % mod;

		inv[i] = fp(fact[i], mod - 2);

	}

}



ll ncr(ll n, ll r)

{

	return ((fact[n] * inv[r]) % mod * inv[n - r]) % mod;

}



ll npr(ll n, ll r)

{

	return (fact[n] * inv[n - r]) % mod;

}











	int dp[N];

int main()

{

	ll n,arr[100005];

	cin >> n;

	for (int i = 0; i < n; ++i) {

		cin >> arr[i];

	}

	dp[0] = 0;//land on 1

	dp[1] = abs(arr[0]-arr[1]);//to land on 2

	for (int i = 2; i < n; ++i) {

		dp[i] = min(abs(arr[i] - arr[i - 1]) + dp[i - 1], abs(arr[i] - arr[i - 2]) + dp[i - 2]);

	}

	cout << dp[n - 1];

	return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>



using namespace std;



#define REP(i,a,b) for(ll i = a; i < b; i++)

#define DB( x ) cout << #x << " = " << x << endl;

#define sz(s) ((int)((s).size()))

#define endl '\n'



typedef long long ll;

typedef unsigned long long ull;

typedef pair<ll, ll> pii;

typedef pair<int, pii> tri;

typedef vector <vector <ll>> matrix;

typedef pair<string, string> pss;



const ll mod = (ll)1e9 + 7;

const int MAX = (int)1e6 + 100;



const ll oo = 0x3f3f3f3f3f3f3f3f;



ll dp[MAX], a[MAX];



int main() {

    ios_base::sync_with_stdio(0);

    cin.tie(0);



    memset(dp, oo, sizeof(dp));



    int n; cin >> n;



    dp[1] = 0;



    for(int i = 1;i <= n;++i){

        cin >> a[i];

    }



    for(int i = 2;i <= n;++i){

        dp[i] = min(dp[i - 1] + abs(a[i] - a[i - 1]), dp[i]);

        if(i - 2 >= 1){

            dp[i] = min(dp[i - 2] + abs(a[i] - a[i - 2]), dp[i]);

        }

    }



    cout << dp[n] << endl;

}


// Faster C++ solution for the same problem (measured speedup: 3.600380521861225x):
#include <bits/stdc++.h>

#define ll long long

using namespace std;



ll n;

std::vector<ll> ara(100010);

ll dp[100010];



ll fnc(ll i){

  if(i+1 == n)return 0;

  if(i == n)return 1<<30;

  if(dp[i] != -1)return dp[i];

  ll res1 = abs(ara[i]-ara[i+1]) + fnc(i+1);

  ll res2 = abs(ara[i]-ara[i+2]) + fnc(i+2);

  dp[i] = min(res1,res2);

  return min(res1,res2);

}



int main(int argc, char const *argv[]) {

  memset(dp,-1,sizeof(dp));

  cin >> n;

  for(ll i = 0;i<n;i++){

    cin >> ara[i];

  }

  cout << fnc(0) << endl;



  return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

#define    sz    100110

#define    ll    long long int

ll n,c=0,p,q;

ll num[sz];

ll vis[sz];

ll dp[sz];



ll tani(int pos)

{



  if(pos>=n)return 0;

  if(vis[pos]==1)return dp[pos];

  vis[pos]=1;



  return  dp[pos]=min(abs(num[pos]-num[pos+1])+tani(pos+1),abs(num[pos]-num[pos+2])+tani(pos+2));



}



int main()

{

    ll t,i;



    for(i=1;i<sz;i++)num[i]=1000000;



    cin>>n;

    for(i=1;i<=n;i++)

    {

        cin>>num[i];

    }

    t=tani(1);

    cout<<t<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.2866936537023717x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

#define     FasterIO        ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

#define     bug(a,b)        cout<<a<<"\t"<<b<<"\n";

#define     Case(a,b)       cout<<"Case "<<a<<": "<<b<<"\n";

#define     watch(a)        cout<<fixed<<setprecision(a)

#define     lp(i,a,b)       for(int i=a;i<b;i++)

#define     Endl            "\n";

#define     tab             "\t"

#define     reset(a,b)      memset(a,b,sizeof(a));

#define     sf(a)           scanf(" %d", &a);

#define     srt(a)          sort(a.begin(),a.end());

#define     pi              2*acos(0.0)

#define     inf             INT_MAX

const int sz=1e5+9;

int n,x,arr[sz],dp[sz];



int clac(int i){

    if(i==n) return 0;

    if(dp[i]!=-1)return dp[i];

    int left=abs(arr[i]-arr[i+1])+clac(i+1);

    int right=inf;

    if(i+2<=n){

        right=abs(arr[i]-arr[i+2])+clac(i+2);

    }

    return dp[i]=min(left,right);

}









void solve(){



    cin>>n;





    reset(dp,-1);

    //bug(n,dp[n]);

    lp(i,1,n+1)cin>>arr[i];

    cout<<clac(1)<<"\n";



}



int main(){

    //FasterIO;

   // #ifndef ONLINE_JUDGE

      //  freopen("1.txt","r",stdin);

   // #endif

    //int t;

    //cin>>t;

    //while(t--)

        solve();

}
// Slower C++ solution:
#include <bits/stdc++.h>



using namespace std;



#define REP(i,a,b) for(ll i = a; i < b; i++)

#define DB( x ) cout << #x << " = " << x << endl;

#define sz(s) ((int)((s).size()))

#define endl '\n'



typedef long long ll;

typedef unsigned long long ull;

typedef pair<ll, ll> pii;

typedef pair<int, pii> tri;

typedef vector <vector <ll>> matrix;

typedef pair<string, string> pss;



const ll mod = (ll)1e9 + 7;

const int MAX = (int)1e6 + 100;



const ll oo = 0x3f3f3f3f3f3f3f3f;



ll dp[MAX], a[MAX];



int main() {

    ios_base::sync_with_stdio(0);

    cin.tie(0);



    memset(dp, oo, sizeof(dp));



    int n; cin >> n;



    dp[1] = 0;



    for(int i = 1;i <= n;++i){

        cin >> a[i];

    }



    for(int i = 2;i <= n;++i){

        dp[i] = min(dp[i - 1] + abs(a[i] - a[i - 1]), dp[i]);

        if(i - 2 >= 1){

            dp[i] = min(dp[i - 2] + abs(a[i] - a[i - 2]), dp[i]);

        }

    }



    cout << dp[n] << endl;

}


// Faster C++ solution for the same problem (measured speedup: 6.720655422454523x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

#define     FasterIO        ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

#define     bug(a,b)        cout<<a<<"\t"<<b<<"\n";

#define     Case(a,b)       cout<<"Case "<<a<<": "<<b<<"\n";

#define     watch(a)        cout<<fixed<<setprecision(a)

#define     lp(i,a,b)       for(int i=a;i<b;i++)

#define     Endl            "\n";

#define     tab             "\t"

#define     reset(a,b)      memset(a,b,sizeof(a));

#define     sf(a)           scanf(" %d", &a);

#define     srt(a)          sort(a.begin(),a.end());

#define     pi              2*acos(0.0)

#define     inf             INT_MAX

const int sz=1e5+9;

int n,x,arr[sz],dp[sz];



int clac(int i){

    if(i==n) return 0;

    if(dp[i]!=-1)return dp[i];

    int left=abs(arr[i]-arr[i+1])+clac(i+1);

    int right=inf;

    if(i+2<=n){

        right=abs(arr[i]-arr[i+2])+clac(i+2);

    }

    return dp[i]=min(left,right);

}









void solve(){



    cin>>n;





    reset(dp,-1);

    //bug(n,dp[n]);

    lp(i,1,n+1)cin>>arr[i];

    cout<<clac(1)<<"\n";



}



int main(){

    //FasterIO;

   // #ifndef ONLINE_JUDGE

      //  freopen("1.txt","r",stdin);

   // #endif

    //int t;

    //cin>>t;

    //while(t--)

        solve();

}
// Slower C++ solution:
#include <bits/stdc++.h>

#include <queue>

typedef long long ll;



using namespace std;



const int N=1e5+5;





int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);

   int n,k=2;

   cin>>n;

   int arr[N]={},dp[N]={};

   for(int i=0;i<n;i++)

    cin>>arr[i];

   memset(dp,-1,sizeof(dp));

   dp[0]=0;

   for(int i=0;i<n;i++)

   {



     for(int j=1;j<=k;j++)

     {

         if(i-j>=0)

         {

             if(dp[i]==-1)

             {

                 dp[i]=dp[i-j]+abs(arr[i]-arr[i-j]);

             }

             else

             {

                 dp[i]=min(dp[i],dp[i-j]+abs(arr[i]-arr[i-j]));





             }





         }



     }









   }

   cout<<dp[n-1];



   return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3301020973829283x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

#define     FasterIO        ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

#define     bug(a,b)        cout<<a<<"\t"<<b<<"\n";

#define     Case(a,b)       cout<<"Case "<<a<<": "<<b<<"\n";

#define     watch(a)        cout<<fixed<<setprecision(a)

#define     lp(i,a,b)       for(int i=a;i<b;i++)

#define     Endl            "\n";

#define     tab             "\t"

#define     reset(a,b)      memset(a,b,sizeof(a));

#define     sf(a)           scanf(" %d", &a);

#define     srt(a)          sort(a.begin(),a.end());

#define     pi              2*acos(0.0)

#define     inf             INT_MAX

const int sz=1e5+9;

int n,x,arr[sz],dp[sz];



int clac(int i){

    if(i==n) return 0;

    if(dp[i]!=-1)return dp[i];

    int left=abs(arr[i]-arr[i+1])+clac(i+1);

    int right=inf;

    if(i+2<=n){

        right=abs(arr[i]-arr[i+2])+clac(i+2);

    }

    return dp[i]=min(left,right);

}









void solve(){



    cin>>n;





    reset(dp,-1);

    //bug(n,dp[n]);

    lp(i,1,n+1)cin>>arr[i];

    cout<<clac(1)<<"\n";



}



int main(){

    //FasterIO;

   // #ifndef ONLINE_JUDGE

      //  freopen("1.txt","r",stdin);

   // #endif

    //int t;

    //cin>>t;

    //while(t--)

        solve();

}
// Slower C++ solution:
/*

				" اللهم صلي و سلم علي سيدنا محمد و علي اله وصحبه اجمعين "

*/





#define _CRT_SECURE_NO_WARNINGS

#include<bits/stdc++.h>

#define f for

#define ll long long

#define ull unsigned long long

#define endl "\n"

using namespace std;

#define BI  acos(-1)

#define all(a) a.begin(), a.end()

void Open()

{

#ifndef ONLINE_JUDGE

	freopen("Input.txt", "r", stdin);

	freopen("Output.txt", "w", stdout);

#endif

}

ll gcd(ll a, ll b)

{

	return b == 0 ? a : gcd(b, a % b);

}

void madesetprobblem()

{

	//$ $

	//less than or equal => $ 1 \leq 100 $

	//  $(1 \le A,B \le 10^5 )$

	// a[i] => $a_i$

	// a[i][j] => $a_{i,j}$

	// double qoutes ''yext''

	/*

		image

		\begine{center}

		\includegraphics{image.jpg}

		\end{center}

	*/

	/*

		list

		\begin{itemize}

			\item first one

			\item second one

		\end{itemize}

	*/

}

int n;

vector<int>a(1e5 + 2),dp(1e5+2,-1);

int ans(int c = 0)

{

	if (c == n-1)

		return 0;

	int ans1 = INT_MAX, ans2 = INT_MAX;

	if (dp[c] != -1)

		ans1 = dp[c];

	else

	{

		if (c + 1 < n)

			ans1 = (abs(a[c] - a[c + 1]) + ans(c + 1));

		if (c + 2 < n)

			ans2 = (abs(a[c] - a[c + 2]) + ans(c + 2));

		dp[c] = min(ans1, ans2);

	}

	//cout <<c<<" "<< ans1 << " " << ans2 << endl;

	return min(ans1,ans2);

}

int main()

{

	// توكلت علي الله 

//Open();

	ios::sync_with_stdio(false);

	cin.tie(NULL);

	cout.tie(NULL);

	cin >> n;

	f(int i = 0; i < n; i++)

	{

		cin >> a[i];

	}

	cout << ans();

}

// Faster C++ solution for the same problem (measured speedup: 1.3083405283811669x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

#define     FasterIO        ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

#define     bug(a,b)        cout<<a<<"\t"<<b<<"\n";

#define     Case(a,b)       cout<<"Case "<<a<<": "<<b<<"\n";

#define     watch(a)        cout<<fixed<<setprecision(a)

#define     lp(i,a,b)       for(int i=a;i<b;i++)

#define     Endl            "\n";

#define     tab             "\t"

#define     reset(a,b)      memset(a,b,sizeof(a));

#define     sf(a)           scanf(" %d", &a);

#define     srt(a)          sort(a.begin(),a.end());

#define     pi              2*acos(0.0)

#define     inf             INT_MAX

const int sz=1e5+9;

int n,x,arr[sz],dp[sz];



int clac(int i){

    if(i==n) return 0;

    if(dp[i]!=-1)return dp[i];

    int left=abs(arr[i]-arr[i+1])+clac(i+1);

    int right=inf;

    if(i+2<=n){

        right=abs(arr[i]-arr[i+2])+clac(i+2);

    }

    return dp[i]=min(left,right);

}









void solve(){



    cin>>n;





    reset(dp,-1);

    //bug(n,dp[n]);

    lp(i,1,n+1)cin>>arr[i];

    cout<<clac(1)<<"\n";



}



int main(){

    //FasterIO;

   // #ifndef ONLINE_JUDGE

      //  freopen("1.txt","r",stdin);

   // #endif

    //int t;

    //cin>>t;

    //while(t--)

        solve();

}
// Slower C++ solution:
#define __USE_MINGW_ANSI_STDIO

#define _CRT_SECURE_NO_WARNINGS

#include <bits/stdc++.h>

#include <unordered_set>

#include <unordered_map>



using namespace std;



typedef long long ll;

typedef long long ull;

typedef pair<ll, ll> ii;



#define all(v)		((v).begin()), ((v).end())

#define sz(v)		((int)((v).size()))

#define endl		"\n"

#define fx(n)		fixed<<setprecision(n)

#define mk			make_pair	



void File()

{

#ifndef ONLINE_JUDGE

	freopen("output.txt", "w", stdout);

	freopen("input.txt", "r", stdin);

#endif

#ifdef ONLINE_JUDGE

	/*freopen("output.txt", "w", stdout);

	freopen("pyramid.in", "r", stdin);*/

#endif

}

void fast()

{

	ios::sync_with_stdio(NULL);

	cout.tie(NULL);

	cin.tie(NULL);

	//File();

}

const double pi = 2 * acos(0.0);

const ll oo = 0x3f3f3f3f;

const int MOD = 1e9 + 7;

const int nn = 1e5 + 15;

int dx[8] = { 1, -1, 0, 0, 1, -1, 1, -1 };

int dy[8] = { 0, 0, 1, -1, 1, -1, -1, 1 };



int n;

ll arr[nn];

ll dp[nn];

int main()

{

	fast();

	cin >> n;

	for (int i = 0; i < n; i++)

		cin >> arr[i];

	memset(dp, oo, sizeof dp);

	dp[0] = 0;

	for (int i = 1; i < n; i++)

	{

		if (i == 1)

			dp[i] = abs(arr[i - 1] - arr[i]) + dp[0];

		else

			dp[i] = min({ dp[i - 1] + abs(arr[i - 1] - arr[i]), dp[i - 2] + abs(arr[i - 2] - arr[i]) });

	}

	cout << dp[n - 1] << endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.2993098258563172x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

#define     FasterIO        ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

#define     bug(a,b)        cout<<a<<"\t"<<b<<"\n";

#define     Case(a,b)       cout<<"Case "<<a<<": "<<b<<"\n";

#define     watch(a)        cout<<fixed<<setprecision(a)

#define     lp(i,a,b)       for(int i=a;i<b;i++)

#define     Endl            "\n";

#define     tab             "\t"

#define     reset(a,b)      memset(a,b,sizeof(a));

#define     sf(a)           scanf(" %d", &a);

#define     srt(a)          sort(a.begin(),a.end());

#define     pi              2*acos(0.0)

#define     inf             INT_MAX

const int sz=1e5+9;

int n,x,arr[sz],dp[sz];



int clac(int i){

    if(i==n) return 0;

    if(dp[i]!=-1)return dp[i];

    int left=abs(arr[i]-arr[i+1])+clac(i+1);

    int right=inf;

    if(i+2<=n){

        right=abs(arr[i]-arr[i+2])+clac(i+2);

    }

    return dp[i]=min(left,right);

}









void solve(){



    cin>>n;





    reset(dp,-1);

    //bug(n,dp[n]);

    lp(i,1,n+1)cin>>arr[i];

    cout<<clac(1)<<"\n";



}



int main(){

    //FasterIO;

   // #ifndef ONLINE_JUDGE

      //  freopen("1.txt","r",stdin);

   // #endif

    //int t;

    //cin>>t;

    //while(t--)

        solve();

}
// Slower C++ solution:
#include "bits/stdc++.h"

    

using namespace std;

vector<long long >dp(100001,-1);

vector<long long>num;

long long  erg(int n){

	if(dp[n]!=-1){

		return dp[n];

	}

	if(n<0){

		return 2000000000000;

	}

	if(n==0){

		return 0;

	}

	long long cmp1=max(num[n],num[n-1])-min(num[n],num[n-1]),cmp2=max(num[n],num[n-2])-min(num[n],num[n-2]);

	dp[n]= min(erg(n-1)+cmp1,erg(n-2)+cmp2);

	return dp[n];

}

int main() {

	int menge,numm;

	cin>>menge;

	for(int i=0;i<menge;i++){

		cin>>numm;

		num.push_back(numm);

	}

cout<<erg(menge-1);

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.2941300341495343x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

#define     FasterIO        ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

#define     bug(a,b)        cout<<a<<"\t"<<b<<"\n";

#define     Case(a,b)       cout<<"Case "<<a<<": "<<b<<"\n";

#define     watch(a)        cout<<fixed<<setprecision(a)

#define     lp(i,a,b)       for(int i=a;i<b;i++)

#define     Endl            "\n";

#define     tab             "\t"

#define     reset(a,b)      memset(a,b,sizeof(a));

#define     sf(a)           scanf(" %d", &a);

#define     srt(a)          sort(a.begin(),a.end());

#define     pi              2*acos(0.0)

#define     inf             INT_MAX

const int sz=1e5+9;

int n,x,arr[sz],dp[sz];



int clac(int i){

    if(i==n) return 0;

    if(dp[i]!=-1)return dp[i];

    int left=abs(arr[i]-arr[i+1])+clac(i+1);

    int right=inf;

    if(i+2<=n){

        right=abs(arr[i]-arr[i+2])+clac(i+2);

    }

    return dp[i]=min(left,right);

}









void solve(){



    cin>>n;





    reset(dp,-1);

    //bug(n,dp[n]);

    lp(i,1,n+1)cin>>arr[i];

    cout<<clac(1)<<"\n";



}



int main(){

    //FasterIO;

   // #ifndef ONLINE_JUDGE

      //  freopen("1.txt","r",stdin);

   // #endif

    //int t;

    //cin>>t;

    //while(t--)

        solve();

}