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

using namespace std;

const int nax = 1e5+3;

int number;

int dyn[nax];

int a[nax];

int op1(int i){

	if (i==number-1) return 0;

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

	int res = abs(a[i]-a[i+1])+op1(i+1);

	if(i!=number-2) res = min(res, abs(a[i]-a[i+2])+op1(i+2));

	return dyn[i]=res;

}

int main(){

	cin>>number;

	for(int i=0; i<number; i++) cin>>a[i];

	memset(dyn, -1, sizeof dyn);

	cout<<op1(0)<<endl;

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.399583711081049x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

int a[100005],dp[100005];

int main()

{

    ll n,i;

	cin>>n; 

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

	cin>>a[i];

	dp[1]=0;

	dp[2]=abs(a[2]-a[1]);

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

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

	cout<<dp[n];

	return 0;

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

using namespace std;

int n, arr[100005], best, memo[100005];



int dp(int x){

	if(x==1 || x == 0){

		return 0;

	}

	if (x == 2){

		return abs(arr[x] - arr[x-1]);

	}

	if (memo[x] != -1){

		return memo[x];

	}

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

	memo[x] = best;

	return best;

}



int main(){

	ios_base::sync_with_stdio(0);

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

	cin.tie();

	cout.tie();

	cin >> n;

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

		cin >> arr[i];

	}

	cout << dp(n) << endl;

}

// Faster C++ solution for the same problem (measured speedup: 1.4154640482081726x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

int a[100005],dp[100005];

int main()

{

    ll n,i;

	cin>>n; 

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

	cin>>a[i];

	dp[1]=0;

	dp[2]=abs(a[2]-a[1]);

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

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

	cout<<dp[n];

	return 0;

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

using namespace std;

int n, arr[100005], best, memo[100005];



int dp(int x){

	if(x==0 or x==1){

		return 0;

	}

	else if(x==2){

		return dp(x-1)+abs(arr[x]-arr[x-1]);

	}

	else if(memo[x]!=-1){

		return memo[x];

	}

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

	memo[x]=best;

	return best;

}



int main(){

	ios_base::sync_with_stdio(0);

	cin.tie();

	cout.tie();

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

	cin >> n;

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

		cin >> arr[i];

	}

	cout << dp(n) << endl;

}


// Faster C++ solution for the same problem (measured speedup: 1.4154640482081726x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

int a[100005],dp[100005];

int main()

{

    ll n,i;

	cin>>n; 

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

	cin>>a[i];

	dp[1]=0;

	dp[2]=abs(a[2]-a[1]);

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

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

	cout<<dp[n];

	return 0;

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

using namespace std;

int arr[100010],n;

int dp[100010];

int f(int pos){

	if(pos>=n-1)return 0;

	int &ret=dp[pos];

	if(ret^-1)return ret;

	if(pos+1==n-1)return ret=abs(arr[pos+1]-arr[pos]);

	return ret=min(f(pos+1)+abs(arr[pos]-arr[pos+1]),f(pos+2)+abs(arr[pos]-arr[pos+2]));

}



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

{

	memset(dp,-1,sizeof dp);

	cin>>n;

	for(int i=0;i<n;i++)cin>>arr[i];

		cout<<f(0)<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3947811512536827x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

int a[100005],dp[100005];

int main()

{

    ll n,i;

	cin>>n; 

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

	cin>>a[i];

	dp[1]=0;

	dp[2]=abs(a[2]-a[1]);

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

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

	cout<<dp[n];

	return 0;

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

#define ll long long

using namespace std;

const int maxn = 1e5+5;

int a[maxn], dp[maxn];



int main(){

	int n; cin >> n;

	memset(dp, 0x3f, sizeof dp);

	for(int i = 1; i <= n; i++) cin >> a[i];

	dp[1] = 0;

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

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

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

	}

	cout << dp[n] << endl;

}

// Faster C++ solution for the same problem (measured speedup: 1.399563953360374x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

int a[100005],dp[100005];

int main()

{

    ll n,i;

	cin>>n; 

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

	cin>>a[i];

	dp[1]=0;

	dp[2]=abs(a[2]-a[1]);

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

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

	cout<<dp[n];

	return 0;

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

/*                   وتزودوا فإن خير الزاد التقوى                    */

/*       وما تدري نفس ماذا تکسب غدا وما تدري نفس باي ارض تموت      */

#include <bits/stdc++.h>

#include<unordered_map>

#include<iomanip>

#include<fstream>

using namespace std;

#include<iostream>

#define endl '\n'

#define ll long long

#define ull unsigned long long

#define dll double

#define PI acos(-1)

#define watch(x) cout<<#x<<" = { "<<x<<" }\n"

#define sz(n) n.size()

#define M_Sayed fast() 

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

#define allr(v) v.rbegin(),v.rend()

#define RT(v) return cout<<v,0

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

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

const double EPS = (1e-7);

ll Ceil(ll x, ll y) { return (x + y - 1) / y; }

void run() {

#ifndef ONLINE_JUDGE

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

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

#else

#endif

}

ll gcd(ll a, ll b) { return !b ? abs(a) : gcd(b, a % b); }

ll lcm(ll a, ll b) { return abs(a / gcd(a, b)) * b; }

ll power(ll x, ll y, ll mod) {

	if (y == 0) return 1;

	if (y == 1) return x % mod;

	ll r = power(x, y >> 1, mod);

	return (((r * r) % mod) * power(x, y & 1, mod)) % mod;

}

void fast() { ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); }

ll n, m, k,su,cnt;

ll vis[100009];

vector<ll>v;

ll Jum(int x) {

	if (x+1 == n)

		return 0;

	ll &ret = vis[x];

	if (~ret)return ret;

	 ret = abs(v[x] - v[x + 1])+Jum(x + 1) ;

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

		if(i+x<n)ret = min(ret,Jum(x + i) + abs(v[x] - v[x + i]));

	}

	return ret;

}

int main()

{

	//run();

	M_Sayed;

	cin >> n ;

	memset(vis, -1, sizeof vis);

	v.resize(n);

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

		cin >> v[i];

	}

	k = 2;

	RT(Jum(0));

}

// Faster C++ solution for the same problem (measured speedup: 1.8155816338972917x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

int a[100005],dp[100005];

int main()

{

    ll n,i;

	cin>>n; 

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

	cin>>a[i];

	dp[1]=0;

	dp[2]=abs(a[2]-a[1]);

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

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

	cout<<dp[n];

	return 0;

}
// Slower C++ solution:
//       Author: Mostafa Mounir Shehab

 

#include <bits/stdc++.h>

using namespace std;

 

#define PI  3.14159265  // PI = acos(-1)

#define EPS (1e-10)

#define endl "\n"

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

typedef long long ll;

typedef long double ld;

//template <class Type>

 

 

void Fast(){

	std::ios_base::sync_with_stdio(0);

	cin.tie(NULL); cout.tie(NULL); cerr.tie(NULL);

}

 

inline int D(){

	int t;

	scanf("%d",&t);

	return t;

}

inline ll llD(){

	ll t;

	scanf("%lld",&t);

	return t;

}

 

 

ll MOD=1e9 + 7; 

const int N=1e5 + 5,M=50 + 1;

int arr[N];

ll dp[N];

ll OO=1e12; 

 

ll n,m,k,cnt,SU,idx,len,MN,MX;

ll l,r,mid;

ll a,b,c,d;



ll Solve(int idx){

	ll &ret = dp[idx];

	if(~ret)

		return ret;

		

	if(idx >= n)

		return ret = (idx == n) ? 0 : OO;

	

	

	ll x = abs(arr[idx] - arr[idx+1]) + Solve(idx+1);

	ll y = abs(arr[idx] - arr[idx+2]) + Solve(idx+2);

	

	return ret = min(x,y);

}







 

 

int main(){

	

	memset(dp, -1 ,sizeof dp);

	cin>>n;

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

		cin>>arr[i];



	cout<<Solve(1)<<endl;

	



	





}

// Faster C++ solution for the same problem (measured speedup: 1.8044778056172777x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

int a[100005],dp[100005];

int main()

{

    ll n,i;

	cin>>n; 

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

	cin>>a[i];

	dp[1]=0;

	dp[2]=abs(a[2]-a[1]);

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

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

	cout<<dp[n];

	return 0;

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

using namespace std;

const int maxn = 1e5 + 5;

int main(){

	int a[maxn],dp[maxn],n;

	memset(dp,0x3f3f,sizeof(dp));

	scanf("%d",&n);

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

		scanf("%d",&a[i]);

	dp[1] = 0;

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

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

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

	}

	printf("%d\n",dp[n]);

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3979609361053946x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

int a[100005],dp[100005];

int main()

{

    ll n,i;

	cin>>n; 

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

	cin>>a[i];

	dp[1]=0;

	dp[2]=abs(a[2]-a[1]);

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

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

	cout<<dp[n];

	return 0;

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



using namespace std;



const int N = 1e5 + 9;



int n, h[N], dp[N][2];



int sol(int i, int prev){

    if(i >= n)  return 1e9;

    if(i == n - 1)  return abs(h[i] - h[i - 1 - (prev == 1)]);



    int &ret = dp[i][prev];

    if(~ret)    return ret;



    return ret = abs(h[i] - h[i - 1 - (prev == 1)]) + min(sol(i + 1, 0), sol(i + 2, 1));

}



int main(){

    cin >> n;

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

        cin >> h[i];



    memset(dp, -1, sizeof dp);



    cout << min(sol(1, 0), sol(2, 1));



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.7998075760707417x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

int a[100005],dp[100005];

int main()

{

    ll n,i;

	cin>>n; 

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

	cin>>a[i];

	dp[1]=0;

	dp[2]=abs(a[2]-a[1]);

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

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

	cout<<dp[n];

	return 0;

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



using namespace std;



int n, h[100002], dp[100002][3];



int sol(int i, int choice){

    if(i==0) return 0;

    if(i<0) return 1E9;



    int &ret = dp[i][choice];

    if(~ret) return ret;



    return ret = min(abs(h[i]-h[i-1]) + sol(i-1, 1), abs(h[i]-h[i-2]) + sol(i-2, 2));

}



int main()

{

    cin >> n;

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

        cin >> h[i];



    memset(dp, -1, sizeof dp);

    cout << min(sol(n-1, 1), sol(n-1, 2));

}


// Faster C++ solution for the same problem (measured speedup: 2.202725248808743x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

int a[100005],dp[100005];

int main()

{

    ll n,i;

	cin>>n; 

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

	cin>>a[i];

	dp[1]=0;

	dp[2]=abs(a[2]-a[1]);

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

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

	cout<<dp[n];

	return 0;

}
// Slower C++ solution:


#include <bits/stdc++.h>

using namespace  std;



int N;

const int SIZE = 100005;

long h [SIZE];

int mem [SIZE];



int dp (int i) {

  if (i == N-1) return 0;

  

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



  long ret = LLONG_MAX;



  if (i+1 < N) 

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

  if (i+2 < N)

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



  return  mem[i] = ret;

}



int main (){ 



  cin  >> N;

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

    cin >> h[i];

  }



  memset(mem, -1, sizeof mem);



  cout << dp(0);



}

// Faster C++ solution for the same problem (measured speedup: 1.3974532494043044x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

int a[100005],dp[100005];

int main()

{

    ll n,i;

	cin>>n; 

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

	cin>>a[i];

	dp[1]=0;

	dp[2]=abs(a[2]-a[1]);

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

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

	cout<<dp[n];

	return 0;

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

#define maxn 100005

using namespace std;



int main(){

    int n;

    cin >> n;

    int h[maxn];

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

        cin >> h[i];

    }

    int dp[maxn];

    fill(dp, dp+maxn, 0x7fffffff);

    dp[0] = 0;

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

        int temp = 0;

        if(i+1 < n){

            temp = abs(h[i+1]-h[i]) + dp[i];

            if(temp < dp[i+1])

                dp[i+1] = temp;

        }

        if(i+2 < n){

            temp = abs(h[i+2]-h[i]) + dp[i];

            if(temp < dp[i+2])

                dp[i+2] = temp;

        }

    }

    cout << dp[n-1];

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.404364902285117x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

int a[100005],dp[100005];

int main()

{

    ll n,i;

	cin>>n; 

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

	cin>>a[i];

	dp[1]=0;

	dp[2]=abs(a[2]-a[1]);

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

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

	cout<<dp[n];

	return 0;

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

#include<bits/stdc++.h>

#define rep(i,a,n) for (int i=a;i<n;i++)

#define per(i,a,n) for (int i=n-1;i>=a;i--)

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

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

#define clr(v, d) memset(v, d, sizeof(v))

#define fi first

#define se second

#define pp push_back

#define pb pop_back

#define pi 3.141592654

#define MOD 1000000007 

#define cons1 1e12

#define cons2 1e6

using namespace std;

typedef long long ll;

typedef vector<int> vi;

typedef pair<int, int> pa;

typedef unsigned long long int ull;

typedef double long dl;

const int N = 1e5+9;

int n;

int v[N], ans[N];

int sol(int i) {

	if (ans[i] == -1) {

		if (i == n - 1)

			ans[i] = 0;

		else if (i == n - 2)

			ans[i] = abs(v[i] - v[i + 1]);

		else 

			ans[i] = min(abs(v[i] - v[i + 1])+sol(i + 1)

				, abs(v[i] - v[i + 2]) + sol(i + 2));

		

	}

	return ans[i];

	

}

void solve() {

	clr(ans, -1);

	cin >> n;

	rep(i, 0, n)

		cin >> v[i];

	cout << sol(0) << endl;

}

int main() {

	ios::sync_with_stdio(0);

	cin.tie(NULL);

	cout.tie(NULL);

	int t = 1;

	//cin >> t;

	while (t--)

		solve();



#ifndef ONLINE_JUDGE

	- system("pause");

#endif



}

// Faster C++ solution for the same problem (measured speedup: 1.411749083917266x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

int a[100005],dp[100005];

int main()

{

    ll n,i;

	cin>>n; 

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

	cin>>a[i];

	dp[1]=0;

	dp[2]=abs(a[2]-a[1]);

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

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

	cout<<dp[n];

	return 0;

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

using namespace std;



typedef long double LD;

typedef long long ll;

typedef map<int,int> mii;

typedef pair<int,int> pii;

typedef vector<pii> vpii;



#define pb              push_back

#define mp              make_pair

#define F               first

#define S               second

#define mod             1000000007

#define INF             2147483647

#define Endl            "\n"

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

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

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

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

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

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



const int MAXN = 1000005;

const int SQRTN = 1003;

const int LOGN = 22;

const double PI=acos(-1);

const int sz=1e5+9;



int strtoint(string str){

    stringstream ss(str);

    int x = 0;

    ss >> x ;

    return x ;

}

string intostr(int x){

    stringstream ss;

    ss << x;

    string str = ss.str();

    return str;

}

void rem(string &s, const string &c) {

	auto pos = s.find(c);

	if (pos == string::npos) {

		cout << -1 << endl;

		exit(0);

	}

	s.erase(0, pos + 1);

}

void println (int n){

    cout<<n<<"\n";

}

void printvec(vector <int> &a){

    for (int i=0;i<a.size();i++){

        cout<<a[i]<<" ";

    }

    cout<<"\n";

}



void printgraph(vector <int> g[],int e){

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

        cout<<i<<" --> ";

        for (int j=0;j<g[i].size();j++){

            cout<<g[i][j]<<" ";

        }

        cout<<"\n";

    }

}



void printmap (map <int,int> &mymap){

    for (auto itr = mymap.begin(); itr != mymap.end(); ++itr) {

        cout << '\t' << itr->first

             << '\t' << itr->second << '\n';

    }

    cout << "\n";

}



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





int rec (int i){

    if (i==n) return 0;

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

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

    int right = INF;

    if (i+2<=n){

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

    }

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

}





void solve(){



    cin>>n;

    reset (dp,-1);

    for (int i=1;i<=n;i++)cin>>arr[i];





    cout << rec (1) << Endl;



    //printvec(v);

}





int main(){

    //FasterIO;



    solve();

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4218088969464568x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

int a[100005],dp[100005];

int main()

{

    ll n,i;

	cin>>n; 

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

	cin>>a[i];

	dp[1]=0;

	dp[2]=abs(a[2]-a[1]);

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

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

	cout<<dp[n];

	return 0;

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



using namespace std;



const int N = 1e5; // Maximum stones as per problem



int n, stones[N];



int memo[N]; // Memoization Table



int jumps(int ind) {

    if (ind == (n - 1)) return 0; // If we are last stone, then the answer will be 0.

    if (ind >= n) return INT_MAX; // Checking the boundary condition





    // If you are done with the whole function, then only read

    // Since we stored the minimum cost for a stone, now we will use the stored cost

    // We are initializing the table to -1, if the value stored is not -1, then we have calculated the minimum cost

    // So just return that cost

    if (memo[ind] != -1) return memo[ind]; // If reading for first time, then ignore this line



    // Calculate the cost for jumping to next stone (ind + 1) and recursively doing it for next stone

    int oneJump = abs(stones[ind] - stones[ind + 1]) + jumps(ind + 1);



    // Initializing the twoJumps because there can be a case in which there can be only one stone.

    int twoJumps = INT_MAX;



    // Checking if there are more than 2 stones

    if (ind + 2 < n) {

        // Calculate the cost for jumping to next to next stone (ind + 2) and recursively doing it for next to next stone

        twoJumps = abs(stones[ind] - stones[ind + 2]) + jumps(ind + 2);

    }



    // Storing the minimum cost in the memoization table and returning the minimum cost

    // After this,  go to the ignored line.

    return memo[ind] = min(oneJump, twoJumps);

}





int main() {

    // Initializing the memoization table to -1

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



    cin >> n;

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

        cin >> stones[i];

    }



    // We will start from first stone

    cout << jumps(0) << endl;



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4028471715846915x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

int a[100005],dp[100005];

int main()

{

    ll n,i;

	cin>>n; 

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

	cin>>a[i];

	dp[1]=0;

	dp[2]=abs(a[2]-a[1]);

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

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

	cout<<dp[n];

	return 0;

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

using namespace std;

#define ll long long

#define ull unsigned long long

#define INF 1e18

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

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

typedef vector<int> vi;

typedef vector<ll> vll;

typedef pair<int,int> pi;



#define M_PI 3.14159265358979323846



int a[100005];

const int oo = 1e9+10;

const int mod = 7901;



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



int dp[100005];

int n;

int memo[100005];



int dpF(int i){

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

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

	

	int ans = abs(a[i] - a[i+1])+dpF(i+1);

	if(i+2 <= n-1){

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

	}



	return memo[i] = ans;

}



void solve(){

	cin>>n;

	for(int i = 0 ; i < n ; ++i)cin>>a[i];

	

	memset(memo, -1 , sizeof memo);

	cout << dpF(0) << '\n';

	return;

	dp[0] = 0;

	dp[1] = abs(a[1] - a[0]);

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

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

	}

	cout << dp[n-1] << '\n';

}

int main(){   

    ios::sync_with_stdio(0);cin.tie(0);

	int t=1;

	while(t--){

		solve();

	}

}

// Faster C++ solution for the same problem (measured speedup: 1.4126790606865465x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

int a[100005],dp[100005];

int main()

{

    ll n,i;

	cin>>n; 

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

	cin>>a[i];

	dp[1]=0;

	dp[2]=abs(a[2]-a[1]);

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

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

	cout<<dp[n];

	return 0;

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

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

    if(dp[idx] != -1) return dp[idx]; // caso dp

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

    ios_base::sync_with_stdio(false); cin.tie(NULL);

    cin >> n;

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

        cin >> h[i];

    }

    memset(dp, -1, sizeof dp);

    cout << Dp(0) << endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4122523997025715x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

int a[100005],dp[100005];

int main()

{

    ll n,i;

	cin>>n; 

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

	cin>>a[i];

	dp[1]=0;

	dp[2]=abs(a[2]-a[1]);

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

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

	cout<<dp[n];

	return 0;

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

using namespace std;



const int inf=0x3f;

int h[100005],dp[100005];



signed main()

{

	int n;

	while(cin>>n){

		for(int i=1;i<=n;i++)cin>>h[i];

		memset(dp,inf,sizeof dp);

		

		dp[1]=0,dp[2]=abs(h[2]-h[1]);

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

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

		}

		cout<<dp[n]<<endl;

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4000582642000154x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

int a[100005],dp[100005];

int main()

{

    ll n,i;

	cin>>n; 

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

	cin>>a[i];

	dp[1]=0;

	dp[2]=abs(a[2]-a[1]);

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

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

	cout<<dp[n];

	return 0;

}