stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 2.1727500857050064x):
#include <bits/stdc++.h>

 

using namespace std;

 

typedef long long ll;

typedef vector<ll> vi;

typedef long double td;

typedef vector<td> vd;

typedef pair<ll,ll> ii;

typedef vector<ii> vii; typedef vector<bool> vb;

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

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

#define forn(i, n) forr(i, 0, n)

#define pb push_back

#define mp make_pair

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

#define DBG(x) cerr << #x << " = " << (x) << endl

#define show(v,n) cerr << #v << " = "; forn(i,n) cerr << v[i] << " "; cerr << endl;

#define esta(x,c) ((c).find(x) != (c).end())

#define RAYA cerr << "===============================" << endl

int MAXN = 500005;





ll solve(vector <int> &v, int n, int k)

{

	ll MAXN = (ll)(1e18);

	vector <ll> dp(n,MAXN);

	dp[0]=0;

	forr(i,1,n)

	{

		forr(j,1,k+1)

		{

			if(i-j < 0) break;

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

		}

	}

	return dp[n-1];

}

 

int main()

{ 	

	FIN;

	#ifdef input

		freopen("test1.txt", "r", stdin);

	#endif



	int n;

	cin >> n;

	vector <int> v(n);

	forn(i,n) cin >> v[i];

	cout << solve(v,n,2) << endl;

    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: 1.3784364023856421x):
#include <bits/stdc++.h>

 

using namespace std;

 

typedef long long ll;

typedef vector<ll> vi;

typedef long double td;

typedef vector<td> vd;

typedef pair<ll,ll> ii;

typedef vector<ii> vii; typedef vector<bool> vb;

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

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

#define forn(i, n) forr(i, 0, n)

#define pb push_back

#define mp make_pair

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

#define DBG(x) cerr << #x << " = " << (x) << endl

#define show(v,n) cerr << #v << " = "; forn(i,n) cerr << v[i] << " "; cerr << endl;

#define esta(x,c) ((c).find(x) != (c).end())

#define RAYA cerr << "===============================" << endl

int MAXN = 500005;





ll solve(vector <int> &v, int n, int k)

{

	ll MAXN = (ll)(1e18);

	vector <ll> dp(n,MAXN);

	dp[0]=0;

	forr(i,1,n)

	{

		forr(j,1,k+1)

		{

			if(i-j < 0) break;

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

		}

	}

	return dp[n-1];

}

 

int main()

{ 	

	FIN;

	#ifdef input

		freopen("test1.txt", "r", stdin);

	#endif



	int n;

	cin >> n;

	vector <int> v(n);

	forn(i,n) cin >> v[i];

	cout << solve(v,n,2) << endl;

    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: 1.3852539999944544x):
#include <bits/stdc++.h>

 

using namespace std;

 

typedef long long ll;

typedef vector<ll> vi;

typedef long double td;

typedef vector<td> vd;

typedef pair<ll,ll> ii;

typedef vector<ii> vii; typedef vector<bool> vb;

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

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

#define forn(i, n) forr(i, 0, n)

#define pb push_back

#define mp make_pair

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

#define DBG(x) cerr << #x << " = " << (x) << endl

#define show(v,n) cerr << #v << " = "; forn(i,n) cerr << v[i] << " "; cerr << endl;

#define esta(x,c) ((c).find(x) != (c).end())

#define RAYA cerr << "===============================" << endl

int MAXN = 500005;





ll solve(vector <int> &v, int n, int k)

{

	ll MAXN = (ll)(1e18);

	vector <ll> dp(n,MAXN);

	dp[0]=0;

	forr(i,1,n)

	{

		forr(j,1,k+1)

		{

			if(i-j < 0) break;

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

		}

	}

	return dp[n-1];

}

 

int main()

{ 	

	FIN;

	#ifdef input

		freopen("test1.txt", "r", stdin);

	#endif



	int n;

	cin >> n;

	vector <int> v(n);

	forn(i,n) cin >> v[i];

	cout << solve(v,n,2) << endl;

    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: 1.3925376960808318x):
#include <bits/stdc++.h>

 

using namespace std;

 

typedef long long ll;

typedef vector<ll> vi;

typedef long double td;

typedef vector<td> vd;

typedef pair<ll,ll> ii;

typedef vector<ii> vii; typedef vector<bool> vb;

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

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

#define forn(i, n) forr(i, 0, n)

#define pb push_back

#define mp make_pair

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

#define DBG(x) cerr << #x << " = " << (x) << endl

#define show(v,n) cerr << #v << " = "; forn(i,n) cerr << v[i] << " "; cerr << endl;

#define esta(x,c) ((c).find(x) != (c).end())

#define RAYA cerr << "===============================" << endl

int MAXN = 500005;





ll solve(vector <int> &v, int n, int k)

{

	ll MAXN = (ll)(1e18);

	vector <ll> dp(n,MAXN);

	dp[0]=0;

	forr(i,1,n)

	{

		forr(j,1,k+1)

		{

			if(i-j < 0) break;

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

		}

	}

	return dp[n-1];

}

 

int main()

{ 	

	FIN;

	#ifdef input

		freopen("test1.txt", "r", stdin);

	#endif



	int n;

	cin >> n;

	vector <int> v(n);

	forn(i,n) cin >> v[i];

	cout << solve(v,n,2) << endl;

    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: 1.3837569229026974x):
#include <bits/stdc++.h>

 

using namespace std;

 

typedef long long ll;

typedef vector<ll> vi;

typedef long double td;

typedef vector<td> vd;

typedef pair<ll,ll> ii;

typedef vector<ii> vii; typedef vector<bool> vb;

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

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

#define forn(i, n) forr(i, 0, n)

#define pb push_back

#define mp make_pair

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

#define DBG(x) cerr << #x << " = " << (x) << endl

#define show(v,n) cerr << #v << " = "; forn(i,n) cerr << v[i] << " "; cerr << endl;

#define esta(x,c) ((c).find(x) != (c).end())

#define RAYA cerr << "===============================" << endl

int MAXN = 500005;





ll solve(vector <int> &v, int n, int k)

{

	ll MAXN = (ll)(1e18);

	vector <ll> dp(n,MAXN);

	dp[0]=0;

	forr(i,1,n)

	{

		forr(j,1,k+1)

		{

			if(i-j < 0) break;

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

		}

	}

	return dp[n-1];

}

 

int main()

{ 	

	FIN;

	#ifdef input

		freopen("test1.txt", "r", stdin);

	#endif



	int n;

	cin >> n;

	vector <int> v(n);

	forn(i,n) cin >> v[i];

	cout << solve(v,n,2) << endl;

    return 0;

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

using namespace std;

int dif(int a, int b)

{

    int ans=a-b;

    if(ans<0) ans*=-1;

    return ans;

}

int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    int N, h[100005],i,j;

    int f1,f2;

    cin>>N;

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

        cin>>h[i];

    int M[100005]={0};

    M[N-2]=dif(h[N-1],h[N-2]);

    for(i=N-3;i>=0;i--)

    {

        f1=dif(h[i+1],h[i])+M[i+1];

        f2=dif(h[i+2],h[i])+M[i+2];

        M[i]=min(f1,f2);

    }

    cout<<M[0]<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4024247158453609x):
#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;



void solve(){

	cin>>n;

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

	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();

	}

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

using namespace std;

int n;

int h[1000005], memo[1000005];

int f(int n) {

    if(n == 1) {                        // bisa juga base case nya n = 0, f(0) harus tinggi, misal 1e9;

        return 0;

    }

    if(n == 2) {

        return abs(h[1] - h[2]);

    } 

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

        return memo[n];

    }

    memo[n] = min(f(n - 1) + abs(h[n] - h[n - 1]), f(n - 2) + abs(h[n] - h[n - 2]));

    return memo[n];

}



int main() {

    scanf("%d", &n);

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

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

    }

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

    f(n);

    printf("%d\n", f(n));

}

// Faster C++ solution for the same problem (measured speedup: 5.001289014447692x):
#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;



void solve(){

	cin>>n;

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

	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();

	}

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

using namespace std;



typedef signed long long ll;

const int MX = 1e5 + 5;

const int INF = 1e9 + 5;

int n, A[MX], dp[MX];



int solve(int pos) {

    if(pos > n) return INF;

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

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

    int ans = INF;

    if(pos + 1 < n) ans = min(ans, abs(A[pos] - A[pos + 1]) + solve(pos + 1));

    if(pos + 2 < n) ans = min(ans, abs(A[pos] - A[pos + 2]) + solve(pos + 2));

    return dp[pos] = ans;

}



int main() {

    scanf("%d", &n);



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

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

    }

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

    printf("%d\n", solve(0));

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.370461980982988x):
#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;



void solve(){

	cin>>n;

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

	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();

	}

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

#include<cmath>

#include<algorithm>

#include<cstring>

using namespace std;



int h[100055],dp[100055];

const int inf = 0x3f3f3f;

int main()

{

	int n;

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

	cin>>n;

	h[1]=inf;

	h[0]=inf;

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

		cin>>h[i];

	for(int i=2;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;

} 

//max(h[i],h[i-2])-min(h[i],h[i-2])

// Faster C++ solution for the same problem (measured speedup: 1.385872514958792x):
#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;



void solve(){

	cin>>n;

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

	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();

	}

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

using namespace std;

#define IO ios::sync_with_stdio(false), cin.tie(0)

typedef long long ll;

const int maxn = 1e6 + 5;

const int inf = 0x3f3f3f3f;

ll a[maxn];

ll dp[maxn];



int main()

{

    IO;

    int n, k=2;

    cin >> n ;

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

        cin >> a[i];

    memset(dp, inf, sizeof(dp));

    dp[1] = 0;

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

    {

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

        {

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

        }

    }

    cout << dp[n] << endl;

}


// Faster C++ solution for the same problem (measured speedup: 9.288175369046765x):
#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;



void solve(){

	cin>>n;

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

	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();

	}

}
// 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: 1.3826127885387292x):
#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;



void solve(){

	cin>>n;

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

	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();

	}

}
// 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: 1.3983005655715928x):
#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;



void solve(){

	cin>>n;

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

	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();

	}

}
// 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: 1.3983005655715928x):
#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;



void solve(){

	cin>>n;

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

	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();

	}

}
// 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: 1.3778684630779032x):
#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;



void solve(){

	cin>>n;

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

	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();

	}

}