stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
/*|In The Name Of Allah|*/



#include <bits/stdc++.h>



using namespace std;



const int N = 1e5+5;



int n;

int h[N];

int mem[N];



int go(int idx){

    if(idx >= n)

        return 0;

    if(mem[idx] != -1)

        return mem[idx];

    int best = 1e9;

    if(idx + 1 <= n)

        best = go(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 <= n)

        best = min(best , go(idx + 2)  + abs(h[idx] - h[idx + 2]));

    return mem[idx] = best;

}



int main(){

    scanf("%d" , &n);

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

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

    }

    memset(mem , -1 , sizeof mem);

    printf("%d\n" , go(1));

}

//1 2 3


// Faster C++ solution for the same problem (measured speedup: 1.3593124235472724x):
#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>

#define pii pair<int,int>

#define vi vector<int>

#define ppii pair<pii>



using namespace std;

const int maxn=1e5+5;



int n;

vi arr;

int dp[maxn];



int dfs(int pos)

{

    if(pos==n){return 0;}



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



    int ans=(1<<29);



    if(pos<n)

        ans=min(ans,dfs(pos+1) + abs(arr[pos]-arr[pos+1]));

    if(pos<n-1)

        ans=min(ans,dfs(pos+2) + abs(arr[pos]-arr[pos+2]));

    dp[pos]=ans;



    return ans;

}



int main()

{

    ios_base::sync_with_stdio(0);

    cin.tie(0);cout.tie(0);



    cin>>n;



    arr=vi(n+1);



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

        cin>>arr[i];



    memset(dp,-1,sizeof dp);



    cout<<dfs(1);



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3853561182850782x):
#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>

#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: 1.8237155974662687x):
#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;



typedef long long ll;

#define mp make_pair

#define f first

#define s second

#define pb push_back



const int inf = 1e7;

const int N  = 1e5 + 5;



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

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



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



int solve(int idx) {



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



  int &ret = dp[idx];



  if(ret!=-1) return ret;



  ret = solve(idx+1) + abs(h[idx]-h[idx+1]);



  if(idx!=n-2)

    ret = min(ret,solve(idx+2) + abs(h[idx]-h[idx+2]));



  return ret;



}



int main() {



    ios_base::sync_with_stdio(false);

    cin.tie(0);



    cin >> n;



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



      cin >> h[i];



    }



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



    cout << solve(0) << endl;



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3843805054440974x):
#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;



const int nax = 1e5+3;

int n;

int dp[nax];

int a[nax];

int f(int i){

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

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

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

	if(i!=n-2) ans = min(ans, abs(a[i]-a[i+2])+f(i+2));

	return dp[i]=ans;

}

int main(){

	cin>>n;

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

	memset(dp, -1, sizeof dp);

	cout<<f(0)<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3713437332516634x):
#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;

int a,b,c,d,dp[100000],cost[100000],weight[100];



void func(int a)

{

    dp[1]=0;

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

    for(int i=3; i<=a; i++)

    {

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

    }

}



int main()

{

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

    cin>>a;

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

    {

        cin>>cost[i];

    }

    func(a);

    cout<<dp[a]<<endl;

    //cout<<nCr(a,b)<<endl;



}


// Faster C++ solution for the same problem (measured speedup: 1.3667584044592085x):
#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>

#define F first

#define S second

#define T second.first

#define C second.second

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

using namespace std;

typedef long long ll;

typedef pair<int, int> pii;

typedef pair<int,pair<int,int>> piii;



const int N= 1e5+5, L = 14, OO = 0x3f3f3f3f;

const ll M=10 ;

const double ESP = 1e-8;



int n;

int A[N], dp[N];

int main(){

    FIO

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

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

    cin>>n;

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

    memset(dp, OO, sizeof dp);

    dp[0] = 0;

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

        if(i+1<n){

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

        }

        if(i+2<n){

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

        }

    }

    cout<<dp[n-1];



}


// Faster C++ solution for the same problem (measured speedup: 1.3833935756751476x):
#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;

typedef long long ll;

typedef double dd;

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

#define endl "\n"

#define clr(n, r) memset(n,r,sizeof(n));



void fast() {



    cin.tie(0);

    cin.sync_with_stdio(0);

}



const int N = 1e5 + 9;

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



int sol(int i){

    if(i<0) return 1e9;

    if(i==0) return 0;



    if(~dp[i]) return dp[i];



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

}





int main() {



    cin >> n;



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



    memset(dp, -1, sizeof dp);



    cout << sol(n - 1);

}

// Faster C++ solution for the same problem (measured speedup: 1.3695772691772987x):
#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;

}