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



using namespace std;

const int maxn = 2e5+2;

const int mod = 1e9+7;

const int inf = 1e9;



typedef long long ll;



int n;

int arr[maxn], dp[maxn];



int solve_dp (int pos){

    if (pos > n) return inf;

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



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



    int ans = inf;

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

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



    return dp[pos] = ans;

}



int main(){



    ios_base::sync_with_stdio(0);

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



    cin >> n;



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

        cin >> arr[i];

    }



    memset(dp, -1, sizeof dp);



    cout << solve_dp(0) << "\n";



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.2995279428431703x):
#include <bits/stdc++.h>



using namespace std;



int N;

const int SIZE = 100005;

int h [SIZE];

int mem [SIZE];



int dp (int i) {



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

  

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

  

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



  int ret = INT_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);



}
// Slower C++ solution:
#pragma GCC optimize ("O3")

#include <bits/stdc++.h>

#define abdelrahman010 ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

typedef long long ll;

using namespace std;

ll n , arr[100005] , memo[100005];

ll solve(ll i)

{

	if(i == n - 1)

		return 0;

	ll &ans = memo[i];

	if(~ans)

		return ans;

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

	if(i < n - 2)

		ans = min(ans , abs(arr[i] - arr[i + 2]) + solve(i + 2));

	return ans;

}

int main()

{

	abdelrahman010

	cin >> n;

	memset(memo , -1 , sizeof memo);

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

		cin >> arr[i];

	cout << solve(0);

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.2974083528149583x):
#include <bits/stdc++.h>



using namespace std;



int N;

const int SIZE = 100005;

int h [SIZE];

int mem [SIZE];



int dp (int i) {



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

  

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

  

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



  int ret = INT_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);



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

#include<stdio.h>

#include<cstring>

#include<cmath>

#include<algorithm>

using namespace std;

#define maxn 100005

int dp[maxn];

int h[maxn];

int n;

int main(){

    scanf("%d",&n);

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

    memset(h,0,sizeof(h));

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

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

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

        if(i-1>=1&&i-2>=1)

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

        else if(i-1>=1) dp[i] = dp[i-1]+abs(h[i]-h[i-1]);

        else if(i-2>=1) dp[i] = dp[i-2]+abs(h[i]-h[i-2]);

    }

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.2795613310244824x):
#include <bits/stdc++.h>



using namespace std;



int N;

const int SIZE = 100005;

int h [SIZE];

int mem [SIZE];



int dp (int i) {



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

  

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

  

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



  int ret = INT_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);



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

#include <bitset>

using namespace std;

#define ll long long

#define mod 1000000009



// يلعن ابو الي يعقدها على نفسه تاني



int main()

{

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



    ll n; cin >> n;

    ll arr[100005];

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

        cin >> arr[i];

    }

    if(n == 2){

        cout << abs(arr[1]-arr[0]);

    }else{

        ll sol[100005] = {};

        sol[1] = abs(arr[1]-arr[0]);

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

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

        }









        cout << sol[n-1];

    }















    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.1486800156010493x):
#include <bits/stdc++.h>



using namespace std;



int N;

const int SIZE = 100005;

int h [SIZE];

int mem [SIZE];



int dp (int i) {



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

  

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

  

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



  int ret = INT_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);



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

#include<bits/stdc++.h>

#include<iostream>

#include<algorithm>

#include<iomanip>

#include<string>

#include<cmath>

#include<vector>

#include<set>

#include<queue>

#include<stack>

#include<fstream>

#include<map>

#include<bitset>

#include<unordered_map>

#include<cstring>

#include<cctype>

using namespace std;

#define  ll long long

#define  ull unsigned long long

#define  ul unsigned long

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

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

#define pb(s) push_back(s);

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

#define finl "\n"

#define Mohammed_Atef_Hassan fast();

#define mod 1000000007

#define INF LLONG_MAX

#define PI acos(-1)

#define clr(dp,n) memset(dp,n,sizeof dp)

#define ex(n) return cout<<n<<finl,0;

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

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

ll fix_mod(ll x, ll y)

{

	return (y + x % y) % y;

}

void fast() {

	std::ios_base::sync_with_stdio(false);

	cin.tie(nullptr); cout.tie(nullptr);





}

int n;

vector<ll>v;

ll mem[(int)1e5 + 10];

ll solve(int i)

{

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

	ll& ret = mem[i];

	if (~ret)return ret;

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

	if (i + 2 < n)

		ret = min(ret, solve(i + 2) + abs(v[i + 2] - v[i]));

	return ret;

}

int main()

{

	Mohammed_Atef_Hassan

		clr(mem, -1);

		cin >> n;

	v = vector<ll>(n);

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

	ex(solve(0));

}

// Faster C++ solution for the same problem (measured speedup: 1.3020048139535316x):
#include <bits/stdc++.h>



using namespace std;



int N;

const int SIZE = 100005;

int h [SIZE];

int mem [SIZE];



int dp (int i) {



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

  

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

  

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



  int ret = INT_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);



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

#include <sstream>

using namespace std;

#include <bits/stdc++.h>

#include<utility>

#include <algorithm>

#include<iterator>

#define  go return 0

#define rep(i,n) for(int i=0;i<n;++i)

#define rep(j,m) for(int j=0;j<m;++j)

#define REP(i,n) for(int i=1;i<=n;++i)

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

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

#define F first

#define S second

#define pb push_back

#define pf push_front

#define mp make_pair

typedef long long ll;

typedef double db;

typedef vector<ll> ve;

typedef vector< vector <int> >vee;

typedef vector<pair<int,int> >vepa;

typedef set<char>se;

typedef map<ll,ll> mpp;

const ll N=1e5+5;

bool f;

ll n,arr[N],dp[N];

ll solve(ll i=0)

{

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

  ll &ret=dp[i];

  if(~ret)return ret;

   return ret=min(solve(i+1)+abs(arr[i]-arr[i+1]) ,i+2<n? solve(i+2)+abs(arr[i]-arr[i+2]):12345 );

}

int main()

{

ios::sync_with_stdio(0);

cin.tie(0);

   cin>>n;

   memset(dp,-1,sizeof dp);

   rep(i,n)

    cin>>arr[i];

   cout<<solve();



go;

}


// Faster C++ solution for the same problem (measured speedup: 1.2974199035991603x):
#include <bits/stdc++.h>



using namespace std;



int N;

const int SIZE = 100005;

int h [SIZE];

int mem [SIZE];



int dp (int i) {



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

  

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

  

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



  int ret = INT_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);



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

using namespace std;



vector<long long int> dp(100005,-1);

long long int n,ara[100005];

int f(long long int i){



    if(i==n) return 0;



    if(i==n-1) return abs(ara[i+1]-ara[i]);



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



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

    return dp[i];





}









int main()

{

    long long int i;

    cin>>n;

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

        scanf("%lld",&ara[i]);

    }



    cout<<f(1)<<endl;



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.297104010586311x):
#include <bits/stdc++.h>



using namespace std;



int N;

const int SIZE = 100005;

int h [SIZE];

int mem [SIZE];



int dp (int i) {



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

  

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

  

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



  int ret = INT_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);



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

#define ll long long

using namespace std;

#define inf 0x3f3f3f3f 

int main(){

	ll t , h[100001] , dp[100001] = {0};

    cin >> t;

	for(int i = 0 ; i < t ; ++i) scanf("%lld",h+i);

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

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

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

        cout<<dp[t-1];

}

// Faster C++ solution for the same problem (measured speedup: 1.3051394593937213x):
#include <bits/stdc++.h>



using namespace std;



int N;

const int SIZE = 100005;

int h [SIZE];

int mem [SIZE];



int dp (int i) {



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

  

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

  

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



  int ret = INT_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);



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

using namespace std;

#define INF 100005

#define maxx 1000000000000



int n;

long long a[INF+5];

long long mem[INF];



long long rec(int pos)

{

    if(pos==n-1)

    {

        return 0;

    }

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

    long long tot=maxx;

    long long left=abs(a[pos]-a[pos+1]);

    long long right=abs(a[pos]-a[pos+2]);

    if(pos+1<n)

    {

        tot=min(tot,left+rec(pos+1));

    }

    if(pos+2<n)

    {

        tot=min(tot,right+rec(pos+2));

    }

    return mem[pos]=tot;

}



int main()

{

    memset(a,0,sizeof(a));

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

    scanf("%d",&n);

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

    {

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

    }

    long long ans=rec(0);

    printf("%lld\n",ans);

}


// Faster C++ solution for the same problem (measured speedup: 1.8489811769206959x):
#include <iostream>

#include <queue>

#include <set>

#include <list>

#include <deque>

#include <vector>

#include <map>

#include <utility>

#include <algorithm>

#include <cmath>

#include <cstdio>

#include <string>

#include <iterator>

#define ll long long

#define pb push_back

#define pf push_front

#define ppb pop_back

#define ppf pop_front

#define powr(x,n,p) for(int axy=0 ; axy<p ; axy++){x=x*n;}

#define YES printf("YES\n")

#define Yes printf("Yes\n")

#define yes printf("yes\n")

#define NO printf("NO\n")

#define No printf("No\n")

#define no printf("no\n")

#define nl printf("\n")

#define fr0(n,q) for(q=0 ; q<n ; q++)

#define fr1(n,q) for(q=1 ; q<=n ; q++)

#define mp make_pair

#define scl(x) scanf("%lld",&x)

#define sci(x) scanf("%d",&x)

#define lst(x) x[x.size()-1]

#define llst(x) x[x.size()-2]

#define md 1000000007

#define check(x,y) (x&(1<<y))

#define set(x,y) x=x|(1<<y)

using namespace std;

int dp[100005];

int arr[100005];

int n;

int solve(int x){

	if(x>=n) return 0;

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

	int ret;

	if(x==n-1) ret=abs(arr[x+1]-arr[x])+solve(x+1);

	else ret=min(abs(arr[x+1]-arr[x])+solve(x+1),abs(arr[x+2]-arr[x])+solve(x+2));

	return dp[x]=ret;

}

int main()

{

	int i;

	cin>>n;

	for(i=0 ; i<100005 ; i++) dp[i]=-1;

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

	cout<<solve(1);

	nl;

	return 0;

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

#include<cstdio>

#include<string>

#include<cstring>

#include<algorithm>

#include<queue>

#include<stack>

#include<map>

#include<vector>

#include<set>

#include<list>

#define ll long long

using namespace std;

ll h[200050],dp[200050];

int main()

{

	ll n;

	cin>>n;

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

		cin>>h[i];

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

	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: 1.8599350672472996x):
#include <iostream>

#include <queue>

#include <set>

#include <list>

#include <deque>

#include <vector>

#include <map>

#include <utility>

#include <algorithm>

#include <cmath>

#include <cstdio>

#include <string>

#include <iterator>

#define ll long long

#define pb push_back

#define pf push_front

#define ppb pop_back

#define ppf pop_front

#define powr(x,n,p) for(int axy=0 ; axy<p ; axy++){x=x*n;}

#define YES printf("YES\n")

#define Yes printf("Yes\n")

#define yes printf("yes\n")

#define NO printf("NO\n")

#define No printf("No\n")

#define no printf("no\n")

#define nl printf("\n")

#define fr0(n,q) for(q=0 ; q<n ; q++)

#define fr1(n,q) for(q=1 ; q<=n ; q++)

#define mp make_pair

#define scl(x) scanf("%lld",&x)

#define sci(x) scanf("%d",&x)

#define lst(x) x[x.size()-1]

#define llst(x) x[x.size()-2]

#define md 1000000007

#define check(x,y) (x&(1<<y))

#define set(x,y) x=x|(1<<y)

using namespace std;

int dp[100005];

int arr[100005];

int n;

int solve(int x){

	if(x>=n) return 0;

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

	int ret;

	if(x==n-1) ret=abs(arr[x+1]-arr[x])+solve(x+1);

	else ret=min(abs(arr[x+1]-arr[x])+solve(x+1),abs(arr[x+2]-arr[x])+solve(x+2));

	return dp[x]=ret;

}

int main()

{

	int i;

	cin>>n;

	for(i=0 ; i<100005 ; i++) dp[i]=-1;

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

	cout<<solve(1);

	nl;

	return 0;

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



using namespace std;

const int maxn = 2e5+2;

const int mod = 1e9+7;

const int inf = 1e9;



typedef long long ll;



int n;

int arr[maxn], dp[maxn];



int solve_dp (int pos){

    if (pos > n) return inf;

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



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



    int ans = inf;

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

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



    return dp[pos] = ans;

}



int main(){



    ios_base::sync_with_stdio(0);

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



    cin >> n;



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

        cin >> arr[i];

    }



    memset(dp, -1, sizeof dp);



    cout << solve_dp(0) << "\n";



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.297762584215265x):
#include <iostream>

#include <queue>

#include <set>

#include <list>

#include <deque>

#include <vector>

#include <map>

#include <utility>

#include <algorithm>

#include <cmath>

#include <cstdio>

#include <string>

#include <iterator>

#define ll long long

#define pb push_back

#define pf push_front

#define ppb pop_back

#define ppf pop_front

#define powr(x,n,p) for(int axy=0 ; axy<p ; axy++){x=x*n;}

#define YES printf("YES\n")

#define Yes printf("Yes\n")

#define yes printf("yes\n")

#define NO printf("NO\n")

#define No printf("No\n")

#define no printf("no\n")

#define nl printf("\n")

#define fr0(n,q) for(q=0 ; q<n ; q++)

#define fr1(n,q) for(q=1 ; q<=n ; q++)

#define mp make_pair

#define scl(x) scanf("%lld",&x)

#define sci(x) scanf("%d",&x)

#define lst(x) x[x.size()-1]

#define llst(x) x[x.size()-2]

#define md 1000000007

#define check(x,y) (x&(1<<y))

#define set(x,y) x=x|(1<<y)

using namespace std;

int dp[100005];

int arr[100005];

int n;

int solve(int x){

	if(x>=n) return 0;

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

	int ret;

	if(x==n-1) ret=abs(arr[x+1]-arr[x])+solve(x+1);

	else ret=min(abs(arr[x+1]-arr[x])+solve(x+1),abs(arr[x+2]-arr[x])+solve(x+2));

	return dp[x]=ret;

}

int main()

{

	int i;

	cin>>n;

	for(i=0 ; i<100005 ; i++) dp[i]=-1;

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

	cout<<solve(1);

	nl;

	return 0;

}
// Slower C++ solution:
#pragma GCC optimize ("O3")

#include <bits/stdc++.h>

#define abdelrahman010 ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

typedef long long ll;

using namespace std;

ll n , arr[100005] , memo[100005];

ll solve(ll i)

{

	if(i == n - 1)

		return 0;

	ll &ans = memo[i];

	if(~ans)

		return ans;

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

	if(i < n - 2)

		ans = min(ans , abs(arr[i] - arr[i + 2]) + solve(i + 2));

	return ans;

}

int main()

{

	abdelrahman010

	cin >> n;

	memset(memo , -1 , sizeof memo);

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

		cin >> arr[i];

	cout << solve(0);

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.2956458735684193x):
#include <iostream>

#include <queue>

#include <set>

#include <list>

#include <deque>

#include <vector>

#include <map>

#include <utility>

#include <algorithm>

#include <cmath>

#include <cstdio>

#include <string>

#include <iterator>

#define ll long long

#define pb push_back

#define pf push_front

#define ppb pop_back

#define ppf pop_front

#define powr(x,n,p) for(int axy=0 ; axy<p ; axy++){x=x*n;}

#define YES printf("YES\n")

#define Yes printf("Yes\n")

#define yes printf("yes\n")

#define NO printf("NO\n")

#define No printf("No\n")

#define no printf("no\n")

#define nl printf("\n")

#define fr0(n,q) for(q=0 ; q<n ; q++)

#define fr1(n,q) for(q=1 ; q<=n ; q++)

#define mp make_pair

#define scl(x) scanf("%lld",&x)

#define sci(x) scanf("%d",&x)

#define lst(x) x[x.size()-1]

#define llst(x) x[x.size()-2]

#define md 1000000007

#define check(x,y) (x&(1<<y))

#define set(x,y) x=x|(1<<y)

using namespace std;

int dp[100005];

int arr[100005];

int n;

int solve(int x){

	if(x>=n) return 0;

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

	int ret;

	if(x==n-1) ret=abs(arr[x+1]-arr[x])+solve(x+1);

	else ret=min(abs(arr[x+1]-arr[x])+solve(x+1),abs(arr[x+2]-arr[x])+solve(x+2));

	return dp[x]=ret;

}

int main()

{

	int i;

	cin>>n;

	for(i=0 ; i<100005 ; i++) dp[i]=-1;

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

	cout<<solve(1);

	nl;

	return 0;

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

#include<stdio.h>

#include<cstring>

#include<cmath>

#include<algorithm>

using namespace std;

#define maxn 100005

int dp[maxn];

int h[maxn];

int n;

int main(){

    scanf("%d",&n);

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

    memset(h,0,sizeof(h));

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

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

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

        if(i-1>=1&&i-2>=1)

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

        else if(i-1>=1) dp[i] = dp[i-1]+abs(h[i]-h[i-1]);

        else if(i-2>=1) dp[i] = dp[i-2]+abs(h[i]-h[i-2]);

    }

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.2778230962692405x):
#include <iostream>

#include <queue>

#include <set>

#include <list>

#include <deque>

#include <vector>

#include <map>

#include <utility>

#include <algorithm>

#include <cmath>

#include <cstdio>

#include <string>

#include <iterator>

#define ll long long

#define pb push_back

#define pf push_front

#define ppb pop_back

#define ppf pop_front

#define powr(x,n,p) for(int axy=0 ; axy<p ; axy++){x=x*n;}

#define YES printf("YES\n")

#define Yes printf("Yes\n")

#define yes printf("yes\n")

#define NO printf("NO\n")

#define No printf("No\n")

#define no printf("no\n")

#define nl printf("\n")

#define fr0(n,q) for(q=0 ; q<n ; q++)

#define fr1(n,q) for(q=1 ; q<=n ; q++)

#define mp make_pair

#define scl(x) scanf("%lld",&x)

#define sci(x) scanf("%d",&x)

#define lst(x) x[x.size()-1]

#define llst(x) x[x.size()-2]

#define md 1000000007

#define check(x,y) (x&(1<<y))

#define set(x,y) x=x|(1<<y)

using namespace std;

int dp[100005];

int arr[100005];

int n;

int solve(int x){

	if(x>=n) return 0;

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

	int ret;

	if(x==n-1) ret=abs(arr[x+1]-arr[x])+solve(x+1);

	else ret=min(abs(arr[x+1]-arr[x])+solve(x+1),abs(arr[x+2]-arr[x])+solve(x+2));

	return dp[x]=ret;

}

int main()

{

	int i;

	cin>>n;

	for(i=0 ; i<100005 ; i++) dp[i]=-1;

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

	cout<<solve(1);

	nl;

	return 0;

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

#include <bitset>

using namespace std;

#define ll long long

#define mod 1000000009



// يلعن ابو الي يعقدها على نفسه تاني



int main()

{

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



    ll n; cin >> n;

    ll arr[100005];

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

        cin >> arr[i];

    }

    if(n == 2){

        cout << abs(arr[1]-arr[0]);

    }else{

        ll sol[100005] = {};

        sol[1] = abs(arr[1]-arr[0]);

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

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

        }









        cout << sol[n-1];

    }















    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.1471195780687808x):
#include <iostream>

#include <queue>

#include <set>

#include <list>

#include <deque>

#include <vector>

#include <map>

#include <utility>

#include <algorithm>

#include <cmath>

#include <cstdio>

#include <string>

#include <iterator>

#define ll long long

#define pb push_back

#define pf push_front

#define ppb pop_back

#define ppf pop_front

#define powr(x,n,p) for(int axy=0 ; axy<p ; axy++){x=x*n;}

#define YES printf("YES\n")

#define Yes printf("Yes\n")

#define yes printf("yes\n")

#define NO printf("NO\n")

#define No printf("No\n")

#define no printf("no\n")

#define nl printf("\n")

#define fr0(n,q) for(q=0 ; q<n ; q++)

#define fr1(n,q) for(q=1 ; q<=n ; q++)

#define mp make_pair

#define scl(x) scanf("%lld",&x)

#define sci(x) scanf("%d",&x)

#define lst(x) x[x.size()-1]

#define llst(x) x[x.size()-2]

#define md 1000000007

#define check(x,y) (x&(1<<y))

#define set(x,y) x=x|(1<<y)

using namespace std;

int dp[100005];

int arr[100005];

int n;

int solve(int x){

	if(x>=n) return 0;

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

	int ret;

	if(x==n-1) ret=abs(arr[x+1]-arr[x])+solve(x+1);

	else ret=min(abs(arr[x+1]-arr[x])+solve(x+1),abs(arr[x+2]-arr[x])+solve(x+2));

	return dp[x]=ret;

}

int main()

{

	int i;

	cin>>n;

	for(i=0 ; i<100005 ; i++) dp[i]=-1;

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

	cout<<solve(1);

	nl;

	return 0;

}