stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
/*

** author:mahmoud magdy

*/

#include <bits/stdc++.h>

using namespace std;



#define ll long long

#define endl "\n"



inline int D()

{

    int m ;

    cin>>m;

    return m ;

}

inline ll lD()

{

    ll m ;

    cin>>m;

    return m ;

}

inline double dD()

{

    double m;

    cin>>m;

    return m;

}

void fast()

{

    ios_base :: sync_with_stdio( 0 ) ;

    cin.tie( 0 ) ;

    cout.tie( 0 ) ;

}

int a[100005];

int dp[100005];

int n;

int solve(int i){

 if(i>=n)return 1e9;

 if(i==n-1)

    return 0;

 if(dp[i]!=-1)

      return dp[i];

 int ch1=abs(a[i]-a[i+1])+solve(i+1);

 int ch2=abs(a[i]-a[i+2])+solve(i+2);

 return dp[i]=min(ch1,ch2);

}

int main()

{

    fast();

    n=D();

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

        a[i]=D();

    }

    memset(dp,-1,sizeof dp);

    cout<<solve(0)<<endl;

}


// Faster C++ solution for the same problem (measured speedup: 1.4141552879099044x):
// Example program

#include <bits/stdc++.h>



using namespace std;



int main()

{

    int n;

    cin >> n;

    

    int a[n];

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

    

    int dp[n];

    for (int i=0; i<n; i++) dp[i] = 1e9;

    

    dp[0] = 0;

    

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

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

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

    }

    

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

    return 0;

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

using namespace std;

const int N = 1e5+1;



vector<int> v(N, 0);

vector<int> memo(N, -1);



int cost(int pos, int n){

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

    else if (pos == n) return 0;

    else if (pos == n-1) return (abs(v[n-1] - v[n]));

    else {

        memo[pos] = min(abs(v[pos] - v[pos + 1]) + cost(pos+1, n), abs(v[pos]-v[pos + 2]) + cost(pos+2, n));

        return memo[pos];

    }

}



int main(){

    int n; cin >> n;

    

    for (int i = 1; i!=n+1; i++) cin >> v[i];

    cout << cost(1, n) << endl;



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.8130577763656506x):
// Example program

#include <bits/stdc++.h>



using namespace std;



int main()

{

    int n;

    cin >> n;

    

    int a[n];

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

    

    int dp[n];

    for (int i=0; i<n; i++) dp[i] = 1e9;

    

    dp[0] = 0;

    

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

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

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

    }

    

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

    return 0;

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

#define eps 1e-3

#define pi acos(-1.0)

#define inf 0x3f

#define INF 0x3f3f3f3f

#define pb push_back

#define debug1 cout<<"&&";

#define debug2 cout<<"**";

#define ms(a, x) memset(a, x, sizeof(a))

#define for0(i, n) for(int i = 0; i < n; ++i)

#define for1(i, n) for(int i = 1; i <= n; ++i)

using namespace std;

typedef double db;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

typedef pair<long long, int> pli;

typedef pair<long long, long long> pll;

const int mod = 1e9 + 7;

const int N = 1e5+10;

const int M = 1e6+10;

inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }

inline ll Pow(ll a, ll b) { ll ans = 1; for( ; b; b >>= 1) { if(b&1) ans = ans * a % mod; a = a * a % mod;} return ans;}

inline ll Mul ( ll a, ll b)  { ll lf = a * ( b >> 25ll ) % mod * ( 1LL << 25 ) % mod; ll rg = a * ( b & ( ( 1ll << 25 ) - 1 ) ) % mod; return ( lf + rg ) % mod;}

//==================================================================================================================================================================//



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



int main() {

   scanf("%d", &n);

   ms(dp, inf); dp[n] = 0;

   for1(i, n) scanf("%d", h+i);

   for(int i = n; i; --i) {

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

      else {

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

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

      }

   }

   printf("%d\n", dp[1]);

   return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3877776628024943x):
// Example program

#include <bits/stdc++.h>



using namespace std;



int main()

{

    int n;

    cin >> n;

    

    int a[n];

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

    

    int dp[n];

    for (int i=0; i<n; i++) dp[i] = 1e9;

    

    dp[0] = 0;

    

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

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

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

    }

    

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

    return 0;

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

using namespace std;



#define clr(arr,val) memset(arr,val ,sizeof arr)

//#define first f

//#define second s



typedef  long long ull;

typedef long long ll;

int const oo=(int)1e9 ;



int const SZ= (int)1e5+1;

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

int arr[SZ];

int dp[SZ];

int n;

//int mn=oo;

int solve(int i){

    

    if(i==n-1)

        return 0;

    if(i>=n)

        return oo;

    

    int &ret=dp[i];

    

    if(ret!=-1)

        return ret;

        

    int ch1=abs(arr[i]-arr[i+1])+solve(i+1);

    int ch2=abs(arr[i]-arr[i+2])+solve(i+2);

    ret=min(ch1,ch2);

    return ret;

    

}

int main() {

//    int t ;

//    cin>>t;

//    while(t--){

        cin>>n;

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

            cin>>arr[i];

        clr(dp,-1);

        cout<<solve(0)<<endl;

//    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4000995550338344x):
// Example program

#include <bits/stdc++.h>



using namespace std;



int main()

{

    int n;

    cin >> n;

    

    int a[n];

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

    

    int dp[n];

    for (int i=0; i<n; i++) dp[i] = 1e9;

    

    dp[0] = 0;

    

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

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

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

    }

    

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

    return 0;

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

using namespace std;

#define    sz    100110

#define    ll    long long int

ll n,c=0,p,q;

ll num[sz];

ll vis[sz];

ll dp[sz];



ll tani(int pos)

{



  if(pos>=n)return 0;

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

  vis[pos]=1;



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



}



int main()

{

    ll t,i;



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



    cin>>n;

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

    {

        cin>>num[i];

    }

    t=tani(1);

    cout<<t<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.8019904468805374x):
#include<bits/stdc++.h>

using namespace std;

typedef		long long       ll;

typedef         long double      dl;

typedef     unsigned long long  ull;

#define		pb              push_back

#define		PB              pop_back

#define		nn              "\n"

#define     O_O             ios_base::sync_with_stdio(false); cin.tie(NULL)

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

#define		zz(v)           (ll)v.size()

#define		ss              ' '

#define	       MEM(a,b)         memset(a,(b),sizeof(a))

#define	       CLR(p)           memset(p,0,sizeof(p))

#define f0(i,b) for(int i=0;i<(b);i++)

#define f1(i,b) for(int i=1;i<=(b);i++)

#define f2(i,a,b) for(int i=(a);i<=(b);i++)

#define fr(i,b,a) for(int i=(b);i>=(a);i--)

#define rep(i,a,b,c) for(int i=(a);i!=(b);i+=(c))

#define         arrsize(a)      (sizeof(a)/sizeof(a[0]))

//#define       arrsize(a)      (sizeof(a)/sizeof(*a))

#define 	S(a)            scanf("%lld",&a)

#define 	SS(a,b)         scanf("%lld %lld",&a,&b)

#define 	SSS(a,b,c)      scanf("%lld %lld %lld",&a,&b,&c)

#define		gcd(a,b)        __gcd(a,b)

#define		lcm(a,b)        (a*b)/gcd(a,b)

#define		pi              acos(-1.0)

#define	       ff               first

#define	       sc              second

typedef        pair <int, int>   pii;

typedef       pair <ll, ll>      pll;

typedef      vector< pair <ll, ll> >      vpll;

typedef		vector<ll>      vll;

typedef		map<string,ll>      msl;

typedef		map<ll,ll>      mll;

#define yes cout << "YES\n"

#define no cout<<"NO\n"

//memset(ar,-1,sizeof(ar));

//#define sort(x) sort(x.begin(), x.end())

//sort(a,a+n,greater<ll>())

//for (auto it = mp.begin(); it != mp.end(); ++it){}

// string x(w.size(),'1');

#define MAX 1000005

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



#define mod 1000000007



ll ar[MAX],ans[MAX];



int main()

{

 //O_O ;

   ll n;

   cin>>n;

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

    cin>>ar[i];

   ans[n]=0,

   ans[n-1]=abs(ar[n-1]-ar[n]);

   for(ll i=n-2;i>=1;i--)

   {

       ans[i]=min((abs(ar[i]-ar[i+1])+ans[i+1]),(abs(ar[i]-ar[i+2])+ans[i+2]));

      // cout<<<<ss<<i<<nn;

       // cout<<(abs(ar[i]-ar[i+2])+ans[i+2])<<ss<<i<<nn;



   }

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

   // cout<<ans[i]<<ss;cout<<nn;

   cout<<ans[1]<<nn;



    return 0;

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

using namespace std;



#define fr(i,n) _back

#define pb	push_back

#define eb	emplace_back

#define mk	make_pair

#define fi	first

#define se	second

#define endl	'\n'



typedef long long ll;

typedef pair<int,int> ii;

typedef vector<ii> vii;

const int INF = 0x3f3f3f3f;

const double PI = acos(-1.0);

const int N = 1e5 + 3;

int h[N];

int dp[N];



int main() {

    ios_base::sync_with_stdio(false);

    memset(dp, INF, sizeof dp);

    int n; cin >> n;

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



    dp[0] = 0;

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

        for(int j = max(0,i-2); j < i; j++)

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



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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4167801498607182x):
#include<bits/stdc++.h>

using namespace std;

typedef		long long       ll;

typedef         long double      dl;

typedef     unsigned long long  ull;

#define		pb              push_back

#define		PB              pop_back

#define		nn              "\n"

#define     O_O             ios_base::sync_with_stdio(false); cin.tie(NULL)

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

#define		zz(v)           (ll)v.size()

#define		ss              ' '

#define	       MEM(a,b)         memset(a,(b),sizeof(a))

#define	       CLR(p)           memset(p,0,sizeof(p))

#define f0(i,b) for(int i=0;i<(b);i++)

#define f1(i,b) for(int i=1;i<=(b);i++)

#define f2(i,a,b) for(int i=(a);i<=(b);i++)

#define fr(i,b,a) for(int i=(b);i>=(a);i--)

#define rep(i,a,b,c) for(int i=(a);i!=(b);i+=(c))

#define         arrsize(a)      (sizeof(a)/sizeof(a[0]))

//#define       arrsize(a)      (sizeof(a)/sizeof(*a))

#define 	S(a)            scanf("%lld",&a)

#define 	SS(a,b)         scanf("%lld %lld",&a,&b)

#define 	SSS(a,b,c)      scanf("%lld %lld %lld",&a,&b,&c)

#define		gcd(a,b)        __gcd(a,b)

#define		lcm(a,b)        (a*b)/gcd(a,b)

#define		pi              acos(-1.0)

#define	       ff               first

#define	       sc              second

typedef        pair <int, int>   pii;

typedef       pair <ll, ll>      pll;

typedef      vector< pair <ll, ll> >      vpll;

typedef		vector<ll>      vll;

typedef		map<string,ll>      msl;

typedef		map<ll,ll>      mll;

#define yes cout << "YES\n"

#define no cout<<"NO\n"

//memset(ar,-1,sizeof(ar));

//#define sort(x) sort(x.begin(), x.end())

//sort(a,a+n,greater<ll>())

//for (auto it = mp.begin(); it != mp.end(); ++it){}

// string x(w.size(),'1');

#define MAX 1000005

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



#define mod 1000000007



ll ar[MAX],ans[MAX];



int main()

{

 //O_O ;

   ll n;

   cin>>n;

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

    cin>>ar[i];

   ans[n]=0,

   ans[n-1]=abs(ar[n-1]-ar[n]);

   for(ll i=n-2;i>=1;i--)

   {

       ans[i]=min((abs(ar[i]-ar[i+1])+ans[i+1]),(abs(ar[i]-ar[i+2])+ans[i+2]));

      // cout<<<<ss<<i<<nn;

       // cout<<(abs(ar[i]-ar[i+2])+ans[i+2])<<ss<<i<<nn;



   }

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

   // cout<<ans[i]<<ss;cout<<nn;

   cout<<ans[1]<<nn;



    return 0;

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

using namespace std;



int N, a[100100], dp[100100];



int f(int n) {

	if (n==1) return 0;

	if (n==2) return abs(a[2] - a[1]);

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

	dp[n] = min(

				f(n-1) + abs(a[n] - a[n-1]),

				f(n-2) + abs(a[n] - a[n-2])

				);

	return dp[n];

}



int main() {

	cin >> N;

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

		cin >> a[i];

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

	cout << f(N) << endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4009394862114382x):
#include<bits/stdc++.h>

using namespace std;

typedef		long long       ll;

typedef         long double      dl;

typedef     unsigned long long  ull;

#define		pb              push_back

#define		PB              pop_back

#define		nn              "\n"

#define     O_O             ios_base::sync_with_stdio(false); cin.tie(NULL)

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

#define		zz(v)           (ll)v.size()

#define		ss              ' '

#define	       MEM(a,b)         memset(a,(b),sizeof(a))

#define	       CLR(p)           memset(p,0,sizeof(p))

#define f0(i,b) for(int i=0;i<(b);i++)

#define f1(i,b) for(int i=1;i<=(b);i++)

#define f2(i,a,b) for(int i=(a);i<=(b);i++)

#define fr(i,b,a) for(int i=(b);i>=(a);i--)

#define rep(i,a,b,c) for(int i=(a);i!=(b);i+=(c))

#define         arrsize(a)      (sizeof(a)/sizeof(a[0]))

//#define       arrsize(a)      (sizeof(a)/sizeof(*a))

#define 	S(a)            scanf("%lld",&a)

#define 	SS(a,b)         scanf("%lld %lld",&a,&b)

#define 	SSS(a,b,c)      scanf("%lld %lld %lld",&a,&b,&c)

#define		gcd(a,b)        __gcd(a,b)

#define		lcm(a,b)        (a*b)/gcd(a,b)

#define		pi              acos(-1.0)

#define	       ff               first

#define	       sc              second

typedef        pair <int, int>   pii;

typedef       pair <ll, ll>      pll;

typedef      vector< pair <ll, ll> >      vpll;

typedef		vector<ll>      vll;

typedef		map<string,ll>      msl;

typedef		map<ll,ll>      mll;

#define yes cout << "YES\n"

#define no cout<<"NO\n"

//memset(ar,-1,sizeof(ar));

//#define sort(x) sort(x.begin(), x.end())

//sort(a,a+n,greater<ll>())

//for (auto it = mp.begin(); it != mp.end(); ++it){}

// string x(w.size(),'1');

#define MAX 1000005

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



#define mod 1000000007



ll ar[MAX],ans[MAX];



int main()

{

 //O_O ;

   ll n;

   cin>>n;

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

    cin>>ar[i];

   ans[n]=0,

   ans[n-1]=abs(ar[n-1]-ar[n]);

   for(ll i=n-2;i>=1;i--)

   {

       ans[i]=min((abs(ar[i]-ar[i+1])+ans[i+1]),(abs(ar[i]-ar[i+2])+ans[i+2]));

      // cout<<<<ss<<i<<nn;

       // cout<<(abs(ar[i]-ar[i+2])+ans[i+2])<<ss<<i<<nn;



   }

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

   // cout<<ans[i]<<ss;cout<<nn;

   cout<<ans[1]<<nn;



    return 0;

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

using namespace std;

typedef long long LL;



int arr[100007];

int dp[100007];



int Solve(int idx, int n){

//	cout<<cost<<" "<<idx<<endl;

	if(idx >= n)return 1e9;

	if(idx == n - 1){

		return 0;	

	}

	

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

	

	

	return dp[idx] = min(abs(arr[idx] - arr[idx+1]) + Solve(idx+1, n), abs(arr[idx] - arr[idx+2]) + Solve(idx+2,n));

}



int main(){

	int n; cin>>n;

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

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

		

	cout<<Solve(0,n)<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4015356127143843x):
#include<bits/stdc++.h>

using namespace std;

typedef		long long       ll;

typedef         long double      dl;

typedef     unsigned long long  ull;

#define		pb              push_back

#define		PB              pop_back

#define		nn              "\n"

#define     O_O             ios_base::sync_with_stdio(false); cin.tie(NULL)

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

#define		zz(v)           (ll)v.size()

#define		ss              ' '

#define	       MEM(a,b)         memset(a,(b),sizeof(a))

#define	       CLR(p)           memset(p,0,sizeof(p))

#define f0(i,b) for(int i=0;i<(b);i++)

#define f1(i,b) for(int i=1;i<=(b);i++)

#define f2(i,a,b) for(int i=(a);i<=(b);i++)

#define fr(i,b,a) for(int i=(b);i>=(a);i--)

#define rep(i,a,b,c) for(int i=(a);i!=(b);i+=(c))

#define         arrsize(a)      (sizeof(a)/sizeof(a[0]))

//#define       arrsize(a)      (sizeof(a)/sizeof(*a))

#define 	S(a)            scanf("%lld",&a)

#define 	SS(a,b)         scanf("%lld %lld",&a,&b)

#define 	SSS(a,b,c)      scanf("%lld %lld %lld",&a,&b,&c)

#define		gcd(a,b)        __gcd(a,b)

#define		lcm(a,b)        (a*b)/gcd(a,b)

#define		pi              acos(-1.0)

#define	       ff               first

#define	       sc              second

typedef        pair <int, int>   pii;

typedef       pair <ll, ll>      pll;

typedef      vector< pair <ll, ll> >      vpll;

typedef		vector<ll>      vll;

typedef		map<string,ll>      msl;

typedef		map<ll,ll>      mll;

#define yes cout << "YES\n"

#define no cout<<"NO\n"

//memset(ar,-1,sizeof(ar));

//#define sort(x) sort(x.begin(), x.end())

//sort(a,a+n,greater<ll>())

//for (auto it = mp.begin(); it != mp.end(); ++it){}

// string x(w.size(),'1');

#define MAX 1000005

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



#define mod 1000000007



ll ar[MAX],ans[MAX];



int main()

{

 //O_O ;

   ll n;

   cin>>n;

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

    cin>>ar[i];

   ans[n]=0,

   ans[n-1]=abs(ar[n-1]-ar[n]);

   for(ll i=n-2;i>=1;i--)

   {

       ans[i]=min((abs(ar[i]-ar[i+1])+ans[i+1]),(abs(ar[i]-ar[i+2])+ans[i+2]));

      // cout<<<<ss<<i<<nn;

       // cout<<(abs(ar[i]-ar[i+2])+ans[i+2])<<ss<<i<<nn;



   }

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

   // cout<<ans[i]<<ss;cout<<nn;

   cout<<ans[1]<<nn;



    return 0;

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

using namespace std;

int dp[100005];

int H[100005];

void memo(int i,int n)

{

	if(i==n) return;

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

	memo(i+1,n);

}

int main()

{

	int N;cin>>N;

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

	{

        cin>>H[i];

	}

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

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

	memo(2,N);

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4008299410005438x):
#include<bits/stdc++.h>

using namespace std;

typedef		long long       ll;

typedef         long double      dl;

typedef     unsigned long long  ull;

#define		pb              push_back

#define		PB              pop_back

#define		nn              "\n"

#define     O_O             ios_base::sync_with_stdio(false); cin.tie(NULL)

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

#define		zz(v)           (ll)v.size()

#define		ss              ' '

#define	       MEM(a,b)         memset(a,(b),sizeof(a))

#define	       CLR(p)           memset(p,0,sizeof(p))

#define f0(i,b) for(int i=0;i<(b);i++)

#define f1(i,b) for(int i=1;i<=(b);i++)

#define f2(i,a,b) for(int i=(a);i<=(b);i++)

#define fr(i,b,a) for(int i=(b);i>=(a);i--)

#define rep(i,a,b,c) for(int i=(a);i!=(b);i+=(c))

#define         arrsize(a)      (sizeof(a)/sizeof(a[0]))

//#define       arrsize(a)      (sizeof(a)/sizeof(*a))

#define 	S(a)            scanf("%lld",&a)

#define 	SS(a,b)         scanf("%lld %lld",&a,&b)

#define 	SSS(a,b,c)      scanf("%lld %lld %lld",&a,&b,&c)

#define		gcd(a,b)        __gcd(a,b)

#define		lcm(a,b)        (a*b)/gcd(a,b)

#define		pi              acos(-1.0)

#define	       ff               first

#define	       sc              second

typedef        pair <int, int>   pii;

typedef       pair <ll, ll>      pll;

typedef      vector< pair <ll, ll> >      vpll;

typedef		vector<ll>      vll;

typedef		map<string,ll>      msl;

typedef		map<ll,ll>      mll;

#define yes cout << "YES\n"

#define no cout<<"NO\n"

//memset(ar,-1,sizeof(ar));

//#define sort(x) sort(x.begin(), x.end())

//sort(a,a+n,greater<ll>())

//for (auto it = mp.begin(); it != mp.end(); ++it){}

// string x(w.size(),'1');

#define MAX 1000005

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



#define mod 1000000007



ll ar[MAX],ans[MAX];



int main()

{

 //O_O ;

   ll n;

   cin>>n;

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

    cin>>ar[i];

   ans[n]=0,

   ans[n-1]=abs(ar[n-1]-ar[n]);

   for(ll i=n-2;i>=1;i--)

   {

       ans[i]=min((abs(ar[i]-ar[i+1])+ans[i+1]),(abs(ar[i]-ar[i+2])+ans[i+2]));

      // cout<<<<ss<<i<<nn;

       // cout<<(abs(ar[i]-ar[i+2])+ans[i+2])<<ss<<i<<nn;



   }

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

   // cout<<ans[i]<<ss;cout<<nn;

   cout<<ans[1]<<nn;



    return 0;

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



using namespace std;



int dp[(int)1e5+5];

int main()

{



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







    fill(dp, dp+(int)1e5+5 , INT_MAX );



    dp[1] = 0;

    int n;

    cin>>n;

    vector <int> v(n+3,0);

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



        cin>>v[i];



    }

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



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

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



    }

    cout<<dp[n]<<"\n";

}


// Faster C++ solution for the same problem (measured speedup: 1.4199683456821561x):
#include<bits/stdc++.h>

using namespace std;

typedef		long long       ll;

typedef         long double      dl;

typedef     unsigned long long  ull;

#define		pb              push_back

#define		PB              pop_back

#define		nn              "\n"

#define     O_O             ios_base::sync_with_stdio(false); cin.tie(NULL)

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

#define		zz(v)           (ll)v.size()

#define		ss              ' '

#define	       MEM(a,b)         memset(a,(b),sizeof(a))

#define	       CLR(p)           memset(p,0,sizeof(p))

#define f0(i,b) for(int i=0;i<(b);i++)

#define f1(i,b) for(int i=1;i<=(b);i++)

#define f2(i,a,b) for(int i=(a);i<=(b);i++)

#define fr(i,b,a) for(int i=(b);i>=(a);i--)

#define rep(i,a,b,c) for(int i=(a);i!=(b);i+=(c))

#define         arrsize(a)      (sizeof(a)/sizeof(a[0]))

//#define       arrsize(a)      (sizeof(a)/sizeof(*a))

#define 	S(a)            scanf("%lld",&a)

#define 	SS(a,b)         scanf("%lld %lld",&a,&b)

#define 	SSS(a,b,c)      scanf("%lld %lld %lld",&a,&b,&c)

#define		gcd(a,b)        __gcd(a,b)

#define		lcm(a,b)        (a*b)/gcd(a,b)

#define		pi              acos(-1.0)

#define	       ff               first

#define	       sc              second

typedef        pair <int, int>   pii;

typedef       pair <ll, ll>      pll;

typedef      vector< pair <ll, ll> >      vpll;

typedef		vector<ll>      vll;

typedef		map<string,ll>      msl;

typedef		map<ll,ll>      mll;

#define yes cout << "YES\n"

#define no cout<<"NO\n"

//memset(ar,-1,sizeof(ar));

//#define sort(x) sort(x.begin(), x.end())

//sort(a,a+n,greater<ll>())

//for (auto it = mp.begin(); it != mp.end(); ++it){}

// string x(w.size(),'1');

#define MAX 1000005

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



#define mod 1000000007



ll ar[MAX],ans[MAX];



int main()

{

 //O_O ;

   ll n;

   cin>>n;

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

    cin>>ar[i];

   ans[n]=0,

   ans[n-1]=abs(ar[n-1]-ar[n]);

   for(ll i=n-2;i>=1;i--)

   {

       ans[i]=min((abs(ar[i]-ar[i+1])+ans[i+1]),(abs(ar[i]-ar[i+2])+ans[i+2]));

      // cout<<<<ss<<i<<nn;

       // cout<<(abs(ar[i]-ar[i+2])+ans[i+2])<<ss<<i<<nn;



   }

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

   // cout<<ans[i]<<ss;cout<<nn;

   cout<<ans[1]<<nn;



    return 0;

}