stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 2.5278752349165923x):
#include<bits/stdc++.h>

using namespace std;

typedef vector<int> vi;

typedef vector<long long> vll;

typedef pair<int,int> pii;

typedef vector< pii > vii;



#define ll          long long

#define ff          first

#define ss          second

#define pb          push_back

#define mkp         make_pair

#define sz          size()



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

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

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

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

#define pi          2*acos(0.0)

#define End         return 0;

#define nl endl



#define scan1(a)    scanf("%d",&a)

#define scan2(a,b)  scanf("%d %d",&a,&b)

#define scan3(a,b,c)  scanf("%d %d %d",&a,&b,&c)

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

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

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



#define print1(a)    printf("%d",a)

#define print2(a,b)  printf("%d %d",a,b)

#define print3(a,b,c)  printf("%d %d %d",a,b,c)

#define printl(a)    printf("%lld",a)

#define printll(a,b) printf("%lld %lld",a,b)

#define printlll(a,b,c) printf("%lld %lld %lld",a,b,c)



#define MAXN        100000007

#define INF 0x3f3f3f3f

int main()

{

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

   int n;

   cin>>n;

   vll v(n+3),vv(n+3);

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



   vv[1]=0;vv[2]=abs(v[2]-v[1]);

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

    vv[i]=min(vv[i-1]+abs(v[i]-v[i-1]), vv[i-2]+abs(v[i]-v[i-2]));

   }

   cout<<vv[n]<<endl;

   return 0;

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

using namespace std;

typedef long long int ll;

const int MAX=1e5+10;

int N,A[MAX],dp[MAX];

int solve(int idx)

{

    if(idx==N)return 0;

    if(idx>N)return 1e9;

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

    return dp[idx]=min(abs(A[idx+1]-A[idx])+solve(idx+1),abs(A[idx+2]-A[idx])+solve(idx+2));

}

int  main ()

{

    memset(dp,-1,sizeof dp);

    cin >> N;

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

        cin >> A[i];

    }

    cout<<solve(1)<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3623966118662825x):
#include<bits/stdc++.h>

using namespace std;

typedef vector<int> vi;

typedef vector<long long> vll;

typedef pair<int,int> pii;

typedef vector< pii > vii;



#define ll          long long

#define ff          first

#define ss          second

#define pb          push_back

#define mkp         make_pair

#define sz          size()



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

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

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

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

#define pi          2*acos(0.0)

#define End         return 0;

#define nl endl



#define scan1(a)    scanf("%d",&a)

#define scan2(a,b)  scanf("%d %d",&a,&b)

#define scan3(a,b,c)  scanf("%d %d %d",&a,&b,&c)

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

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

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



#define print1(a)    printf("%d",a)

#define print2(a,b)  printf("%d %d",a,b)

#define print3(a,b,c)  printf("%d %d %d",a,b,c)

#define printl(a)    printf("%lld",a)

#define printll(a,b) printf("%lld %lld",a,b)

#define printlll(a,b,c) printf("%lld %lld %lld",a,b,c)



#define MAXN        100000007

#define INF 0x3f3f3f3f

int main()

{

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

   int n;

   cin>>n;

   vll v(n+3),vv(n+3);

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



   vv[1]=0;vv[2]=abs(v[2]-v[1]);

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

    vv[i]=min(vv[i-1]+abs(v[i]-v[i-1]), vv[i-2]+abs(v[i]-v[i-2]));

   }

   cout<<vv[n]<<endl;

   return 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: 2.590287842960429x):
#include<bits/stdc++.h>

using namespace std;



int arr[100010];

int dp[100010];



int main (){

	int n;

	cin >> n;

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

		cin >> arr[i];

	}

	dp[1] = 0;

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

	

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

		dp[i] = min(abs(arr[i] - arr[i - 1]) + dp[i - 1], abs(arr[i] - arr[i - 2]) + dp[i - 2]);		//nyari dp

	}

	

	cout << dp[n] << endl;

}
// Slower C++ solution:
// Created by ahmed_drawy on 10/09/19.

///dp educational contest atcoder - vjudge

#include <bits/stdc++.h>

#include <string.h>

#pragma clang diagnostic push

#pragma ide diagnostic ignored "OCUnusedMacroInspection"

using namespace std;

#define lp(i,start , end)       for(int i = start ; i<end ; ++i)

#define lllp(i,start , end)       for(ll i = start ; i<end ; ++i)

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

#define Rlp(i,start , end)      for(int i = start ; i>end ; --i)

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

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

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

#define inf 0x3f3f3f3f

//#define dbg (v)    for(auto it : v)cout<<it<<' ';

typedef long long               ll;

typedef vector<int>             vi;

typedef vector  <ll>            vll;

typedef vector<vector<int> >    adj;

typedef pair<int ,int>          pii;

const double EPS =1e-7;

const int OO = 1e6;

#define endl '\n'

#define cint(c) (int)c -'0';

#define itc(c) (char)c +'A';

bool sortpair( const pair<int , int> & x, const pair<int , int> & y){

    return  x.first != y.first ? x.first < y.first : x.second > y.second ;

}

int dcomp(double x , double y){

    return fabs(x-y) <= EPS? 0: x>y?1:-1;

}



void smile() {

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

#ifndef ONLINE_JUDGE

//    freopen("/home/tw3/Desktop/Desktop/training/in.txt", "r", stdin);

//    freopen("/home/tw3/Desktop/Desktop/training/out.txt" , "w" , stdout);

#endif

}

#define fastin smile()



int n ;

const int N = 1e5 +3;

ll dp[N];

int h[N];

//ll solve(int indx , int state){

//    if(indx == n-1 )return 0 ;

//    ll &ret = dp[indx  ] [ state];

//    if(~ret )

//        return ret;

//    ret=  INT_MAX;

//    for (int i = 1; i <=2 && i+indx < n ; ++i) {

//        ret = min(ret , abs(h[indx]  - h[indx+ i])+solve(indx + i , i ));

//    }

//    return ret ;

//}

int main() {

    fastin;

    cin >> n;

    inN(h,  n );

    clr(dp , 0);

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

        if(j>=2){

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

        }

        else if(j ==1){

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

        }

    }

    cout << dp[n-1];

//

//    clr(dp , -1);

//    cout << solve(0, 1)<<endl;

}

// Faster C++ solution for the same problem (measured speedup: 1.8150911085196635x):
#include<bits/stdc++.h>

using namespace std;



int arr[100010];

int dp[100010];



int main (){

	int n;

	cin >> n;

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

		cin >> arr[i];

	}

	dp[1] = 0;

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

	

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

		dp[i] = min(abs(arr[i] - arr[i - 1]) + dp[i - 1], abs(arr[i] - arr[i - 2]) + dp[i - 2]);		//nyari dp

	}

	

	cout << dp[n] << endl;

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

using namespace std;

typedef long long int ll;

const int MAX=1e5+10;

int N,A[MAX],dp[MAX];

int solve(int idx)

{

    if(idx==N)return 0;

    if(idx>N)return 1e9;

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

    return dp[idx]=min(abs(A[idx+1]-A[idx])+solve(idx+1),abs(A[idx+2]-A[idx])+solve(idx+2));

}

int  main ()

{

    memset(dp,-1,sizeof dp);

    cin >> N;

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

        cin >> A[i];

    }

    cout<<solve(1)<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3960338438634017x):
#include<bits/stdc++.h>

using namespace std;



int arr[100010];

int dp[100010];



int main (){

	int n;

	cin >> n;

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

		cin >> arr[i];

	}

	dp[1] = 0;

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

	

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

		dp[i] = min(abs(arr[i] - arr[i - 1]) + dp[i - 1], abs(arr[i] - arr[i - 2]) + dp[i - 2]);		//nyari dp

	}

	

	cout << dp[n] << endl;

}
// 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: 2.5935012456577433x):
#include<bits/stdc++.h>

using namespace std;



int n,h[100005],c[100005];



int main()

{

    cin>>n;

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

        cin>>h[i];

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

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

    {

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

    }

    cout<<c[n-1];

}
// Slower C++ solution:
// Created by ahmed_drawy on 10/09/19.

///dp educational contest atcoder - vjudge

#include <bits/stdc++.h>

#include <string.h>

#pragma clang diagnostic push

#pragma ide diagnostic ignored "OCUnusedMacroInspection"

using namespace std;

#define lp(i,start , end)       for(int i = start ; i<end ; ++i)

#define lllp(i,start , end)       for(ll i = start ; i<end ; ++i)

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

#define Rlp(i,start , end)      for(int i = start ; i>end ; --i)

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

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

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

#define inf 0x3f3f3f3f

//#define dbg (v)    for(auto it : v)cout<<it<<' ';

typedef long long               ll;

typedef vector<int>             vi;

typedef vector  <ll>            vll;

typedef vector<vector<int> >    adj;

typedef pair<int ,int>          pii;

const double EPS =1e-7;

const int OO = 1e6;

#define endl '\n'

#define cint(c) (int)c -'0';

#define itc(c) (char)c +'A';

bool sortpair( const pair<int , int> & x, const pair<int , int> & y){

    return  x.first != y.first ? x.first < y.first : x.second > y.second ;

}

int dcomp(double x , double y){

    return fabs(x-y) <= EPS? 0: x>y?1:-1;

}



void smile() {

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

#ifndef ONLINE_JUDGE

//    freopen("/home/tw3/Desktop/Desktop/training/in.txt", "r", stdin);

//    freopen("/home/tw3/Desktop/Desktop/training/out.txt" , "w" , stdout);

#endif

}

#define fastin smile()



int n ;

const int N = 1e5 +3;

ll dp[N];

int h[N];

//ll solve(int indx , int state){

//    if(indx == n-1 )return 0 ;

//    ll &ret = dp[indx  ] [ state];

//    if(~ret )

//        return ret;

//    ret=  INT_MAX;

//    for (int i = 1; i <=2 && i+indx < n ; ++i) {

//        ret = min(ret , abs(h[indx]  - h[indx+ i])+solve(indx + i , i ));

//    }

//    return ret ;

//}

int main() {

    fastin;

    cin >> n;

    inN(h,  n );

    clr(dp , 0);

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

        if(j>=2){

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

        }

        else if(j ==1){

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

        }

    }

    cout << dp[n-1];

//

//    clr(dp , -1);

//    cout << solve(0, 1)<<endl;

}

// Faster C++ solution for the same problem (measured speedup: 1.8173428345893508x):
#include<bits/stdc++.h>

using namespace std;



int n,h[100005],c[100005];



int main()

{

    cin>>n;

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

        cin>>h[i];

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

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

    {

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

    }

    cout<<c[n-1];

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

using namespace std;

typedef long long int ll;

const int MAX=1e5+10;

int N,A[MAX],dp[MAX];

int solve(int idx)

{

    if(idx==N)return 0;

    if(idx>N)return 1e9;

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

    return dp[idx]=min(abs(A[idx+1]-A[idx])+solve(idx+1),abs(A[idx+2]-A[idx])+solve(idx+2));

}

int  main ()

{

    memset(dp,-1,sizeof dp);

    cin >> N;

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

        cin >> A[i];

    }

    cout<<solve(1)<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3977657050276362x):
#include<bits/stdc++.h>

using namespace std;



int n,h[100005],c[100005];



int main()

{

    cin>>n;

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

        cin>>h[i];

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

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

    {

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

    }

    cout<<c[n-1];

}
// 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: 1.828696530101189x):
#include <bits/stdc++.h>

using namespace std;



#define mp make_pair

#define pb push_back

#define st first

#define nd second

#define lb lower_bound

#define ub upper_bound



typedef long long ll;

typedef long double ld;

typedef pair<int,int> pi;

typedef pair<ll,ll> pll;

typedef vector<int> vi;

typedef vector<ll> vll;

typedef vector<pi> vpi;

typedef vector<pll> vpll;



const ll oo=1e18;

const int MOD=1000000007;

const int INF=1000000000;

const ld EPS=0.000000001;



int dp[100005],h[100005],n,k=2;



signed main() {

    ios_base::sync_with_stdio(NULL);

    cin.tie(NULL);

    cout.tie(NULL);

    memset(dp,0x3f3f,sizeof dp);

    dp[1]=0;

    h[0]=0;

    cin>>n;

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

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

        for(int j=1;j<=min(k,i);++j) {

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

        }

    }

    cout<<dp[n];

    return 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: 2.623554605221572x):
#include<bits/stdc++.h>

using namespace std;

long long n,h[100010],dp[100010];



int main()

{

	scanf("%lld",&n);

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

	{

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

	}

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

	{

		if(i-1>=0) dp[i] = dp[i-1]+abs(h[i-1]-h[i]);

		if(i>2)

		{

			

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

		}

	}

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

	return 0;

}
// Slower C++ solution:
// Created by ahmed_drawy on 10/09/19.

///dp educational contest atcoder - vjudge

#include <bits/stdc++.h>

#include <string.h>

#pragma clang diagnostic push

#pragma ide diagnostic ignored "OCUnusedMacroInspection"

using namespace std;

#define lp(i,start , end)       for(int i = start ; i<end ; ++i)

#define lllp(i,start , end)       for(ll i = start ; i<end ; ++i)

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

#define Rlp(i,start , end)      for(int i = start ; i>end ; --i)

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

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

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

#define inf 0x3f3f3f3f

//#define dbg (v)    for(auto it : v)cout<<it<<' ';

typedef long long               ll;

typedef vector<int>             vi;

typedef vector  <ll>            vll;

typedef vector<vector<int> >    adj;

typedef pair<int ,int>          pii;

const double EPS =1e-7;

const int OO = 1e6;

#define endl '\n'

#define cint(c) (int)c -'0';

#define itc(c) (char)c +'A';

bool sortpair( const pair<int , int> & x, const pair<int , int> & y){

    return  x.first != y.first ? x.first < y.first : x.second > y.second ;

}

int dcomp(double x , double y){

    return fabs(x-y) <= EPS? 0: x>y?1:-1;

}



void smile() {

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

#ifndef ONLINE_JUDGE

//    freopen("/home/tw3/Desktop/Desktop/training/in.txt", "r", stdin);

//    freopen("/home/tw3/Desktop/Desktop/training/out.txt" , "w" , stdout);

#endif

}

#define fastin smile()



int n ;

const int N = 1e5 +3;

ll dp[N];

int h[N];

//ll solve(int indx , int state){

//    if(indx == n-1 )return 0 ;

//    ll &ret = dp[indx  ] [ state];

//    if(~ret )

//        return ret;

//    ret=  INT_MAX;

//    for (int i = 1; i <=2 && i+indx < n ; ++i) {

//        ret = min(ret , abs(h[indx]  - h[indx+ i])+solve(indx + i , i ));

//    }

//    return ret ;

//}

int main() {

    fastin;

    cin >> n;

    inN(h,  n );

    clr(dp , 0);

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

        if(j>=2){

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

        }

        else if(j ==1){

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

        }

    }

    cout << dp[n-1];

//

//    clr(dp , -1);

//    cout << solve(0, 1)<<endl;

}

// Faster C++ solution for the same problem (measured speedup: 1.8384021102500456x):
#include<bits/stdc++.h>

using namespace std;

long long n,h[100010],dp[100010];



int main()

{

	scanf("%lld",&n);

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

	{

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

	}

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

	{

		if(i-1>=0) dp[i] = dp[i-1]+abs(h[i-1]-h[i]);

		if(i>2)

		{

			

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

		}

	}

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

	return 0;

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

using namespace std;

typedef long long int ll;

const int MAX=1e5+10;

int N,A[MAX],dp[MAX];

int solve(int idx)

{

    if(idx==N)return 0;

    if(idx>N)return 1e9;

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

    return dp[idx]=min(abs(A[idx+1]-A[idx])+solve(idx+1),abs(A[idx+2]-A[idx])+solve(idx+2));

}

int  main ()

{

    memset(dp,-1,sizeof dp);

    cin >> N;

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

        cin >> A[i];

    }

    cout<<solve(1)<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.413962942406842x):
#include<bits/stdc++.h>

using namespace std;

long long n,h[100010],dp[100010];



int main()

{

	scanf("%lld",&n);

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

	{

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

	}

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

	{

		if(i-1>=0) dp[i] = dp[i-1]+abs(h[i-1]-h[i]);

		if(i>2)

		{

			

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

		}

	}

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

	return 0;

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

using namespace std;



#define mp make_pair

#define pb push_back

#define st first

#define nd second

#define lb lower_bound

#define ub upper_bound



typedef long long ll;

typedef long double ld;

typedef pair<int,int> pi;

typedef pair<ll,ll> pll;

typedef vector<int> vi;

typedef vector<ll> vll;

typedef vector<pi> vpi;

typedef vector<pll> vpll;



const ll oo=1e18;

const int MOD=1000000007;

const int INF=1000000000;

const ld EPS=0.000000001;



int dp[100005],h[100005],n,k=2;



signed main() {

    ios_base::sync_with_stdio(NULL);

    cin.tie(NULL);

    cout.tie(NULL);

    memset(dp,0x3f3f,sizeof dp);

    dp[1]=0;

    h[0]=0;

    cin>>n;

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

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

        for(int j=1;j<=min(k,i);++j) {

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

        }

    }

    cout<<dp[n];

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.434658272729593x):
#include<bits/stdc++.h>

using namespace std;

long long n,h[100010],dp[100010];



int main()

{

	scanf("%lld",&n);

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

	{

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

	}

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

	{

		if(i-1>=0) dp[i] = dp[i-1]+abs(h[i-1]-h[i]);

		if(i>2)

		{

			

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

		}

	}

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

	return 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: 1.8615051177594149x):
#include<cstdio>

#include<iostream>

#include<cmath>

#include<stack>

#include<cstring>

#include<algorithm>

using namespace std;

int n;

int h[100005];

int dp[100005];



int main()

{

memset(dp, 0x3f ,sizeof dp);

scanf("%d",&n);

for( int i = 1 ; i <= n ; i++) scanf("%d",&h[i]);

dp[1] = 0;

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

{

	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",dp[n]);



	return 0;

}