stablegradients/pie-gem5-pairs huggingface dataset:
// 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 == n - 1)  return 0;

    if(i == n)  return 1e9;



    int &ret = dp[i];

    if(~ret)    return ret;



    return ret = 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(0);

}

// Faster C++ solution for the same problem (measured speedup: 1.4004377447122085x):
#include <bits/stdc++.h>



using namespace std;



int main()

{

    int N;

    cin >> N;



    int h[N], dp[N];



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

    {

        cin >> h[i];

    }

    dp[N-1]=0;

    dp[N-2]=abs(h[N-1]-h[N-2]);



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

    {

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

    }

    cout << dp[0];

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



using namespace std;



const int N=1e5;

int n;

int arr[N];

int mem[N];



int solve(int i){

    if(i==n-1)

        return 0;

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

    int op1=solve(i+1)+abs(arr[i]-arr[i+1]);

    int op2=i+2<=n-1?solve(i+2)+abs(arr[i]-arr[i+2]):1000000000;

    return mem[i]=min(op1,op2);

}



int main() {

    scanf("%d",&n);

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

        scanf("%d",arr+i);

    memset(mem,-1,sizeof mem);

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3894572619743624x):
#include <bits/stdc++.h>



using namespace std;



int main()

{

    int N;

    cin >> N;



    int h[N], dp[N];



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

    {

        cin >> h[i];

    }

    dp[N-1]=0;

    dp[N-2]=abs(h[N-1]-h[N-2]);



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

    {

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

    }

    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: 1.6088557746050687x):
#include <bits/stdc++.h>



using namespace std;



int main()

{

    int N;

    cin >> N;



    int h[N], dp[N];



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

    {

        cin >> h[i];

    }

    dp[N-1]=0;

    dp[N-2]=abs(h[N-1]-h[N-2]);



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

    {

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

    }

    cout << dp[0];

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

 

using namespace std;

 

#define sf(n) scanf("%d", &n)

#define pd(n) printf("%.12lf", n)

#define pf(n) printf("%d", n)

#define p() printf("\n")

#define ps() printf(" ")

#define ll long long

#define ull unsigned long long

#define MAX 1e9

const int mod = 1000 * 1000 * 1000 + 7;

#define ex 1e18

#define dbl 2e09

#define qu(q) queue<int> q

#define pqu(q) priority_queue<int> q

#define len(s) s.length()

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

#define pb(x) push_back(x)

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

#define rall(x) x.rbegin(), x.rend()

#define F first

#define S second

#define clr(x) x.clear()

#define LOG(x) std::cout << x << std::endl;

 

const int INF = 0x3f3f3f3f;

const ll LL_INF = 0x3f3f3f3f3f3f3f3f;

const double PI = 3.14159265359;

 

int dx[6] = {-1,-1,0,0,1,1};

int dy[6] = {-1,0,-1,1,0,1};



int a[100005];

int dp[100005];

int n;

int solve(int idx){

    if(idx==(n-1))

        return 0;

    if(idx>=n)

        return INF;

    int &ret = dp[idx];

    if(~ret)

        return ret;

    ret = INF;

    ret=min(ret,solve(idx+1)+abs(a[idx+1]-a[idx]));

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

    return ret;

}







int main(){



    cin>>n;

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

        cin>>a[i];

    }

    memset(dp,-1,sizeof dp);

    cout<<solve(0)<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4021518699267659x):
#include <bits/stdc++.h>



using namespace std;



int main()

{

    int N;

    cin >> N;



    int h[N], dp[N];



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

    {

        cin >> h[i];

    }

    dp[N-1]=0;

    dp[N-2]=abs(h[N-1]-h[N-2]);



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

    {

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

    }

    cout << dp[0];

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

using namespace std;

int abssub(int a,int b){

	if(a>b)return a-b;

	return b-a;

}

int main(){

	int n,h[100001],bp[100001]={0};

	cin>>n; 

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

		cin>>h[i];

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

		if(i==2)bp[i]=bp[i-1]+abssub(h[i],h[i-1]);

		else{

			bp[i]=min(bp[i-1]+abssub(h[i],h[i-1]),bp[i-2]+abssub(h[i],h[i-2]));

		}

	}

	cout<<bp[n]<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4133687277934612x):
#include <bits/stdc++.h>



using namespace std;



int main()

{

    int N;

    cin >> N;



    int h[N], dp[N];



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

    {

        cin >> h[i];

    }

    dp[N-1]=0;

    dp[N-2]=abs(h[N-1]-h[N-2]);



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

    {

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

    }

    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: 1.823604428598561x):
#include <bits/stdc++.h>



using namespace std;



int main()

{

    int N;

    cin >> N;



    int h[N], dp[N];



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

    {

        cin >> h[i];

    }

    dp[N-1]=0;

    dp[N-2]=abs(h[N-1]-h[N-2]);



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

    {

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

    }

    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: 1.8171827451014235x):
#include <bits/stdc++.h>



using namespace std;



int main()

{

    int N;

    cin >> N;



    int h[N], dp[N];



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

    {

        cin >> h[i];

    }

    dp[N-1]=0;

    dp[N-2]=abs(h[N-1]-h[N-2]);



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

    {

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

    }

    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: 1.8167403013477434x):
#include <bits/stdc++.h>



using namespace std;



int main()

{

    int N;

    cin >> N;



    int h[N], dp[N];



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

    {

        cin >> h[i];

    }

    dp[N-1]=0;

    dp[N-2]=abs(h[N-1]-h[N-2]);



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

    {

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

    }

    cout << dp[0];

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

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

#define ii          pair<int,int>

#define F           first

#define S           second

#define pb          push_back

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

#define rall(x)     (x).rbegin(),(x).rend()

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

#define forr(i,a)   for(int i = a; i >= 0; i--)

#define fori(i,m)   for(auto i = m.begin(); i != m.end(); i++) 

#define w(t)        while(t--)

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

#define cls(a,car)  memset(a,car,sizeof (a))

#define db(x)       cerr << #x << " is " << x << '\n'

#define FL          freopen("in", "r", stdin),freopen("out", "w", stdout);

#define EjecuteTime cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n"

#define angle(x)    double(x * acos(-1) / 180.0)

using namespace std;

typedef long long   ll;

typedef vector<int> vi;

typedef vector<ii>  vii;

const int N = 1 * 1e5 + 5;

const ll mod = 1e9 + 7;

const double E = 1e-9;

//int dx[] = {1, 0}, dy[] = {0, 1};

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

//int dx[] = {2, 1, -1, -2, -2, -1, 1, 2}, dy[] = {1, 2, 2, 1, -1, -2, -2, -1};

// fflush(stdout)

// cout << flush

// a + b = a ^ b + 2 * (a & b)

int n, k;

int v[N];

int dp[N];

int f(int i){

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

    int &ans = dp[i];

    if(ans != -1)return ans;

    ans = 1e9;

    for(int j = 1; j <= 2; j++){

        if(i + j < n)

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

    }

    return ans;

}

int main(){

    //FL;

    FIN;

    cin >> n;

    fore(i, 0, n)cin >> v[i];

    cls(dp, -1);

    cout << f(0) << '\n';

    EjecuteTime;

    return 0;

}       

// Faster C++ solution for the same problem (measured speedup: 1.4421335948842557x):
#include <bits/stdc++.h>



using namespace std;



int main()

{

    int N;

    cin >> N;



    int h[N], dp[N];



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

    {

        cin >> h[i];

    }

    dp[N-1]=0;

    dp[N-2]=abs(h[N-1]-h[N-2]);



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

    {

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

    }

    cout << dp[0];

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



#define clr(x) memset((x), 0, sizeof(x))

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

#define pb push_back

#define mp make_pair

#define For(i, st, en) for(int i=(st); i<=(int)(en); i++)

#define Ford(i, st, en) for(int i=(st); i>=(int)(en); i--)

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

#define forn2(i, n) for(int i=0; i<=(int)(n); i++)

#define ford(i, n) for(int i=(n)-1; i>=0; i--)

#define fori(it, x) for (__typeof((x).begin()) it = (x).begin(); it != (x).end(); it++)

#define in(x) int (x); input((x));

#define x first

#define y second

#define less(a,b) ((a) < (b) - EPS)

#define more(a,b) ((a) > (b) + EPS)

#define eq(a,b) (fabs((a) - (b)) < EPS)

#define remax(a, b) ((a) = (b) > (a) ? (b) : (a))

#define remin(a, b) ((a) = (b) < (a) ? (b) : (a))

#define salfinal(x) cout<<x<<endl;

using namespace std;

const int MAX =1e5+1;

int v[MAX];

int memo[MAX];

int c;



int f(int i)

{



    if(memo[i]!=-1)

        return memo[i];



    if(i == 0)

        return memo[0] = 0;



    if(i == 1)

        return memo[1] = abs(v[0] - v[1]);



    return memo[i] = min(abs(v[i] - v[i-1]) + f(i-1), abs(v[i] - v[i-2]) + f(i-2));

}



int main()

{

    ios_base::sync_with_stdio(false);

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

    cin>>c;



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

    {

        cin>>v[i];

    }



    cout<<f(c-1)<<endl;



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4183443611732343x):
#include <bits/stdc++.h>



using namespace std;



int main()

{

    int N;

    cin >> N;



    int h[N], dp[N];



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

    {

        cin >> h[i];

    }

    dp[N-1]=0;

    dp[N-2]=abs(h[N-1]-h[N-2]);



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

    {

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

    }

    cout << dp[0];

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





using namespace std;

int n;

vector<int> vec;



int meme[100007];



int solve(int index){



    if(index==n-1){

        return 0;

    }

    if(meme[index]!=-1){

        return meme[index];

    }

    

    int ret1 = 100000000;

    int ret2 = 100000000;

    if(index+1<n){

         ret1 = abs (vec[index]- vec[index+1]) + solve(index+1); 

    }

    if(index+2<n){

         ret2 = abs (vec[index]- vec[index+2]) + solve(index+2); 

    }





    return meme[index] = min(ret1,ret2);

}



int main(){

    cin>>n;

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

    int num;

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

        cin>>num;

        vec.push_back(num);

    }

    cout<<solve(0)<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4050506899952466x):
#include <bits/stdc++.h>



using namespace std;



int main()

{

    int N;

    cin >> N;



    int h[N], dp[N];



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

    {

        cin >> h[i];

    }

    dp[N-1]=0;

    dp[N-2]=abs(h[N-1]-h[N-2]);



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

    {

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

    }

    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: 1.8279948526934295x):
#include <bits/stdc++.h>



using namespace std;



int main()

{

    int N;

    cin >> N;



    int h[N], dp[N];



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

    {

        cin >> h[i];

    }

    dp[N-1]=0;

    dp[N-2]=abs(h[N-1]-h[N-2]);



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

    {

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

    }

    cout << dp[0];

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

using namespace std;



typedef long long ll;

typedef double dd;

typedef vector<int> vi;

typedef vector<ll> vl;

typedef pair<int,int> pii;



const int mx=1e5+11;

const double PI = acos(-1);

#define MOD 1000000007





#define nl '\n'

#define pb push_back

#define F first

#define S second

#define I insert

#define mp make_pair



#define sz(x) x.size()

#define sa(v) sort(v.begin(),v.end())

#define sd(v) sort(v.begin(),v.end(),greater<>())

#define rev(s) reverse(s.begin(),s.end())

#define tt int t; cin>>t; while(t--)

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

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

#define rbe(a) (a).rbegin(),(a).rend()



#define sqr(a) ((a) * (a))

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

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

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



#define yes cout<<"YES"<<nl;

#define no cout<<"NO"<<nl;



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

#define file freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);

#define frac cout.unsetf(ios::floatfield); cout.precision(10); cout.setf(ios::fixed,ios::floatfield);



#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)

void faltu ()

{

    cerr << endl;

}

template < typename T, typename ... hello>void faltu( T arg, const hello &... rest)

{

    cerr << arg << ' ';

    faltu(rest...);

}



int dp[mx],n,a[mx];



int solve (int i)

{

    if ( i >= n )

        return 0;

    if ( dp[i]!= -1 )

        return dp[i];



    int ret1 = INT_MAX, ret2 = INT_MAX;



    if (i+1<=n )

        ret1 = abs(a[i]-a[i+1])+ solve (i+1);



    if(i+2<=n)

        ret2 = abs(a[i]-a[i+2])+solve (i+2);

    ///dbg(i);

    dp[i]= min ( ret1, ret2 );



    return dp[i];

}



int main()

{

    su;



    mem ( dp, -1 );



    cin>>n;



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

    {

        cin >>a[i];

    }



    cout << solve( 1 )<< nl;





    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.419087567253854x):
#include <bits/stdc++.h>



using namespace std;



int main()

{

    int N;

    cin >> N;



    int h[N], dp[N];



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

    {

        cin >> h[i];

    }

    dp[N-1]=0;

    dp[N-2]=abs(h[N-1]-h[N-2]);



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

    {

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

    }

    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: 2.5498898184075323x):
#include <bits/stdc++.h>

#include <set>

using namespace std;

#define mp             make_pair

#define pb             push_back

#define fi             first

#define se             second

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

#define fill(x, y)     memset(x, y, sizeof(y))

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

typedef long long      ll;

typedef pair<int, int> ii;

typedef pair<ii, int>  iii;

typedef vector<int>    vi;

typedef vector<ii>     vii;

typedef vector<iii>    viii;



const int INF = 1e9 + 5;



int main() {

  ios::sync_with_stdio(false);

  cin.tie(0);

  int N;

  cin >> N;

  vi h(N+1);

  h[0] = INF;

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

    cin >> h[i];

  }

  vi dp(N, 0);

  dp[0] = 0;

  dp[1] = 0;

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

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

  }

  cout << dp[N] << "\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: 1.374261356433351x):
#include <bits/stdc++.h>

#include <set>

using namespace std;

#define mp             make_pair

#define pb             push_back

#define fi             first

#define se             second

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

#define fill(x, y)     memset(x, y, sizeof(y))

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

typedef long long      ll;

typedef pair<int, int> ii;

typedef pair<ii, int>  iii;

typedef vector<int>    vi;

typedef vector<ii>     vii;

typedef vector<iii>    viii;



const int INF = 1e9 + 5;



int main() {

  ios::sync_with_stdio(false);

  cin.tie(0);

  int N;

  cin >> N;

  vi h(N+1);

  h[0] = INF;

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

    cin >> h[i];

  }

  vi dp(N, 0);

  dp[0] = 0;

  dp[1] = 0;

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

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

  }

  cout << dp[N] << "\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: 1.3943755983757657x):
#include <bits/stdc++.h>

#include <set>

using namespace std;

#define mp             make_pair

#define pb             push_back

#define fi             first

#define se             second

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

#define fill(x, y)     memset(x, y, sizeof(y))

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

typedef long long      ll;

typedef pair<int, int> ii;

typedef pair<ii, int>  iii;

typedef vector<int>    vi;

typedef vector<ii>     vii;

typedef vector<iii>    viii;



const int INF = 1e9 + 5;



int main() {

  ios::sync_with_stdio(false);

  cin.tie(0);

  int N;

  cin >> N;

  vi h(N+1);

  h[0] = INF;

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

    cin >> h[i];

  }

  vi dp(N, 0);

  dp[0] = 0;

  dp[1] = 0;

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

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

  }

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

  

  return 0;

}