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: 1.3807153989510825x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    return 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: 1.3698895542026779x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    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: 1.586198280556259x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    return 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: 1.3824053841634965x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    return 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: 1.3934642751730175x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    return 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: 1.797922632168737x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    return 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: 1.791591385153176x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    return 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: 1.791155172328877x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    return 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: 1.4218240470307824x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    return 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: 1.3983698367754003x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << 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: 2.589173543323327x):
#include <bits/stdc++.h>

using namespace std;



#define oo (1L<<62)

#define ll long long int



int N;

int h[100005];

ll memo[100005];

char vis[100005];



ll elJoshua(int n){

	if(n < 0) return oo;

	if(!n) return 0;

	if(n == 1) return memo[1]=abs(h[0]-h[1]);

	if(vis[n]) return memo[n];

	vis[n] = 1;

	return memo[n] = min(elJoshua(n-1)+abs(h[n]-h[n-1]), 

						 elJoshua(n-2)+abs(h[n]-h[n-2]));

}



int main() {

	cin>>N;

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

		cin>>h[i];

	cout<<elJoshua(N-1)<<"\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: 1.814310285890616x):
#include <bits/stdc++.h>

using namespace std;



#define oo (1L<<62)

#define ll long long int



int N;

int h[100005];

ll memo[100005];

char vis[100005];



ll elJoshua(int n){

	if(n < 0) return oo;

	if(!n) return 0;

	if(n == 1) return memo[1]=abs(h[0]-h[1]);

	if(vis[n]) return memo[n];

	vis[n] = 1;

	return memo[n] = min(elJoshua(n-1)+abs(h[n]-h[n-1]), 

						 elJoshua(n-2)+abs(h[n]-h[n-2]));

}



int main() {

	cin>>N;

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

		cin>>h[i];

	cout<<elJoshua(N-1)<<"\n";

	return 0;

}