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

using namespace std;

#define Me3za() ios_base::sync_with_stdio(0);cout.tie(0);cin.tie(0);srand(time(NULL))

#define endl "\n"

int n;

int arr[100005];

int mem[100005];

long long solve(int i,int cur){

if(i>=n)return 1e9;

//cout<<i<<" "<<cur<<endl;

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

long long ch1=solve(i+1,arr[i+1])+abs(cur-arr[i+1]);

long long ch2=0;

if(i+2<=n)ch2=solve(i+2,arr[i+2])+abs(cur-arr[i+2]);

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

}

int main() {

Me3za();

cin>>n;

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

    cin>>arr[i];

}

memset(mem,-1,sizeof mem);

cout<<solve(0,arr[0]);

}




// Faster C++ solution for the same problem (measured speedup: 1.3793646681105163x):
//TINY_OUNCE911666

// You are not prepared!



#include <bits/stdc++.h>

using namespace std;



template<class TH> void _dbg(const char *sdbg, TH h){ cerr<<sdbg<<" = "<<h<<endl; }

template<class TH, class... TA> void _dbg(const char *sdbg, TH h, TA... a) {

  while(*sdbg!=',')cerr<<*sdbg++;

  cerr<<" = "<<h<<", "; _dbg(sdbg+1, a...);

}

#define dbg(...) _dbg(#__VA_ARGS__, __VA_ARGS__)

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

typedef long long ll;



const int INF = 1e9+7;



int32_t main() {

	_upgrade

	

	int n, k;

	cin >> n;

	

	vector<int> a(n), dp(n, INF);

	for (int& x : a) {

		cin >> x;

	}

	

	dp[0] = 0;

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

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

			if (j < n) {

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

			}

		}

	}

	

	cout << dp[n-1] << "\n";

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

using namespace std;



int const N=1e5+5, M=1e9+7, OO = 0x3f3f3f3f;



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<=n-3) ? solve(i+2)+abs(arr[i]-arr[i+2]) : OO;

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

}





int main()

{

//    freopen("input.txt","rt",stdin);

//    freopen("output.txt","wt",stdout);

    scanf("%d",&n);

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

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

    memset(mem, -1, sizeof mem);

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



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3560651388935994x):
//TINY_OUNCE911666

// You are not prepared!



#include <bits/stdc++.h>

using namespace std;



template<class TH> void _dbg(const char *sdbg, TH h){ cerr<<sdbg<<" = "<<h<<endl; }

template<class TH, class... TA> void _dbg(const char *sdbg, TH h, TA... a) {

  while(*sdbg!=',')cerr<<*sdbg++;

  cerr<<" = "<<h<<", "; _dbg(sdbg+1, a...);

}

#define dbg(...) _dbg(#__VA_ARGS__, __VA_ARGS__)

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

typedef long long ll;



const int INF = 1e9+7;



int32_t main() {

	_upgrade

	

	int n, k;

	cin >> n;

	

	vector<int> a(n), dp(n, INF);

	for (int& x : a) {

		cin >> x;

	}

	

	dp[0] = 0;

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

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

			if (j < n) {

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

			}

		}

	}

	

	cout << dp[n-1] << "\n";

}
// 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: 1.3664452638110722x):
//TINY_OUNCE911666

// You are not prepared!



#include <bits/stdc++.h>

using namespace std;



template<class TH> void _dbg(const char *sdbg, TH h){ cerr<<sdbg<<" = "<<h<<endl; }

template<class TH, class... TA> void _dbg(const char *sdbg, TH h, TA... a) {

  while(*sdbg!=',')cerr<<*sdbg++;

  cerr<<" = "<<h<<", "; _dbg(sdbg+1, a...);

}

#define dbg(...) _dbg(#__VA_ARGS__, __VA_ARGS__)

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

typedef long long ll;



const int INF = 1e9+7;



int32_t main() {

	_upgrade

	

	int n, k;

	cin >> n;

	

	vector<int> a(n), dp(n, INF);

	for (int& x : a) {

		cin >> x;

	}

	

	dp[0] = 0;

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

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

			if (j < n) {

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

			}

		}

	}

	

	cout << dp[n-1] << "\n";

}
// 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: 1.3557313076297697x):
//TINY_OUNCE911666

// You are not prepared!



#include <bits/stdc++.h>

using namespace std;



template<class TH> void _dbg(const char *sdbg, TH h){ cerr<<sdbg<<" = "<<h<<endl; }

template<class TH, class... TA> void _dbg(const char *sdbg, TH h, TA... a) {

  while(*sdbg!=',')cerr<<*sdbg++;

  cerr<<" = "<<h<<", "; _dbg(sdbg+1, a...);

}

#define dbg(...) _dbg(#__VA_ARGS__, __VA_ARGS__)

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

typedef long long ll;



const int INF = 1e9+7;



int32_t main() {

	_upgrade

	

	int n, k;

	cin >> n;

	

	vector<int> a(n), dp(n, INF);

	for (int& x : a) {

		cin >> x;

	}

	

	dp[0] = 0;

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

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

			if (j < n) {

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

			}

		}

	}

	

	cout << dp[n-1] << "\n";

}
// 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: 1.5698044141305023x):
//TINY_OUNCE911666

// You are not prepared!



#include <bits/stdc++.h>

using namespace std;



template<class TH> void _dbg(const char *sdbg, TH h){ cerr<<sdbg<<" = "<<h<<endl; }

template<class TH, class... TA> void _dbg(const char *sdbg, TH h, TA... a) {

  while(*sdbg!=',')cerr<<*sdbg++;

  cerr<<" = "<<h<<", "; _dbg(sdbg+1, a...);

}

#define dbg(...) _dbg(#__VA_ARGS__, __VA_ARGS__)

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

typedef long long ll;



const int INF = 1e9+7;



int32_t main() {

	_upgrade

	

	int n, k;

	cin >> n;

	

	vector<int> a(n), dp(n, INF);

	for (int& x : a) {

		cin >> x;

	}

	

	dp[0] = 0;

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

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

			if (j < n) {

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

			}

		}

	}

	

	cout << dp[n-1] << "\n";

}
// 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: 1.3681177824859332x):
//TINY_OUNCE911666

// You are not prepared!



#include <bits/stdc++.h>

using namespace std;



template<class TH> void _dbg(const char *sdbg, TH h){ cerr<<sdbg<<" = "<<h<<endl; }

template<class TH, class... TA> void _dbg(const char *sdbg, TH h, TA... a) {

  while(*sdbg!=',')cerr<<*sdbg++;

  cerr<<" = "<<h<<", "; _dbg(sdbg+1, a...);

}

#define dbg(...) _dbg(#__VA_ARGS__, __VA_ARGS__)

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

typedef long long ll;



const int INF = 1e9+7;



int32_t main() {

	_upgrade

	

	int n, k;

	cin >> n;

	

	vector<int> a(n), dp(n, INF);

	for (int& x : a) {

		cin >> x;

	}

	

	dp[0] = 0;

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

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

			if (j < n) {

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

			}

		}

	}

	

	cout << dp[n-1] << "\n";

}
// 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: 1.3790623763207257x):
//TINY_OUNCE911666

// You are not prepared!



#include <bits/stdc++.h>

using namespace std;



template<class TH> void _dbg(const char *sdbg, TH h){ cerr<<sdbg<<" = "<<h<<endl; }

template<class TH, class... TA> void _dbg(const char *sdbg, TH h, TA... a) {

  while(*sdbg!=',')cerr<<*sdbg++;

  cerr<<" = "<<h<<", "; _dbg(sdbg+1, a...);

}

#define dbg(...) _dbg(#__VA_ARGS__, __VA_ARGS__)

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

typedef long long ll;



const int INF = 1e9+7;



int32_t main() {

	_upgrade

	

	int n, k;

	cin >> n;

	

	vector<int> a(n), dp(n, INF);

	for (int& x : a) {

		cin >> x;

	}

	

	dp[0] = 0;

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

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

			if (j < n) {

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

			}

		}

	}

	

	cout << dp[n-1] << "\n";

}
// 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: 1.779340526868961x):
//TINY_OUNCE911666

// You are not prepared!



#include <bits/stdc++.h>

using namespace std;



template<class TH> void _dbg(const char *sdbg, TH h){ cerr<<sdbg<<" = "<<h<<endl; }

template<class TH, class... TA> void _dbg(const char *sdbg, TH h, TA... a) {

  while(*sdbg!=',')cerr<<*sdbg++;

  cerr<<" = "<<h<<", "; _dbg(sdbg+1, a...);

}

#define dbg(...) _dbg(#__VA_ARGS__, __VA_ARGS__)

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

typedef long long ll;



const int INF = 1e9+7;



int32_t main() {

	_upgrade

	

	int n, k;

	cin >> n;

	

	vector<int> a(n), dp(n, INF);

	for (int& x : a) {

		cin >> x;

	}

	

	dp[0] = 0;

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

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

			if (j < n) {

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

			}

		}

	}

	

	cout << dp[n-1] << "\n";

}
// 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: 2.5471110247132276x):
/*

██╗  ██╗ █████╗ ██████╗ ███████╗███████╗███╗   ███╗        ██╗  ██╗██╗  ██╗ █████╗ ██╗     ███████╗██████╗

██║ ██╔╝██╔══██╗██╔══██╗██╔════╝██╔════╝████╗ ████║        ██║ ██╔╝██║  ██║██╔══██╗██║     ██╔════╝██╔══██╗

█████╔╝ ███████║██████╔╝█████╗  █████╗  ██╔████╔██║        █████╔╝ ███████║███████║██║     █████╗  ██║  ██║

██╔═██╗ ██╔══██║██╔══██╗██╔══╝  ██╔══╝  ██║╚██╔╝██║        ██╔═██╗ ██╔══██║██╔══██║██║     ██╔══╝  ██║  ██║

██║  ██╗██║  ██║██║  ██║███████╗███████╗██║ ╚═╝ ██║███████╗██║  ██╗██║  ██║██║  ██║███████╗███████╗██████╔╝

╚═╝  ╚═╝╚═╝  ╚═╝╚═╝  ╚═╝╚══════╝╚══════╝╚═╝     ╚═╝╚══════╝╚═╝  ╚═╝╚═╝  ╚═╝╚═╝  ╚═╝╚══════╝╚══════╝╚═════╝

Believe in yourself ,( try and try and then try ), You can do it  

*/

#define _CRT_SECURE_NO_WARNINGS

#include<bits/stdc++.h>

#include<unordered_map>

#define endl "\n"

#define Ceil(x,y) ((x+y-1)/y)

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

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

#define allr(s) s.rbegin(),s.rend()

#define rt(s) return cout<<s,0

#define watch(x) cout<<(#x)<<" = "<<x<<endl

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

#define OO 0x3f3f3f3f3f3f3f3fLL

const int oo = 0x3f3f3f3f, mod = 1e9 + 7;

using namespace std;

typedef long long ll;

typedef long double ld;

typedef unsigned long long ull;

//x = xDegrees*3.14159265 / 180;

//ef1tn=(( (n / 2)*( (n / 2) + 1) / 2) * 2);

//of1tn=(n * (n + 1) / 2 - 2 * ((n / 2) * (n / 2 + 1) / 2));

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

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

void K_K(){

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

}

ll gcd(ll a, ll b) {

	return b == 0 ? abs(a) : gcd(b, a % b);

}

ll lcm(ll a, ll b) {

	return a / gcd(a, b) * b;

}

int n;

vector<int>arr;

map < pair<int, int>, int > dp;

int sol(int idx, int cur){

	if (idx >= n){

		if (cur == arr[n - 1])

			return 0;

		return oo;

	}

	if (dp.find({ idx, cur }) != dp.end())

		return dp[{idx, cur}];

	int ch1 = sol(idx + 1, arr[idx]) + abs(cur - arr[idx]);

	int ch2 = sol(idx + 2, arr[idx]) + abs(cur - arr[idx]);

	return dp[{idx, cur}] = min(ch1, ch2);

}

int main(){

	K_K();

	cin >> n;

	arr.resize(n);

	for (auto &it : arr)

		cin >> it;

	cout << sol(0, arr[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: 1.372763727491106x):
/*

██╗  ██╗ █████╗ ██████╗ ███████╗███████╗███╗   ███╗        ██╗  ██╗██╗  ██╗ █████╗ ██╗     ███████╗██████╗

██║ ██╔╝██╔══██╗██╔══██╗██╔════╝██╔════╝████╗ ████║        ██║ ██╔╝██║  ██║██╔══██╗██║     ██╔════╝██╔══██╗

█████╔╝ ███████║██████╔╝█████╗  █████╗  ██╔████╔██║        █████╔╝ ███████║███████║██║     █████╗  ██║  ██║

██╔═██╗ ██╔══██║██╔══██╗██╔══╝  ██╔══╝  ██║╚██╔╝██║        ██╔═██╗ ██╔══██║██╔══██║██║     ██╔══╝  ██║  ██║

██║  ██╗██║  ██║██║  ██║███████╗███████╗██║ ╚═╝ ██║███████╗██║  ██╗██║  ██║██║  ██║███████╗███████╗██████╔╝

╚═╝  ╚═╝╚═╝  ╚═╝╚═╝  ╚═╝╚══════╝╚══════╝╚═╝     ╚═╝╚══════╝╚═╝  ╚═╝╚═╝  ╚═╝╚═╝  ╚═╝╚══════╝╚══════╝╚═════╝

Believe in yourself ,( try and try and then try ), You can do it  

*/

#define _CRT_SECURE_NO_WARNINGS

#include<bits/stdc++.h>

#include<unordered_map>

#define endl "\n"

#define Ceil(x,y) ((x+y-1)/y)

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

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

#define allr(s) s.rbegin(),s.rend()

#define rt(s) return cout<<s,0

#define watch(x) cout<<(#x)<<" = "<<x<<endl

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

#define OO 0x3f3f3f3f3f3f3f3fLL

const int oo = 0x3f3f3f3f, mod = 1e9 + 7;

using namespace std;

typedef long long ll;

typedef long double ld;

typedef unsigned long long ull;

//x = xDegrees*3.14159265 / 180;

//ef1tn=(( (n / 2)*( (n / 2) + 1) / 2) * 2);

//of1tn=(n * (n + 1) / 2 - 2 * ((n / 2) * (n / 2 + 1) / 2));

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

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

void K_K(){

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

}

ll gcd(ll a, ll b) {

	return b == 0 ? abs(a) : gcd(b, a % b);

}

ll lcm(ll a, ll b) {

	return a / gcd(a, b) * b;

}

int n;

vector<int>arr;

map < pair<int, int>, int > dp;

int sol(int idx, int cur){

	if (idx >= n){

		if (cur == arr[n - 1])

			return 0;

		return oo;

	}

	if (dp.find({ idx, cur }) != dp.end())

		return dp[{idx, cur}];

	int ch1 = sol(idx + 1, arr[idx]) + abs(cur - arr[idx]);

	int ch2 = sol(idx + 2, arr[idx]) + abs(cur - arr[idx]);

	return dp[{idx, cur}] = min(ch1, ch2);

}

int main(){

	K_K();

	cin >> n;

	arr.resize(n);

	for (auto &it : arr)

		cin >> it;

	cout << sol(0, arr[0]);

}
// Slower C++ solution:
#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;

}






// Faster C++ solution for the same problem (measured speedup: 1.368307290918994x):
/*

██╗  ██╗ █████╗ ██████╗ ███████╗███████╗███╗   ███╗        ██╗  ██╗██╗  ██╗ █████╗ ██╗     ███████╗██████╗

██║ ██╔╝██╔══██╗██╔══██╗██╔════╝██╔════╝████╗ ████║        ██║ ██╔╝██║  ██║██╔══██╗██║     ██╔════╝██╔══██╗

█████╔╝ ███████║██████╔╝█████╗  █████╗  ██╔████╔██║        █████╔╝ ███████║███████║██║     █████╗  ██║  ██║

██╔═██╗ ██╔══██║██╔══██╗██╔══╝  ██╔══╝  ██║╚██╔╝██║        ██╔═██╗ ██╔══██║██╔══██║██║     ██╔══╝  ██║  ██║

██║  ██╗██║  ██║██║  ██║███████╗███████╗██║ ╚═╝ ██║███████╗██║  ██╗██║  ██║██║  ██║███████╗███████╗██████╔╝

╚═╝  ╚═╝╚═╝  ╚═╝╚═╝  ╚═╝╚══════╝╚══════╝╚═╝     ╚═╝╚══════╝╚═╝  ╚═╝╚═╝  ╚═╝╚═╝  ╚═╝╚══════╝╚══════╝╚═════╝

Believe in yourself ,( try and try and then try ), You can do it  

*/

#define _CRT_SECURE_NO_WARNINGS

#include<bits/stdc++.h>

#include<unordered_map>

#define endl "\n"

#define Ceil(x,y) ((x+y-1)/y)

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

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

#define allr(s) s.rbegin(),s.rend()

#define rt(s) return cout<<s,0

#define watch(x) cout<<(#x)<<" = "<<x<<endl

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

#define OO 0x3f3f3f3f3f3f3f3fLL

const int oo = 0x3f3f3f3f, mod = 1e9 + 7;

using namespace std;

typedef long long ll;

typedef long double ld;

typedef unsigned long long ull;

//x = xDegrees*3.14159265 / 180;

//ef1tn=(( (n / 2)*( (n / 2) + 1) / 2) * 2);

//of1tn=(n * (n + 1) / 2 - 2 * ((n / 2) * (n / 2 + 1) / 2));

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

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

void K_K(){

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

}

ll gcd(ll a, ll b) {

	return b == 0 ? abs(a) : gcd(b, a % b);

}

ll lcm(ll a, ll b) {

	return a / gcd(a, b) * b;

}

int n;

vector<int>arr;

map < pair<int, int>, int > dp;

int sol(int idx, int cur){

	if (idx >= n){

		if (cur == arr[n - 1])

			return 0;

		return oo;

	}

	if (dp.find({ idx, cur }) != dp.end())

		return dp[{idx, cur}];

	int ch1 = sol(idx + 1, arr[idx]) + abs(cur - arr[idx]);

	int ch2 = sol(idx + 2, arr[idx]) + abs(cur - arr[idx]);

	return dp[{idx, cur}] = min(ch1, ch2);

}

int main(){

	K_K();

	cin >> n;

	arr.resize(n);

	for (auto &it : arr)

		cin >> it;

	cout << sol(0, arr[0]);

}