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

using namespace std;



const int mx=1e5+1;

int n,k;

int h[mx],dp[mx];

int solve(int s) //s->no of stone

{

    if(s==n) return 0;

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

    int ret=INT_MAX;

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

        if((i+s)<=n) ret=min(ret,abs(h[s+i]-h[s])+solve(s+i));

    }

    return dp[s]=ret;

}

int main()

{

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

    //freopen("input.txt","r",stdin);    return 0;



    cin>>n;

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

    memset(dp,-1,sizeof dp);

    cout<<solve(1);



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4127028103775123x):
#include <iostream>

using namespace std;

const int maxn = 1e5;

int h[maxn];

int dp[maxn];

int N;



int main(int argc, char const *argv[])

{

    cin >> N;

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

        cin >> h[i];

    }

    dp[0] = 0;

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

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

        int t1 = dp[i-2] + abs(h[i-2] - h[i]);

        int t2 = dp[i-1] + abs(h[i-1] - h[i]);

        dp[i] = min(t1, t2);

    }

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

}
// Slower C++ solution:
#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

const int maxn=100000+10;

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

int main()

{

	cin>>n;

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

	memset(dp,0x3f3f3f3f,sizeof(dp));

	dp[1]=0;

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

	{

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

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

	}

	cout<<dp[n]<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4005887611857455x):
#include <iostream>

using namespace std;

const int maxn = 1e5;

int h[maxn];

int dp[maxn];

int N;



int main(int argc, char const *argv[])

{

    cin >> N;

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

        cin >> h[i];

    }

    dp[0] = 0;

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

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

        int t1 = dp[i-2] + abs(h[i-2] - h[i]);

        int t2 = dp[i-1] + abs(h[i-1] - h[i]);

        dp[i] = min(t1, t2);

    }

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

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

#define N 100007

#define inf 999999999

using namespace std;



long long mem[N];

long long ara[N];

int n;



long long dp(int i, int j)

{

    if(i >= n)

        return inf;

    if(i == j)

        return 0;

    if(mem[i] != -1)

        return mem[i];

    long long ans = min(abs(ara[i + 1] - ara[i]) + dp(i + 1, j), abs(ara[i + 2] - ara[i]) + dp(i + 2, j));

    mem[i] = ans;

    return ans;

}

int main()

{

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

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

    cin >> n;

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

        cin >> ara[i];

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

    cout << dp(0, n - 1) << endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.805312351677041x):
#include <iostream>

using namespace std;

const int maxn = 1e5;

int h[maxn];

int dp[maxn];

int N;



int main(int argc, char const *argv[])

{

    cin >> N;

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

        cin >> h[i];

    }

    dp[0] = 0;

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

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

        int t1 = dp[i-2] + abs(h[i-2] - h[i]);

        int t2 = dp[i-1] + abs(h[i-1] - h[i]);

        dp[i] = min(t1, t2);

    }

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

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

using namespace std;

using ll = long long;

using ii = pair<int, int>;

using iii = pair<ii, int>;

#define vt vector

#define pq priority_queue

#define pb push_back

#define pf push_front

#define mp make_pair

#define fi first

#define se second

#define __lcm(a, b) a * b / __gcd(a, b)

const int MOD = 1e9 + 7;

const int INF = 2e9;

const int N = 1e5 + 5;



int n;

int a[N], dp[N];



int f(int idx) {

    if(idx > n) return INF;

    if(idx == n) return 0;

    int &sol = dp[idx];

    if(sol != -1) return sol;

    return sol = min(f(idx + 1) + abs(a[idx] - a[idx + 1]), 

                     f(idx + 2) + abs(a[idx] - a[idx + 2]));

}



void solve() {

    cin >> n;

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

        cin >> a[i];

    }

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

    cout << f(1) << "\n";

}



int main() {

 

    ios_base::sync_with_stdio(0);

    cin.tie(0); cout.tie(0);

    

    solve();

    

}

// Faster C++ solution for the same problem (measured speedup: 1.4136162238323753x):
#include <iostream>

using namespace std;

const int maxn = 1e5;

int h[maxn];

int dp[maxn];

int N;



int main(int argc, char const *argv[])

{

    cin >> N;

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

        cin >> h[i];

    }

    dp[0] = 0;

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

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

        int t1 = dp[i-2] + abs(h[i-2] - h[i]);

        int t2 = dp[i-1] + abs(h[i-1] - h[i]);

        dp[i] = min(t1, t2);

    }

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

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

using namespace std;

#define N (int)2e5

int n,memo[N],c[N];



int dp(int curr) {

    if (curr == n) return 0;

    if (curr > n) return 1e9;

    int &sol = memo[curr];

    if (sol != -1) return sol;

    sol = min(

        dp(curr + 1) + abs(c[curr] - c[curr + 1]) ,

        dp(curr + 2) + abs(c[curr] - c[curr + 2])

    );

    return sol;

}



int main() {

    scanf("%d" , &n);

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

        scanf("%d" , &c[i]);

    memset (memo , -1 , sizeof memo);

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

}


// Faster C++ solution for the same problem (measured speedup: 1.8044320285900952x):
#include<iostream>

#include<cstdio>

#include<fstream>

#include<algorithm>

#include<cmath>

#include<deque>

#include<vector>

#include<queue>

#include<string>

#include<cstring>

#include<map>

#include<stack>

#include<set>

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

typedef long long ll;

typedef unsigned long long ull;

using namespace std;

const int maxn =1200100;

typedef pair<ll,ll>p1;

typedef pair<pair<ll,ll>,ll>p2;

map<pair<pair<ll,ll>,ll>,ll>m1;

map<pair<ll,ll>,ll>m2;

const int MOD = 1e9;

const int N=351;

const int inf = 1000010;

typedef pair<char, int> PAIR;

map<string,int>mm;

ll a[maxn];

ll dp[maxn];

int main()

{

    int n;

    scanf("%d",&n);

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

    {

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

    }

    dp[1] = 0;

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

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

    {

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

    }

    cout<<dp[n]<<endl;

}
// Slower C++ solution:
#include<set>

#include<map>

#include<cmath>

#include<ctime>

#include<queue>

#include<stack>

#include<cstdio>

#include<string>

#include<vector>

#include<random>

#include<cstdlib>

#include<cstring>

#include<iomanip>

#include<iostream>

#include<algorithm>

#include<unordered_map>

#define endl "\n"

#define fi first

#define se second

#define db double

#define gcd __gcd

#define pb push_back

#define mp make_pair

#define lowbit(x) (x & (-x))

#define PII  pair<int, int> 

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

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

#define rep(i, a, b) for(__typeof(b) i = a; i <= (b); i++)

#define Rep(i, a, b) for(__typeof(a) i = a; i >= (b); i--)

#define FAST ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)

template<class T> inline T qmin(T a, T b) { return a < b ? a : b; }

template<class T> inline T qmax(T a, T b) { return a > b ? a : b; }

typedef long long ll;

typedef unsigned long long ull;

const db eps = 1e-9;

const db PI = acos(-1);

const int inf = 0x3f3f3f3f;

const int mod = (int)1e9 + 7;

const int maxn = (int)1e5 + 5;//remember to modify it, No RE or MLE

const ll INF = 0x3f3f3f3f3f3f3f3f;

using namespace std;

ll qpow(ll x, ll y) { ll ret = 1; for( ; y; y >>= 1, x = (x * x) % mod) if(y & 1) ret = (ret * x) % mod; return ret; }



inline int read(){

	char ch = getchar();

	int x = 0, f = 1;

	while(ch < '0' || ch > '9'){

		if(ch == '-') f = -1;

		ch = getchar();

	}

	while('0' <= ch && ch <= '9'){

		x = x * 10 + ch - '0';

		ch = getchar();

	}

	return x * f;

}



int n;

int h[maxn], dp[maxn];



int main()

{

	n = read();

	rep(i, 1, n) h[i] = read();

	memset(dp, 0x3f, sizeof dp);

	dp[1] = 0;

	rep(i, 1, n){

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

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

	}

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3965585531370885x):
#include<iostream>

#include<cstdio>

#include<fstream>

#include<algorithm>

#include<cmath>

#include<deque>

#include<vector>

#include<queue>

#include<string>

#include<cstring>

#include<map>

#include<stack>

#include<set>

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

typedef long long ll;

typedef unsigned long long ull;

using namespace std;

const int maxn =1200100;

typedef pair<ll,ll>p1;

typedef pair<pair<ll,ll>,ll>p2;

map<pair<pair<ll,ll>,ll>,ll>m1;

map<pair<ll,ll>,ll>m2;

const int MOD = 1e9;

const int N=351;

const int inf = 1000010;

typedef pair<char, int> PAIR;

map<string,int>mm;

ll a[maxn];

ll dp[maxn];

int main()

{

    int n;

    scanf("%d",&n);

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

    {

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

    }

    dp[1] = 0;

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

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

    {

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

    }

    cout<<dp[n]<<endl;

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

using namespace std;

#define ll long long int

#define dbl double

#define pi acos(-1)

#define sc1(a) scanf("%lld",&a)

#define sc2(a,b) scanf("%lld%lld",&a,&b)

#define sc3(a,b,c) scanf("%lld%lld%lld",&a,&b,&c)

#define pb push_back



ll ara[1000005];



map<ll,ll>vis;

map<ll,ll>vis1;

map<ll,ll>pos;

map<ll,ll>high;

map<ll,ll>low;

map<string,ll>mp;

vector<ll>vec;

vector<ll>v;

stack<ll>q;

string str1[100];

string str2[100];

priority_queue<ll>pq;

ll dp[1000005],t;

ll mn=1000000;

ll call(ll id)

{

    if(id==t )return 0;

    if(dp[id]!=-1)

        return dp[id];

    ll p=call(id+1)+abs(ara[id+1]-ara[id]);

    ll q=100000000;

    if(id+2!=t+1)

        q=call(id+2)+abs(ara[id+2]-ara[id]);

    dp[id]=min(p,q);

    return dp[id];



}

int main()

{



    // ll t;

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

    sc1(t);

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

    {

        sc1(ara[i]);

    }

    cout<<call(1)<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 9.486587997696867x):
#include<iostream>

#include<cstdio>

#include<fstream>

#include<algorithm>

#include<cmath>

#include<deque>

#include<vector>

#include<queue>

#include<string>

#include<cstring>

#include<map>

#include<stack>

#include<set>

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

typedef long long ll;

typedef unsigned long long ull;

using namespace std;

const int maxn =1200100;

typedef pair<ll,ll>p1;

typedef pair<pair<ll,ll>,ll>p2;

map<pair<pair<ll,ll>,ll>,ll>m1;

map<pair<ll,ll>,ll>m2;

const int MOD = 1e9;

const int N=351;

const int inf = 1000010;

typedef pair<char, int> PAIR;

map<string,int>mm;

ll a[maxn];

ll dp[maxn];

int main()

{

    int n;

    scanf("%d",&n);

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

    {

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

    }

    dp[1] = 0;

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

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

    {

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

    }

    cout<<dp[n]<<endl;

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



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

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

#define fore(i, l, r) for (int i = int(l); i <= int(r); ++i)

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

#define pb push_back

#define eb emplace_back

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

#define F first

#define S second

#define endl '\n'

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

#define de2(x,y) cout<<#x<<" = "<<(x) <<' '<< #y<<" = "<<y<<endl;



using namespace std;



typedef long long ll;

typedef pair<int, int> pii;

const int INF = 0x3f3f3f3f;

mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());



const int N=1e5+5;

int dp[N];

int main() {

    ios::sync_with_stdio(false);

    cin.tie(nullptr);

	int n;

	cin>>n;

	vector<int> a(n);

	for(auto& it:a) cin>>it;

	ms(dp,INF);

	dp[0]=0;

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

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

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

	}

	cout<<dp[n-1];

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.4418118703684695x):
#include<iostream>

#include<cstdio>

#include<fstream>

#include<algorithm>

#include<cmath>

#include<deque>

#include<vector>

#include<queue>

#include<string>

#include<cstring>

#include<map>

#include<stack>

#include<set>

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

typedef long long ll;

typedef unsigned long long ull;

using namespace std;

const int maxn =1200100;

typedef pair<ll,ll>p1;

typedef pair<pair<ll,ll>,ll>p2;

map<pair<pair<ll,ll>,ll>,ll>m1;

map<pair<ll,ll>,ll>m2;

const int MOD = 1e9;

const int N=351;

const int inf = 1000010;

typedef pair<char, int> PAIR;

map<string,int>mm;

ll a[maxn];

ll dp[maxn];

int main()

{

    int n;

    scanf("%d",&n);

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

    {

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

    }

    dp[1] = 0;

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

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

    {

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

    }

    cout<<dp[n]<<endl;

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

using namespace std;



int a[1000001];

int dp[1000001];

int it;



int fr(int p){

	if (p==it-1){

      return 0;

    }

	if(dp[p]!=-1){

      return dp[p];

    }

	int re=abs(a[p]-a[p+1])+fr(p+1);

	if(p!=it-2){

      re=min(re,abs(a[p]-a[p+2])+fr(p+2));

    }

	return dp[p]=re;

}

int main(){

	cin>>it;

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

    cin>>a[i];

    }



	memset(dp,-1,sizeof dp);

	cout<<fr(0)<<endl;

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.1260593317428125x):
#include<iostream>

#include<cstdio>

#include<fstream>

#include<algorithm>

#include<cmath>

#include<deque>

#include<vector>

#include<queue>

#include<string>

#include<cstring>

#include<map>

#include<stack>

#include<set>

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

typedef long long ll;

typedef unsigned long long ull;

using namespace std;

const int maxn =1200100;

typedef pair<ll,ll>p1;

typedef pair<pair<ll,ll>,ll>p2;

map<pair<pair<ll,ll>,ll>,ll>m1;

map<pair<ll,ll>,ll>m2;

const int MOD = 1e9;

const int N=351;

const int inf = 1000010;

typedef pair<char, int> PAIR;

map<string,int>mm;

ll a[maxn];

ll dp[maxn];

int main()

{

    int n;

    scanf("%d",&n);

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

    {

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

    }

    dp[1] = 0;

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

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

    {

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

    }

    cout<<dp[n]<<endl;

}
// Slower C++ solution:


#include <iostream>

#include <bits/stdc++.h>





using namespace std;





int a[100005], p[100005],n, i;

int respuesta(int x)

{

    if(x==n) return 0;

    if(x>n) return 1e9;

    if(p[x]!=-1)return p[x];

    return p[x]=min(abs(a[x+1]-a[x])+respuesta(x+1), abs(a[x+2]-a[x])+respuesta(x+2));

}





int main()

{

    memset(p,-1,sizeof p);

    cin>>n;

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

    {

        cin>>a[i];

    }

    cout<<respuesta(1)<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4154258833799374x):
#include<iostream>

#include<cstdio>

#include<fstream>

#include<algorithm>

#include<cmath>

#include<deque>

#include<vector>

#include<queue>

#include<string>

#include<cstring>

#include<map>

#include<stack>

#include<set>

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

typedef long long ll;

typedef unsigned long long ull;

using namespace std;

const int maxn =1200100;

typedef pair<ll,ll>p1;

typedef pair<pair<ll,ll>,ll>p2;

map<pair<pair<ll,ll>,ll>,ll>m1;

map<pair<ll,ll>,ll>m2;

const int MOD = 1e9;

const int N=351;

const int inf = 1000010;

typedef pair<char, int> PAIR;

map<string,int>mm;

ll a[maxn];

ll dp[maxn];

int main()

{

    int n;

    scanf("%d",&n);

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

    {

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

    }

    dp[1] = 0;

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

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

    {

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

    }

    cout<<dp[n]<<endl;

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

using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9;

const int mod = 1e9+9;

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

const int nax = 1e5+3;

int n;

int dp[nax];

int a[nax];

int f(int i){

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

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

	int ans = abs(a[i]-a[i+1])+f(i+1);

	if(i!=n-2) ans = min(ans, abs(a[i]-a[i+2])+f(i+2));

	return dp[i]=ans;

}

int main(){

	cin>>n;

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

	memset(dp, -1, sizeof dp);

	cout<<f(0)<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.412306088146062x):
#include<iostream>

#include<cstdio>

#include<fstream>

#include<algorithm>

#include<cmath>

#include<deque>

#include<vector>

#include<queue>

#include<string>

#include<cstring>

#include<map>

#include<stack>

#include<set>

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

typedef long long ll;

typedef unsigned long long ull;

using namespace std;

const int maxn =1200100;

typedef pair<ll,ll>p1;

typedef pair<pair<ll,ll>,ll>p2;

map<pair<pair<ll,ll>,ll>,ll>m1;

map<pair<ll,ll>,ll>m2;

const int MOD = 1e9;

const int N=351;

const int inf = 1000010;

typedef pair<char, int> PAIR;

map<string,int>mm;

ll a[maxn];

ll dp[maxn];

int main()

{

    int n;

    scanf("%d",&n);

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

    {

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

    }

    dp[1] = 0;

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

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

    {

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

    }

    cout<<dp[n]<<endl;

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

using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9;

const int mod = 1e9+9;

/*ll gcd(ll x, ll y){return (y ? gcd(y, x%y) : x); }

*ll lcm(ll x, ll y){return x*(y/gcd(x,y));}

*void exEuclid(int a, int b){

	if(b==0){x = 1; y=0; d=a; return ;}

	exEuclid(b, a%b);

	int x1=y;

	int y1=x-(a/b)*y;

	x=x1;

	y=y1;

} 

* */

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

const int nax = 1e5+3;

int n;

int dp[nax];

int a[nax];

int f(int i){

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

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

	int ans = abs(a[i]-a[i+1])+f(i+1);

	if(i!=n-2) ans = min(ans, abs(a[i]-a[i+2])+f(i+2));

	return dp[i]=ans;

}

int main(){

	cin>>n;

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

	memset(dp, -1, sizeof dp);

	cout<<f(0)<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.412306088146062x):
#include<iostream>

#include<cstdio>

#include<fstream>

#include<algorithm>

#include<cmath>

#include<deque>

#include<vector>

#include<queue>

#include<string>

#include<cstring>

#include<map>

#include<stack>

#include<set>

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

typedef long long ll;

typedef unsigned long long ull;

using namespace std;

const int maxn =1200100;

typedef pair<ll,ll>p1;

typedef pair<pair<ll,ll>,ll>p2;

map<pair<pair<ll,ll>,ll>,ll>m1;

map<pair<ll,ll>,ll>m2;

const int MOD = 1e9;

const int N=351;

const int inf = 1000010;

typedef pair<char, int> PAIR;

map<string,int>mm;

ll a[maxn];

ll dp[maxn];

int main()

{

    int n;

    scanf("%d",&n);

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

    {

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

    }

    dp[1] = 0;

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

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

    {

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

    }

    cout<<dp[n]<<endl;

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

using namespace std;

typedef long long ll;

ll dp[100009];

int arr[100009];

ll sum = 0;

long long Frog1(int n){

 

    if(n==0) return 0;

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

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

    else {

            ll x = Frog1(n-1);

            ll y = Frog1(n-2);

        dp[n] = min(abs (arr[n-1] - arr[n]) + x,abs(arr[n-2] - arr[n]) + y);

        //cout << n << " = " << dp[n] << endl;

         return dp[n];

    }

}

 

 

int main()

{

    int n;

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

    cin >> n;

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

    cout << Frog1(n-1) << endl;

}

// Faster C++ solution for the same problem (measured speedup: 1.817658893824893x):
#include<iostream>

#include<cstdio>

#include<fstream>

#include<algorithm>

#include<cmath>

#include<deque>

#include<vector>

#include<queue>

#include<string>

#include<cstring>

#include<map>

#include<stack>

#include<set>

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

typedef long long ll;

typedef unsigned long long ull;

using namespace std;

const int maxn =1200100;

typedef pair<ll,ll>p1;

typedef pair<pair<ll,ll>,ll>p2;

map<pair<pair<ll,ll>,ll>,ll>m1;

map<pair<ll,ll>,ll>m2;

const int MOD = 1e9;

const int N=351;

const int inf = 1000010;

typedef pair<char, int> PAIR;

map<string,int>mm;

ll a[maxn];

ll dp[maxn];

int main()

{

    int n;

    scanf("%d",&n);

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

    {

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

    }

    dp[1] = 0;

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

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

    {

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

    }

    cout<<dp[n]<<endl;

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

using namespace std;

#define ll long long

#define ld long double

#define sf scanf

#define pf printf

#define pb push_back

#define llu unsigned long long

#define U unsigned int

#define SIZE 1000002

#define pie 3.14159265358979323

#define minuss 1e-6

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

 

#define BIG 100000000

 

ll  lcm(ll a,ll b)

{

    ll GCD=__gcd(a,b);

    ll lcm=(a*b)/GCD;

    return lcm;

}

ll dp[100005];

ll weight[1000005],value[1000005];

ll total,val;

ll call(ll n)

{

    if(n>=total)

        return 0;

 

    if(dp[n]!=-1)

        return dp[n];

 

    ll res1=BIG,res2=BIG;

 

   // if(weight[n]+c<=val)

    res1=abs(weight[n]-weight[n+1]) + call(n+1);

    if(n+2<=total) res2=abs(weight[n]-weight[n+2]) + call(n+2);

    //cout<<n<<" "<<res1<<" "<<res2<<endl;

    dp[n]=min(res1,res2);

    return dp[n];

 

}

int main()

{

    cin>>total;

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

        cin>>weight[i];

 

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

    ll ret=call(1);

    cout<<ret<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.8169773979273507x):
#include<iostream>

#include<cstdio>

#include<fstream>

#include<algorithm>

#include<cmath>

#include<deque>

#include<vector>

#include<queue>

#include<string>

#include<cstring>

#include<map>

#include<stack>

#include<set>

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

typedef long long ll;

typedef unsigned long long ull;

using namespace std;

const int maxn =1200100;

typedef pair<ll,ll>p1;

typedef pair<pair<ll,ll>,ll>p2;

map<pair<pair<ll,ll>,ll>,ll>m1;

map<pair<ll,ll>,ll>m2;

const int MOD = 1e9;

const int N=351;

const int inf = 1000010;

typedef pair<char, int> PAIR;

map<string,int>mm;

ll a[maxn];

ll dp[maxn];

int main()

{

    int n;

    scanf("%d",&n);

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

    {

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

    }

    dp[1] = 0;

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

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

    {

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

    }

    cout<<dp[n]<<endl;

}