stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
//       Author: Mostafa Mounir Shehab

 

#include <bits/stdc++.h>

using namespace std;

 

#define PI  3.14159265  // PI = acos(-1)

#define EPS (1e-10)

#define endl "\n"

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

typedef long long ll;

typedef long double ld;

//template <class Type>

 

 

void Fast(){

	std::ios_base::sync_with_stdio(0);

	cin.tie(NULL); cout.tie(NULL); cerr.tie(NULL);

}

 

inline int D(){

	int t;

	scanf("%d",&t);

	return t;

}

inline ll llD(){

	ll t;

	scanf("%lld",&t);

	return t;

}

 

 

ll MOD=1e9 + 7; 

const int N=1e5 + 5,M=50 + 1;

int arr[N];

ll dp[N];

ll OO=1e12; 

 

ll n,m,k,cnt,SU,idx,len,MN,MX;

ll l,r,mid;

ll a,b,c,d;



ll Solve(int idx){

	ll &ret = dp[idx];

	if(~ret)

		return ret;

		

	if(idx >= n)

		return ret = (idx == n) ? 0 : OO;

	

	

	ll x = abs(arr[idx] - arr[idx+1]) + Solve(idx+1);

	ll y = abs(arr[idx] - arr[idx+2]) + Solve(idx+2);

	

	return ret = min(x,y);

}







 

 

int main(){

	

	memset(dp, -1 ,sizeof dp);

	cin>>n;

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

		cin>>arr[i];



	cout<<Solve(1)<<endl;

	



	





}

// Faster C++ solution for the same problem (measured speedup: 1.782339653821436x):
#include <iostream>

#include <vector>

#include <cmath>

#include <set>

#include <queue>

#include <map>

#include <cstdio>

#include <cstring>

#include <string>

#include <stack>

#include <algorithm>

#include <limits.h>

#include <bits/stdc++.h>

#define fw(x) freopen("x.txt","w",stdout)

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

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

#define REP1(i,n) for(int i = 1; i<=n;i++)

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

#define fast ios_base::sync_with_stdio(0);cin.tie(NULL)

#define pf printf

#define sc scanf

#define ll long long

#define pii pair<int,int>

#define pll pair<ll,ll>

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

#define vii vector<int>

#define vll vector<ll>

#define vss vector<string>

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

#define pb push_back





using namespace std;





// moves



//int dx[]= {0,0,1,-1};/*4 side move*/

//int dy[]= {-1,1,0,0};/*4 side move*/

//int dx[]= {1,1,0,-1,-1,-1,0,1};/*8 side move*/

//int dy[]= {0,1,1,1,0,-1,-1,-1};/*8 side move*/

//int dx[]={1,1,2,2,-1,-1,-2,-2};/*knight move*/

//int dy[]={2,-2,1,-1,2,-2,1,-1};/*knight move*/





//big_mod



//ll bigmod(ll a,ll b,ll m)

//{if(b == 0) return 1%m;ll x = bigmod(a,b/2,m);x = (x * x) % m;if(b % 2 == 1) x = (x * a)% m;return x;}

//ll BigMod(ll B,ll P,ll M){ ll R=1%M; while(P>0) {if(P%2==1){R=(R*B)%M;}P/=2;B=(B*B)%M;} return R;} /// (B^P)%M



//ll getBit(ll num, int idx) {return ((num >> idx) & 1ll) == 1ll;}

//ll setBit1(ll num, int idx) {return num or (1ll<<idx);}

//ll setBit0(ll num, int idx) {return num & ~(1ll<<idx);}

//ll flipBit(ll num, int idx) {return num ^ (1ll<<idx);}



const int mod = 1000000007;



int main()

{

    fast;

    int n;

    cin>>n;

    int arr[n];

    REP(i, n) cin>>arr[i];

    int cost[n] = {0};

    cost[1] = abs(arr[1]-arr[0]);

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

      cost[i] = min(cost[i-1]+abs(arr[i-1]-arr[i]), cost[i-2]+abs(arr[i-2]-arr[i]));

    }

    cout<<cost[n-1]<<'\n';

    return 0;

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

using namespace std;

const int maxn = 1e5 + 5;

int main(){

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

	memset(dp,0x3f3f,sizeof(dp));

	scanf("%d",&n);

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

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

	dp[1] = 0;

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

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

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

	}

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3808101175628682x):
#include <iostream>

#include <vector>

#include <cmath>

#include <set>

#include <queue>

#include <map>

#include <cstdio>

#include <cstring>

#include <string>

#include <stack>

#include <algorithm>

#include <limits.h>

#include <bits/stdc++.h>

#define fw(x) freopen("x.txt","w",stdout)

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

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

#define REP1(i,n) for(int i = 1; i<=n;i++)

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

#define fast ios_base::sync_with_stdio(0);cin.tie(NULL)

#define pf printf

#define sc scanf

#define ll long long

#define pii pair<int,int>

#define pll pair<ll,ll>

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

#define vii vector<int>

#define vll vector<ll>

#define vss vector<string>

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

#define pb push_back





using namespace std;





// moves



//int dx[]= {0,0,1,-1};/*4 side move*/

//int dy[]= {-1,1,0,0};/*4 side move*/

//int dx[]= {1,1,0,-1,-1,-1,0,1};/*8 side move*/

//int dy[]= {0,1,1,1,0,-1,-1,-1};/*8 side move*/

//int dx[]={1,1,2,2,-1,-1,-2,-2};/*knight move*/

//int dy[]={2,-2,1,-1,2,-2,1,-1};/*knight move*/





//big_mod



//ll bigmod(ll a,ll b,ll m)

//{if(b == 0) return 1%m;ll x = bigmod(a,b/2,m);x = (x * x) % m;if(b % 2 == 1) x = (x * a)% m;return x;}

//ll BigMod(ll B,ll P,ll M){ ll R=1%M; while(P>0) {if(P%2==1){R=(R*B)%M;}P/=2;B=(B*B)%M;} return R;} /// (B^P)%M



//ll getBit(ll num, int idx) {return ((num >> idx) & 1ll) == 1ll;}

//ll setBit1(ll num, int idx) {return num or (1ll<<idx);}

//ll setBit0(ll num, int idx) {return num & ~(1ll<<idx);}

//ll flipBit(ll num, int idx) {return num ^ (1ll<<idx);}



const int mod = 1000000007;



int main()

{

    fast;

    int n;

    cin>>n;

    int arr[n];

    REP(i, n) cin>>arr[i];

    int cost[n] = {0};

    cost[1] = abs(arr[1]-arr[0]);

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

      cost[i] = min(cost[i-1]+abs(arr[i-1]-arr[i]), cost[i-2]+abs(arr[i-2]-arr[i]));

    }

    cout<<cost[n-1]<<'\n';

    return 0;

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



using namespace std;



const int N = 1e5 + 9;



int n, h[N], dp[N][2];



int sol(int i, int prev){

    if(i >= n)  return 1e9;

    if(i == n - 1)  return abs(h[i] - h[i - 1 - (prev == 1)]);



    int &ret = dp[i][prev];

    if(~ret)    return ret;



    return ret = abs(h[i] - h[i - 1 - (prev == 1)]) + min(sol(i + 1, 0), sol(i + 2, 1));

}



int main(){

    cin >> n;

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

        cin >> h[i];



    memset(dp, -1, sizeof dp);



    cout << min(sol(1, 0), sol(2, 1));



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.7777267207682683x):
#include <iostream>

#include <vector>

#include <cmath>

#include <set>

#include <queue>

#include <map>

#include <cstdio>

#include <cstring>

#include <string>

#include <stack>

#include <algorithm>

#include <limits.h>

#include <bits/stdc++.h>

#define fw(x) freopen("x.txt","w",stdout)

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

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

#define REP1(i,n) for(int i = 1; i<=n;i++)

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

#define fast ios_base::sync_with_stdio(0);cin.tie(NULL)

#define pf printf

#define sc scanf

#define ll long long

#define pii pair<int,int>

#define pll pair<ll,ll>

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

#define vii vector<int>

#define vll vector<ll>

#define vss vector<string>

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

#define pb push_back





using namespace std;





// moves



//int dx[]= {0,0,1,-1};/*4 side move*/

//int dy[]= {-1,1,0,0};/*4 side move*/

//int dx[]= {1,1,0,-1,-1,-1,0,1};/*8 side move*/

//int dy[]= {0,1,1,1,0,-1,-1,-1};/*8 side move*/

//int dx[]={1,1,2,2,-1,-1,-2,-2};/*knight move*/

//int dy[]={2,-2,1,-1,2,-2,1,-1};/*knight move*/





//big_mod



//ll bigmod(ll a,ll b,ll m)

//{if(b == 0) return 1%m;ll x = bigmod(a,b/2,m);x = (x * x) % m;if(b % 2 == 1) x = (x * a)% m;return x;}

//ll BigMod(ll B,ll P,ll M){ ll R=1%M; while(P>0) {if(P%2==1){R=(R*B)%M;}P/=2;B=(B*B)%M;} return R;} /// (B^P)%M



//ll getBit(ll num, int idx) {return ((num >> idx) & 1ll) == 1ll;}

//ll setBit1(ll num, int idx) {return num or (1ll<<idx);}

//ll setBit0(ll num, int idx) {return num & ~(1ll<<idx);}

//ll flipBit(ll num, int idx) {return num ^ (1ll<<idx);}



const int mod = 1000000007;



int main()

{

    fast;

    int n;

    cin>>n;

    int arr[n];

    REP(i, n) cin>>arr[i];

    int cost[n] = {0};

    cost[1] = abs(arr[1]-arr[0]);

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

      cost[i] = min(cost[i-1]+abs(arr[i-1]-arr[i]), cost[i-2]+abs(arr[i-2]-arr[i]));

    }

    cout<<cost[n-1]<<'\n';

    return 0;

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



using namespace std;



int n, h[100002], dp[100002][3];



int sol(int i, int choice){

    if(i==0) return 0;

    if(i<0) return 1E9;



    int &ret = dp[i][choice];

    if(~ret) return ret;



    return ret = min(abs(h[i]-h[i-1]) + sol(i-1, 1), abs(h[i]-h[i-2]) + sol(i-2, 2));

}



int main()

{

    cin >> n;

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

        cin >> h[i];



    memset(dp, -1, sizeof dp);

    cout << min(sol(n-1, 1), sol(n-1, 2));

}


// Faster C++ solution for the same problem (measured speedup: 2.175701216830705x):
#include <iostream>

#include <vector>

#include <cmath>

#include <set>

#include <queue>

#include <map>

#include <cstdio>

#include <cstring>

#include <string>

#include <stack>

#include <algorithm>

#include <limits.h>

#include <bits/stdc++.h>

#define fw(x) freopen("x.txt","w",stdout)

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

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

#define REP1(i,n) for(int i = 1; i<=n;i++)

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

#define fast ios_base::sync_with_stdio(0);cin.tie(NULL)

#define pf printf

#define sc scanf

#define ll long long

#define pii pair<int,int>

#define pll pair<ll,ll>

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

#define vii vector<int>

#define vll vector<ll>

#define vss vector<string>

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

#define pb push_back





using namespace std;





// moves



//int dx[]= {0,0,1,-1};/*4 side move*/

//int dy[]= {-1,1,0,0};/*4 side move*/

//int dx[]= {1,1,0,-1,-1,-1,0,1};/*8 side move*/

//int dy[]= {0,1,1,1,0,-1,-1,-1};/*8 side move*/

//int dx[]={1,1,2,2,-1,-1,-2,-2};/*knight move*/

//int dy[]={2,-2,1,-1,2,-2,1,-1};/*knight move*/





//big_mod



//ll bigmod(ll a,ll b,ll m)

//{if(b == 0) return 1%m;ll x = bigmod(a,b/2,m);x = (x * x) % m;if(b % 2 == 1) x = (x * a)% m;return x;}

//ll BigMod(ll B,ll P,ll M){ ll R=1%M; while(P>0) {if(P%2==1){R=(R*B)%M;}P/=2;B=(B*B)%M;} return R;} /// (B^P)%M



//ll getBit(ll num, int idx) {return ((num >> idx) & 1ll) == 1ll;}

//ll setBit1(ll num, int idx) {return num or (1ll<<idx);}

//ll setBit0(ll num, int idx) {return num & ~(1ll<<idx);}

//ll flipBit(ll num, int idx) {return num ^ (1ll<<idx);}



const int mod = 1000000007;



int main()

{

    fast;

    int n;

    cin>>n;

    int arr[n];

    REP(i, n) cin>>arr[i];

    int cost[n] = {0};

    cost[1] = abs(arr[1]-arr[0]);

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

      cost[i] = min(cost[i-1]+abs(arr[i-1]-arr[i]), cost[i-2]+abs(arr[i-2]-arr[i]));

    }

    cout<<cost[n-1]<<'\n';

    return 0;

}
// Slower C++ solution:


#include <bits/stdc++.h>

using namespace  std;



int N;

const int SIZE = 100005;

long h [SIZE];

int mem [SIZE];



int dp (int i) {

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

  

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



  long ret = LLONG_MAX;



  if (i+1 < N) 

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

  if (i+2 < N)

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



  return  mem[i] = ret;

}



int main (){ 



  cin  >> N;

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

    cin >> h[i];

  }



  memset(mem, -1, sizeof mem);



  cout << dp(0);



}

// Faster C++ solution for the same problem (measured speedup: 1.3803086593923912x):
#include <iostream>

#include <vector>

#include <cmath>

#include <set>

#include <queue>

#include <map>

#include <cstdio>

#include <cstring>

#include <string>

#include <stack>

#include <algorithm>

#include <limits.h>

#include <bits/stdc++.h>

#define fw(x) freopen("x.txt","w",stdout)

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

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

#define REP1(i,n) for(int i = 1; i<=n;i++)

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

#define fast ios_base::sync_with_stdio(0);cin.tie(NULL)

#define pf printf

#define sc scanf

#define ll long long

#define pii pair<int,int>

#define pll pair<ll,ll>

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

#define vii vector<int>

#define vll vector<ll>

#define vss vector<string>

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

#define pb push_back





using namespace std;





// moves



//int dx[]= {0,0,1,-1};/*4 side move*/

//int dy[]= {-1,1,0,0};/*4 side move*/

//int dx[]= {1,1,0,-1,-1,-1,0,1};/*8 side move*/

//int dy[]= {0,1,1,1,0,-1,-1,-1};/*8 side move*/

//int dx[]={1,1,2,2,-1,-1,-2,-2};/*knight move*/

//int dy[]={2,-2,1,-1,2,-2,1,-1};/*knight move*/





//big_mod



//ll bigmod(ll a,ll b,ll m)

//{if(b == 0) return 1%m;ll x = bigmod(a,b/2,m);x = (x * x) % m;if(b % 2 == 1) x = (x * a)% m;return x;}

//ll BigMod(ll B,ll P,ll M){ ll R=1%M; while(P>0) {if(P%2==1){R=(R*B)%M;}P/=2;B=(B*B)%M;} return R;} /// (B^P)%M



//ll getBit(ll num, int idx) {return ((num >> idx) & 1ll) == 1ll;}

//ll setBit1(ll num, int idx) {return num or (1ll<<idx);}

//ll setBit0(ll num, int idx) {return num & ~(1ll<<idx);}

//ll flipBit(ll num, int idx) {return num ^ (1ll<<idx);}



const int mod = 1000000007;



int main()

{

    fast;

    int n;

    cin>>n;

    int arr[n];

    REP(i, n) cin>>arr[i];

    int cost[n] = {0};

    cost[1] = abs(arr[1]-arr[0]);

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

      cost[i] = min(cost[i-1]+abs(arr[i-1]-arr[i]), cost[i-2]+abs(arr[i-2]-arr[i]));

    }

    cout<<cost[n-1]<<'\n';

    return 0;

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

#define maxn 100005

using namespace std;



int main(){

    int n;

    cin >> n;

    int h[maxn];

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

        cin >> h[i];

    }

    int dp[maxn];

    fill(dp, dp+maxn, 0x7fffffff);

    dp[0] = 0;

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

        int temp = 0;

        if(i+1 < n){

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

            if(temp < dp[i+1])

                dp[i+1] = temp;

        }

        if(i+2 < n){

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

            if(temp < dp[i+2])

                dp[i+2] = temp;

        }

    }

    cout << dp[n-1];

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.387135516982201x):
#include <iostream>

#include <vector>

#include <cmath>

#include <set>

#include <queue>

#include <map>

#include <cstdio>

#include <cstring>

#include <string>

#include <stack>

#include <algorithm>

#include <limits.h>

#include <bits/stdc++.h>

#define fw(x) freopen("x.txt","w",stdout)

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

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

#define REP1(i,n) for(int i = 1; i<=n;i++)

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

#define fast ios_base::sync_with_stdio(0);cin.tie(NULL)

#define pf printf

#define sc scanf

#define ll long long

#define pii pair<int,int>

#define pll pair<ll,ll>

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

#define vii vector<int>

#define vll vector<ll>

#define vss vector<string>

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

#define pb push_back





using namespace std;





// moves



//int dx[]= {0,0,1,-1};/*4 side move*/

//int dy[]= {-1,1,0,0};/*4 side move*/

//int dx[]= {1,1,0,-1,-1,-1,0,1};/*8 side move*/

//int dy[]= {0,1,1,1,0,-1,-1,-1};/*8 side move*/

//int dx[]={1,1,2,2,-1,-1,-2,-2};/*knight move*/

//int dy[]={2,-2,1,-1,2,-2,1,-1};/*knight move*/





//big_mod



//ll bigmod(ll a,ll b,ll m)

//{if(b == 0) return 1%m;ll x = bigmod(a,b/2,m);x = (x * x) % m;if(b % 2 == 1) x = (x * a)% m;return x;}

//ll BigMod(ll B,ll P,ll M){ ll R=1%M; while(P>0) {if(P%2==1){R=(R*B)%M;}P/=2;B=(B*B)%M;} return R;} /// (B^P)%M



//ll getBit(ll num, int idx) {return ((num >> idx) & 1ll) == 1ll;}

//ll setBit1(ll num, int idx) {return num or (1ll<<idx);}

//ll setBit0(ll num, int idx) {return num & ~(1ll<<idx);}

//ll flipBit(ll num, int idx) {return num ^ (1ll<<idx);}



const int mod = 1000000007;



int main()

{

    fast;

    int n;

    cin>>n;

    int arr[n];

    REP(i, n) cin>>arr[i];

    int cost[n] = {0};

    cost[1] = abs(arr[1]-arr[0]);

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

      cost[i] = min(cost[i-1]+abs(arr[i-1]-arr[i]), cost[i-2]+abs(arr[i-2]-arr[i]));

    }

    cout<<cost[n-1]<<'\n';

    return 0;

}
// Slower C++ solution:
#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;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.394926247005469x):
#include <iostream>

#include <vector>

#include <cmath>

#include <set>

#include <queue>

#include <map>

#include <cstdio>

#include <cstring>

#include <string>

#include <stack>

#include <algorithm>

#include <limits.h>

#include <bits/stdc++.h>

#define fw(x) freopen("x.txt","w",stdout)

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

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

#define REP1(i,n) for(int i = 1; i<=n;i++)

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

#define fast ios_base::sync_with_stdio(0);cin.tie(NULL)

#define pf printf

#define sc scanf

#define ll long long

#define pii pair<int,int>

#define pll pair<ll,ll>

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

#define vii vector<int>

#define vll vector<ll>

#define vss vector<string>

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

#define pb push_back





using namespace std;





// moves



//int dx[]= {0,0,1,-1};/*4 side move*/

//int dy[]= {-1,1,0,0};/*4 side move*/

//int dx[]= {1,1,0,-1,-1,-1,0,1};/*8 side move*/

//int dy[]= {0,1,1,1,0,-1,-1,-1};/*8 side move*/

//int dx[]={1,1,2,2,-1,-1,-2,-2};/*knight move*/

//int dy[]={2,-2,1,-1,2,-2,1,-1};/*knight move*/





//big_mod



//ll bigmod(ll a,ll b,ll m)

//{if(b == 0) return 1%m;ll x = bigmod(a,b/2,m);x = (x * x) % m;if(b % 2 == 1) x = (x * a)% m;return x;}

//ll BigMod(ll B,ll P,ll M){ ll R=1%M; while(P>0) {if(P%2==1){R=(R*B)%M;}P/=2;B=(B*B)%M;} return R;} /// (B^P)%M



//ll getBit(ll num, int idx) {return ((num >> idx) & 1ll) == 1ll;}

//ll setBit1(ll num, int idx) {return num or (1ll<<idx);}

//ll setBit0(ll num, int idx) {return num & ~(1ll<<idx);}

//ll flipBit(ll num, int idx) {return num ^ (1ll<<idx);}



const int mod = 1000000007;



int main()

{

    fast;

    int n;

    cin>>n;

    int arr[n];

    REP(i, n) cin>>arr[i];

    int cost[n] = {0};

    cost[1] = abs(arr[1]-arr[0]);

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

      cost[i] = min(cost[i-1]+abs(arr[i-1]-arr[i]), cost[i-2]+abs(arr[i-2]-arr[i]));

    }

    cout<<cost[n-1]<<'\n';

    return 0;

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

using namespace std;



const int inf=0x3f;

int h[100005],dp[100005];



signed main()

{

	int n;

	while(cin>>n){

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

		memset(dp,inf,sizeof dp);

		

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

		for(int i=3;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]<<endl;

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3828817146855812x):
#include <iostream>

#include <vector>

#include <cmath>

#include <set>

#include <queue>

#include <map>

#include <cstdio>

#include <cstring>

#include <string>

#include <stack>

#include <algorithm>

#include <limits.h>

#include <bits/stdc++.h>

#define fw(x) freopen("x.txt","w",stdout)

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

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

#define REP1(i,n) for(int i = 1; i<=n;i++)

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

#define fast ios_base::sync_with_stdio(0);cin.tie(NULL)

#define pf printf

#define sc scanf

#define ll long long

#define pii pair<int,int>

#define pll pair<ll,ll>

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

#define vii vector<int>

#define vll vector<ll>

#define vss vector<string>

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

#define pb push_back





using namespace std;





// moves



//int dx[]= {0,0,1,-1};/*4 side move*/

//int dy[]= {-1,1,0,0};/*4 side move*/

//int dx[]= {1,1,0,-1,-1,-1,0,1};/*8 side move*/

//int dy[]= {0,1,1,1,0,-1,-1,-1};/*8 side move*/

//int dx[]={1,1,2,2,-1,-1,-2,-2};/*knight move*/

//int dy[]={2,-2,1,-1,2,-2,1,-1};/*knight move*/





//big_mod



//ll bigmod(ll a,ll b,ll m)

//{if(b == 0) return 1%m;ll x = bigmod(a,b/2,m);x = (x * x) % m;if(b % 2 == 1) x = (x * a)% m;return x;}

//ll BigMod(ll B,ll P,ll M){ ll R=1%M; while(P>0) {if(P%2==1){R=(R*B)%M;}P/=2;B=(B*B)%M;} return R;} /// (B^P)%M



//ll getBit(ll num, int idx) {return ((num >> idx) & 1ll) == 1ll;}

//ll setBit1(ll num, int idx) {return num or (1ll<<idx);}

//ll setBit0(ll num, int idx) {return num & ~(1ll<<idx);}

//ll flipBit(ll num, int idx) {return num ^ (1ll<<idx);}



const int mod = 1000000007;



int main()

{

    fast;

    int n;

    cin>>n;

    int arr[n];

    REP(i, n) cin>>arr[i];

    int cost[n] = {0};

    cost[1] = abs(arr[1]-arr[0]);

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

      cost[i] = min(cost[i-1]+abs(arr[i-1]-arr[i]), cost[i-2]+abs(arr[i-2]-arr[i]));

    }

    cout<<cost[n-1]<<'\n';

    return 0;

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

using namespace std;

vector<int> dp(100001,-1);



int back(vector<int> &N, int index){

	if(index+2==N.size()){

		return abs(N[index]-N[index+1]);

	}

	if(index+1==N.size()){

		return 0;

	}

	if(dp[N.size()-index]!=-1)return dp[N.size()-index];

	int a = abs(N[index]-N[index+1]) + back(N,index+1);

	int b = abs(N[index]-N[index+2]) + back(N,index+2);

	

	dp[N.size()-index] = min(a,b);

	return dp[N.size()-index];

}



int main(){

//	10 30 40 20

	int n;

	cin>>n;

	vector<int> N(n);

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

		cin>>N[i];

	}

	dp[0]=0;

	dp[1]=abs(N[0]-N[1]);



	cout<<back(N,0);	

}

// Faster C++ solution for the same problem (measured speedup: 1.3935848767636858x):
#include <iostream>

#include <vector>

#include <cmath>

#include <set>

#include <queue>

#include <map>

#include <cstdio>

#include <cstring>

#include <string>

#include <stack>

#include <algorithm>

#include <limits.h>

#include <bits/stdc++.h>

#define fw(x) freopen("x.txt","w",stdout)

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

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

#define REP1(i,n) for(int i = 1; i<=n;i++)

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

#define fast ios_base::sync_with_stdio(0);cin.tie(NULL)

#define pf printf

#define sc scanf

#define ll long long

#define pii pair<int,int>

#define pll pair<ll,ll>

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

#define vii vector<int>

#define vll vector<ll>

#define vss vector<string>

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

#define pb push_back





using namespace std;





// moves



//int dx[]= {0,0,1,-1};/*4 side move*/

//int dy[]= {-1,1,0,0};/*4 side move*/

//int dx[]= {1,1,0,-1,-1,-1,0,1};/*8 side move*/

//int dy[]= {0,1,1,1,0,-1,-1,-1};/*8 side move*/

//int dx[]={1,1,2,2,-1,-1,-2,-2};/*knight move*/

//int dy[]={2,-2,1,-1,2,-2,1,-1};/*knight move*/





//big_mod



//ll bigmod(ll a,ll b,ll m)

//{if(b == 0) return 1%m;ll x = bigmod(a,b/2,m);x = (x * x) % m;if(b % 2 == 1) x = (x * a)% m;return x;}

//ll BigMod(ll B,ll P,ll M){ ll R=1%M; while(P>0) {if(P%2==1){R=(R*B)%M;}P/=2;B=(B*B)%M;} return R;} /// (B^P)%M



//ll getBit(ll num, int idx) {return ((num >> idx) & 1ll) == 1ll;}

//ll setBit1(ll num, int idx) {return num or (1ll<<idx);}

//ll setBit0(ll num, int idx) {return num & ~(1ll<<idx);}

//ll flipBit(ll num, int idx) {return num ^ (1ll<<idx);}



const int mod = 1000000007;



int main()

{

    fast;

    int n;

    cin>>n;

    int arr[n];

    REP(i, n) cin>>arr[i];

    int cost[n] = {0};

    cost[1] = abs(arr[1]-arr[0]);

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

      cost[i] = min(cost[i-1]+abs(arr[i-1]-arr[i]), cost[i-2]+abs(arr[i-2]-arr[i]));

    }

    cout<<cost[n-1]<<'\n';

    return 0;

}