stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 2.5544386554109293x):
#define _CRT_SECURE_NO_WARNINGS

#include<iostream>

#include<functional>

#include <bitset>

#include<algorithm>

#include<fstream>

#include<string>

#include <cstring>

#include<stdio.h>

#include<iomanip>

#include<sstream>

#include<cmath>

#include<vector>

#include<assert.h>

#include<queue>

#include<deque>

#include<stack>

#include<set>

#include<map>

// never lose hope ..

typedef unsigned long long ull;

typedef long long ll;

typedef long double ld;

#define eps 1e-8

using namespace std;

double const  PI = 3.141592653;

const int OO = 1000000000 + 9;



void fast() {

	ios_base::sync_with_stdio(0);

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

}





int main()

{

	fast();



	int n;

	cin >> n;

	vector<int>v(n + 1, 0), dp(n + 1, OO);

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

		cin >> v[i];

	}

	dp[1] = 0;

	dp[2] = dp[1] + abs(v[1] - v[2]);

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

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

	}

	cout << dp[n];





	return 0;

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

using namespace std;

typedef long long int ll;

const int MAX=1e5+10;

int N,A[MAX],dp[MAX];

int solve(int idx)

{

    if(idx==N)return 0;

    if(idx>N)return 1e9;

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

    return dp[idx]=min(abs(A[idx+1]-A[idx])+solve(idx+1),abs(A[idx+2]-A[idx])+solve(idx+2));

}

int  main ()

{

    memset(dp,-1,sizeof dp);

    cin >> N;

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

        cin >> A[i];

    }

    cout<<solve(1)<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3767129489944707x):
#define _CRT_SECURE_NO_WARNINGS

#include<iostream>

#include<functional>

#include <bitset>

#include<algorithm>

#include<fstream>

#include<string>

#include <cstring>

#include<stdio.h>

#include<iomanip>

#include<sstream>

#include<cmath>

#include<vector>

#include<assert.h>

#include<queue>

#include<deque>

#include<stack>

#include<set>

#include<map>

// never lose hope ..

typedef unsigned long long ull;

typedef long long ll;

typedef long double ld;

#define eps 1e-8

using namespace std;

double const  PI = 3.141592653;

const int OO = 1000000000 + 9;



void fast() {

	ios_base::sync_with_stdio(0);

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

}





int main()

{

	fast();



	int n;

	cin >> n;

	vector<int>v(n + 1, 0), dp(n + 1, OO);

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

		cin >> v[i];

	}

	dp[1] = 0;

	dp[2] = dp[1] + abs(v[1] - v[2]);

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

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

	}

	cout << dp[n];





	return 0;

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

using namespace std;



#define mp make_pair

#define pb push_back

#define st first

#define nd second

#define lb lower_bound

#define ub upper_bound



typedef long long ll;

typedef long double ld;

typedef pair<int,int> pi;

typedef pair<ll,ll> pll;

typedef vector<int> vi;

typedef vector<ll> vll;

typedef vector<pi> vpi;

typedef vector<pll> vpll;



const ll oo=1e18;

const int MOD=1000000007;

const int INF=1000000000;

const ld EPS=0.000000001;



int dp[100005],h[100005],n,k=2;



signed main() {

    ios_base::sync_with_stdio(NULL);

    cin.tie(NULL);

    cout.tie(NULL);

    memset(dp,0x3f3f,sizeof dp);

    dp[1]=0;

    h[0]=0;

    cin>>n;

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

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

        for(int j=1;j<=min(k,i);++j) {

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

        }

    }

    cout<<dp[n];

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3968630734316436x):
#define _CRT_SECURE_NO_WARNINGS

#include<iostream>

#include<functional>

#include <bitset>

#include<algorithm>

#include<fstream>

#include<string>

#include <cstring>

#include<stdio.h>

#include<iomanip>

#include<sstream>

#include<cmath>

#include<vector>

#include<assert.h>

#include<queue>

#include<deque>

#include<stack>

#include<set>

#include<map>

// never lose hope ..

typedef unsigned long long ull;

typedef long long ll;

typedef long double ld;

#define eps 1e-8

using namespace std;

double const  PI = 3.141592653;

const int OO = 1000000000 + 9;



void fast() {

	ios_base::sync_with_stdio(0);

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

}





int main()

{

	fast();



	int n;

	cin >> n;

	vector<int>v(n + 1, 0), dp(n + 1, OO);

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

		cin >> v[i];

	}

	dp[1] = 0;

	dp[2] = dp[1] + abs(v[1] - v[2]);

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

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

	}

	cout << dp[n];





	return 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: 1.3722436919676901x):
#define _CRT_SECURE_NO_WARNINGS

#include<iostream>

#include<functional>

#include <bitset>

#include<algorithm>

#include<fstream>

#include<string>

#include <cstring>

#include<stdio.h>

#include<iomanip>

#include<sstream>

#include<cmath>

#include<vector>

#include<assert.h>

#include<queue>

#include<deque>

#include<stack>

#include<set>

#include<map>

// never lose hope ..

typedef unsigned long long ull;

typedef long long ll;

typedef long double ld;

#define eps 1e-8

using namespace std;

double const  PI = 3.141592653;

const int OO = 1000000000 + 9;



void fast() {

	ios_base::sync_with_stdio(0);

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

}





int main()

{

	fast();



	int n;

	cin >> n;

	vector<int>v(n + 1, 0), dp(n + 1, OO);

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

		cin >> v[i];

	}

	dp[1] = 0;

	dp[2] = dp[1] + abs(v[1] - v[2]);

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

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

	}

	cout << dp[n];





	return 0;

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

#include<cstdio>

#include<string>

#include<cstring>

#include<algorithm>

#include<queue>

#include<stack>

#include<map>

#include<vector>

#include<set>

#include<list>

#define ll long long

using namespace std;

ll h[200050],dp[200050];

int main()

{

	ll n;

	cin>>n;

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

		cin>>h[i];

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

	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: 2.5695718764662177x):
#define _CRT_SECURE_NO_WARNINGS

#include<iostream>

#include<functional>

#include <bitset>

#include<algorithm>

#include<fstream>

#include<string>

#include <cstring>

#include<stdio.h>

#include<iomanip>

#include<sstream>

#include<cmath>

#include<vector>

#include<assert.h>

#include<queue>

#include<deque>

#include<stack>

#include<set>

#include<map>

// never lose hope ..

typedef unsigned long long ull;

typedef long long ll;

typedef long double ld;

#define eps 1e-8

using namespace std;

double const  PI = 3.141592653;

const int OO = 1000000000 + 9;



void fast() {

	ios_base::sync_with_stdio(0);

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

}





int main()

{

	fast();



	int n;

	cin >> n;

	vector<int>v(n + 1, 0), dp(n + 1, OO);

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

		cin >> v[i];

	}

	dp[1] = 0;

	dp[2] = dp[1] + abs(v[1] - v[2]);

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

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

	}

	cout << dp[n];





	return 0;

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



using namespace std;

const int maxn = 2e5+2;

const int mod = 1e9+7;

const int inf = 1e9;



typedef long long ll;



int n;

int arr[maxn], dp[maxn];



int solve_dp (int pos){

    if (pos > n) return inf;

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



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



    int ans = inf;

    ans = min(ans, solve_dp(pos + 1) + abs(arr[pos + 1] - arr[pos]));

    ans = min(ans, solve_dp(pos + 2) + abs(arr[pos + 2] - arr[pos]));



    return dp[pos] = ans;

}



int main(){



    ios_base::sync_with_stdio(0);

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



    cin >> n;



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

        cin >> arr[i];

    }



    memset(dp, -1, sizeof dp);



    cout << solve_dp(0) << "\n";



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.792908955507252x):
#define _CRT_SECURE_NO_WARNINGS

#include<iostream>

#include<functional>

#include <bitset>

#include<algorithm>

#include<fstream>

#include<string>

#include <cstring>

#include<stdio.h>

#include<iomanip>

#include<sstream>

#include<cmath>

#include<vector>

#include<assert.h>

#include<queue>

#include<deque>

#include<stack>

#include<set>

#include<map>

// never lose hope ..

typedef unsigned long long ull;

typedef long long ll;

typedef long double ld;

#define eps 1e-8

using namespace std;

double const  PI = 3.141592653;

const int OO = 1000000000 + 9;



void fast() {

	ios_base::sync_with_stdio(0);

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

}





int main()

{

	fast();



	int n;

	cin >> n;

	vector<int>v(n + 1, 0), dp(n + 1, OO);

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

		cin >> v[i];

	}

	dp[1] = 0;

	dp[2] = dp[1] + abs(v[1] - v[2]);

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

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

	}

	cout << dp[n];





	return 0;

}
// Slower C++ solution:
#pragma GCC optimize ("O3")

#include <bits/stdc++.h>

#define abdelrahman010 ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

typedef long long ll;

using namespace std;

ll n , arr[100005] , memo[100005];

ll solve(ll i)

{

	if(i == n - 1)

		return 0;

	ll &ans = memo[i];

	if(~ans)

		return ans;

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

	if(i < n - 2)

		ans = min(ans , abs(arr[i] - arr[i + 2]) + solve(i + 2));

	return ans;

}

int main()

{

	abdelrahman010

	cin >> n;

	memset(memo , -1 , sizeof memo);

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

		cin >> arr[i];

	cout << solve(0);

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.7899846382853604x):
#define _CRT_SECURE_NO_WARNINGS

#include<iostream>

#include<functional>

#include <bitset>

#include<algorithm>

#include<fstream>

#include<string>

#include <cstring>

#include<stdio.h>

#include<iomanip>

#include<sstream>

#include<cmath>

#include<vector>

#include<assert.h>

#include<queue>

#include<deque>

#include<stack>

#include<set>

#include<map>

// never lose hope ..

typedef unsigned long long ull;

typedef long long ll;

typedef long double ld;

#define eps 1e-8

using namespace std;

double const  PI = 3.141592653;

const int OO = 1000000000 + 9;



void fast() {

	ios_base::sync_with_stdio(0);

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

}





int main()

{

	fast();



	int n;

	cin >> n;

	vector<int>v(n + 1, 0), dp(n + 1, OO);

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

		cin >> v[i];

	}

	dp[1] = 0;

	dp[2] = dp[1] + abs(v[1] - v[2]);

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

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

	}

	cout << dp[n];





	return 0;

}
// 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: 1.3813662510860825x):
#define _CRT_SECURE_NO_WARNINGS

#include<iostream>

#include<functional>

#include <bitset>

#include<algorithm>

#include<fstream>

#include<string>

#include <cstring>

#include<stdio.h>

#include<iomanip>

#include<sstream>

#include<cmath>

#include<vector>

#include<assert.h>

#include<queue>

#include<deque>

#include<stack>

#include<set>

#include<map>

// never lose hope ..

typedef unsigned long long ull;

typedef long long ll;

typedef long double ld;

#define eps 1e-8

using namespace std;

double const  PI = 3.141592653;

const int OO = 1000000000 + 9;



void fast() {

	ios_base::sync_with_stdio(0);

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

}





int main()

{

	fast();



	int n;

	cin >> n;

	vector<int>v(n + 1, 0), dp(n + 1, OO);

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

		cin >> v[i];

	}

	dp[1] = 0;

	dp[2] = dp[1] + abs(v[1] - v[2]);

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

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

	}

	cout << dp[n];





	return 0;

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

#include<stdio.h>

#include<cstring>

#include<cmath>

#include<algorithm>

using namespace std;

#define maxn 100005

int dp[maxn];

int h[maxn];

int n;

int main(){

    scanf("%d",&n);

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

    memset(h,0,sizeof(h));

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

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

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

        if(i-1>=1&&i-2>=1)

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

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

        else if(i-2>=1) dp[i] = dp[i-2]+abs(h[i]-h[i-2]);

    }

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.7653617855229413x):
#define _CRT_SECURE_NO_WARNINGS

#include<iostream>

#include<functional>

#include <bitset>

#include<algorithm>

#include<fstream>

#include<string>

#include <cstring>

#include<stdio.h>

#include<iomanip>

#include<sstream>

#include<cmath>

#include<vector>

#include<assert.h>

#include<queue>

#include<deque>

#include<stack>

#include<set>

#include<map>

// never lose hope ..

typedef unsigned long long ull;

typedef long long ll;

typedef long double ld;

#define eps 1e-8

using namespace std;

double const  PI = 3.141592653;

const int OO = 1000000000 + 9;



void fast() {

	ios_base::sync_with_stdio(0);

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

}





int main()

{

	fast();



	int n;

	cin >> n;

	vector<int>v(n + 1, 0), dp(n + 1, OO);

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

		cin >> v[i];

	}

	dp[1] = 0;

	dp[2] = dp[1] + abs(v[1] - v[2]);

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

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

	}

	cout << dp[n];





	return 0;

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



using namespace std;

int v[100010];

int memo[100010];

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: 1.39721677754652x):
#define _CRT_SECURE_NO_WARNINGS

#include<iostream>

#include<functional>

#include <bitset>

#include<algorithm>

#include<fstream>

#include<string>

#include <cstring>

#include<stdio.h>

#include<iomanip>

#include<sstream>

#include<cmath>

#include<vector>

#include<assert.h>

#include<queue>

#include<deque>

#include<stack>

#include<set>

#include<map>

// never lose hope ..

typedef unsigned long long ull;

typedef long long ll;

typedef long double ld;

#define eps 1e-8

using namespace std;

double const  PI = 3.141592653;

const int OO = 1000000000 + 9;



void fast() {

	ios_base::sync_with_stdio(0);

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

}





int main()

{

	fast();



	int n;

	cin >> n;

	vector<int>v(n + 1, 0), dp(n + 1, OO);

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

		cin >> v[i];

	}

	dp[1] = 0;

	dp[2] = dp[1] + abs(v[1] - v[2]);

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

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

	}

	cout << dp[n];





	return 0;

}
// 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: 1.392532151623487x):
#define _CRT_SECURE_NO_WARNINGS

#include<iostream>

#include<functional>

#include <bitset>

#include<algorithm>

#include<fstream>

#include<string>

#include <cstring>

#include<stdio.h>

#include<iomanip>

#include<sstream>

#include<cmath>

#include<vector>

#include<assert.h>

#include<queue>

#include<deque>

#include<stack>

#include<set>

#include<map>

// never lose hope ..

typedef unsigned long long ull;

typedef long long ll;

typedef long double ld;

#define eps 1e-8

using namespace std;

double const  PI = 3.141592653;

const int OO = 1000000000 + 9;



void fast() {

	ios_base::sync_with_stdio(0);

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

}





int main()

{

	fast();



	int n;

	cin >> n;

	vector<int>v(n + 1, 0), dp(n + 1, OO);

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

		cin >> v[i];

	}

	dp[1] = 0;

	dp[2] = dp[1] + abs(v[1] - v[2]);

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

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

	}

	cout << dp[n];





	return 0;

}
// 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: 1.369010203945436x):
#define _CRT_SECURE_NO_WARNINGS

#include<iostream>

#include<functional>

#include <bitset>

#include<algorithm>

#include<fstream>

#include<string>

#include <cstring>

#include<stdio.h>

#include<iomanip>

#include<sstream>

#include<cmath>

#include<vector>

#include<assert.h>

#include<queue>

#include<deque>

#include<stack>

#include<set>

#include<map>

// never lose hope ..

typedef unsigned long long ull;

typedef long long ll;

typedef long double ld;

#define eps 1e-8

using namespace std;

double const  PI = 3.141592653;

const int OO = 1000000000 + 9;



void fast() {

	ios_base::sync_with_stdio(0);

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

}





int main()

{

	fast();



	int n;

	cin >> n;

	vector<int>v(n + 1, 0), dp(n + 1, OO);

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

		cin >> v[i];

	}

	dp[1] = 0;

	dp[2] = dp[1] + abs(v[1] - v[2]);

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

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

	}

	cout << dp[n];





	return 0;

}
// 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: 1.379489418050035x):
#define _CRT_SECURE_NO_WARNINGS

#include<iostream>

#include<functional>

#include <bitset>

#include<algorithm>

#include<fstream>

#include<string>

#include <cstring>

#include<stdio.h>

#include<iomanip>

#include<sstream>

#include<cmath>

#include<vector>

#include<assert.h>

#include<queue>

#include<deque>

#include<stack>

#include<set>

#include<map>

// never lose hope ..

typedef unsigned long long ull;

typedef long long ll;

typedef long double ld;

#define eps 1e-8

using namespace std;

double const  PI = 3.141592653;

const int OO = 1000000000 + 9;



void fast() {

	ios_base::sync_with_stdio(0);

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

}





int main()

{

	fast();



	int n;

	cin >> n;

	vector<int>v(n + 1, 0), dp(n + 1, OO);

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

		cin >> v[i];

	}

	dp[1] = 0;

	dp[2] = dp[1] + abs(v[1] - v[2]);

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

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

	}

	cout << dp[n];





	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: 1.36867318591181x):
#define _CRT_SECURE_NO_WARNINGS

#include<iostream>

#include<functional>

#include <bitset>

#include<algorithm>

#include<fstream>

#include<string>

#include <cstring>

#include<stdio.h>

#include<iomanip>

#include<sstream>

#include<cmath>

#include<vector>

#include<assert.h>

#include<queue>

#include<deque>

#include<stack>

#include<set>

#include<map>

// never lose hope ..

typedef unsigned long long ull;

typedef long long ll;

typedef long double ld;

#define eps 1e-8

using namespace std;

double const  PI = 3.141592653;

const int OO = 1000000000 + 9;



void fast() {

	ios_base::sync_with_stdio(0);

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

}





int main()

{

	fast();



	int n;

	cin >> n;

	vector<int>v(n + 1, 0), dp(n + 1, OO);

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

		cin >> v[i];

	}

	dp[1] = 0;

	dp[2] = dp[1] + abs(v[1] - v[2]);

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

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

	}

	cout << dp[n];





	return 0;

}