stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 1.370823651939799x):
#include <bits/stdc++.h>

#include <algorithm>

#include <climits>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <iostream>

#include <math.h>

using namespace __gnu_pbds;

using namespace std;

#define ll long long

typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> pbds;

#define vll vector<ll> 

#define nl cout << endl

#define fr(a,b) for(auto i=a;i<b;i++)

#define pll pair<ll,ll>

#define f first

#define s second

#define vpll vector<pll>

#define ump unordered_map

#define YES cout << "YES" << endl

#define NO cout << "NO" << endl

#define IN(a,b) ((a).find(b)!=(a).end())

#define pb push_back

#define mkp make_pair

#define PI M_PI

#define INF 1e18

#define deb(x) if(ON) cout << #x <<" : " << x << endl

#define fast ios_base::sync_with_stdio(false)

#define io cin.tie(NULL),cout.tie(NULL)

#define ON !false

ll dx[]={0,-1,0,1};

ll dy[]={1,0,-1,0};

const ll mod=1e9+7;

ll i,j;

ll add(ll a,ll b){

    return  ((a%mod)+(b%mod)+mod)%mod;

}

ll mul(ll a,ll b){

    return ((a%mod)*(b%mod)+mod)%mod;

}

void pause(ll id){

    cout << "Waiting " << id << endl;

    cin >> id;

}



const int mxn=1e5+10;

ll n,t;





signed main(){

    fast;

    io;

    if(!true){

        freopen("in.txt","r",stdin);

        freopen("out.txt","w",stdout);

    }



    cin >> n;

    ll h[n];

    vll ans(n,0);

    ll i,j;

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

        cin >> h[i];

    }

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

        ll mn=abs(h[i]-h[i-1])+ans[i-1];

        if(i>=2){

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

        }

        ans[i]+=mn;

    }

    cout << ans[n-1] << endl;

   

    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;

/*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: 1.370823651939799x):
#include <bits/stdc++.h>

#include <algorithm>

#include <climits>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <iostream>

#include <math.h>

using namespace __gnu_pbds;

using namespace std;

#define ll long long

typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> pbds;

#define vll vector<ll> 

#define nl cout << endl

#define fr(a,b) for(auto i=a;i<b;i++)

#define pll pair<ll,ll>

#define f first

#define s second

#define vpll vector<pll>

#define ump unordered_map

#define YES cout << "YES" << endl

#define NO cout << "NO" << endl

#define IN(a,b) ((a).find(b)!=(a).end())

#define pb push_back

#define mkp make_pair

#define PI M_PI

#define INF 1e18

#define deb(x) if(ON) cout << #x <<" : " << x << endl

#define fast ios_base::sync_with_stdio(false)

#define io cin.tie(NULL),cout.tie(NULL)

#define ON !false

ll dx[]={0,-1,0,1};

ll dy[]={1,0,-1,0};

const ll mod=1e9+7;

ll i,j;

ll add(ll a,ll b){

    return  ((a%mod)+(b%mod)+mod)%mod;

}

ll mul(ll a,ll b){

    return ((a%mod)*(b%mod)+mod)%mod;

}

void pause(ll id){

    cout << "Waiting " << id << endl;

    cin >> id;

}



const int mxn=1e5+10;

ll n,t;





signed main(){

    fast;

    io;

    if(!true){

        freopen("in.txt","r",stdin);

        freopen("out.txt","w",stdout);

    }



    cin >> n;

    ll h[n];

    vll ans(n,0);

    ll i,j;

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

        cin >> h[i];

    }

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

        ll mn=abs(h[i]-h[i-1])+ans[i-1];

        if(i>=2){

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

        }

        ans[i]+=mn;

    }

    cout << ans[n-1] << endl;

   

    return 0;

}
// 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: 1.7642703828351707x):
#include <bits/stdc++.h>

#include <algorithm>

#include <climits>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <iostream>

#include <math.h>

using namespace __gnu_pbds;

using namespace std;

#define ll long long

typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> pbds;

#define vll vector<ll> 

#define nl cout << endl

#define fr(a,b) for(auto i=a;i<b;i++)

#define pll pair<ll,ll>

#define f first

#define s second

#define vpll vector<pll>

#define ump unordered_map

#define YES cout << "YES" << endl

#define NO cout << "NO" << endl

#define IN(a,b) ((a).find(b)!=(a).end())

#define pb push_back

#define mkp make_pair

#define PI M_PI

#define INF 1e18

#define deb(x) if(ON) cout << #x <<" : " << x << endl

#define fast ios_base::sync_with_stdio(false)

#define io cin.tie(NULL),cout.tie(NULL)

#define ON !false

ll dx[]={0,-1,0,1};

ll dy[]={1,0,-1,0};

const ll mod=1e9+7;

ll i,j;

ll add(ll a,ll b){

    return  ((a%mod)+(b%mod)+mod)%mod;

}

ll mul(ll a,ll b){

    return ((a%mod)*(b%mod)+mod)%mod;

}

void pause(ll id){

    cout << "Waiting " << id << endl;

    cin >> id;

}



const int mxn=1e5+10;

ll n,t;





signed main(){

    fast;

    io;

    if(!true){

        freopen("in.txt","r",stdin);

        freopen("out.txt","w",stdout);

    }



    cin >> n;

    ll h[n];

    vll ans(n,0);

    ll i,j;

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

        cin >> h[i];

    }

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

        ll mn=abs(h[i]-h[i-1])+ans[i-1];

        if(i>=2){

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

        }

        ans[i]+=mn;

    }

    cout << ans[n-1] << endl;

   

    return 0;

}
// 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: 1.7636089039228497x):
#include <bits/stdc++.h>

#include <algorithm>

#include <climits>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <iostream>

#include <math.h>

using namespace __gnu_pbds;

using namespace std;

#define ll long long

typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> pbds;

#define vll vector<ll> 

#define nl cout << endl

#define fr(a,b) for(auto i=a;i<b;i++)

#define pll pair<ll,ll>

#define f first

#define s second

#define vpll vector<pll>

#define ump unordered_map

#define YES cout << "YES" << endl

#define NO cout << "NO" << endl

#define IN(a,b) ((a).find(b)!=(a).end())

#define pb push_back

#define mkp make_pair

#define PI M_PI

#define INF 1e18

#define deb(x) if(ON) cout << #x <<" : " << x << endl

#define fast ios_base::sync_with_stdio(false)

#define io cin.tie(NULL),cout.tie(NULL)

#define ON !false

ll dx[]={0,-1,0,1};

ll dy[]={1,0,-1,0};

const ll mod=1e9+7;

ll i,j;

ll add(ll a,ll b){

    return  ((a%mod)+(b%mod)+mod)%mod;

}

ll mul(ll a,ll b){

    return ((a%mod)*(b%mod)+mod)%mod;

}

void pause(ll id){

    cout << "Waiting " << id << endl;

    cin >> id;

}



const int mxn=1e5+10;

ll n,t;





signed main(){

    fast;

    io;

    if(!true){

        freopen("in.txt","r",stdin);

        freopen("out.txt","w",stdout);

    }



    cin >> n;

    ll h[n];

    vll ans(n,0);

    ll i,j;

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

        cin >> h[i];

    }

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

        ll mn=abs(h[i]-h[i-1])+ans[i-1];

        if(i>=2){

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

        }

        ans[i]+=mn;

    }

    cout << ans[n-1] << endl;

   

    return 0;

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

using namespace std;



int num[100001], memo[100001];



int dp(int n){

    if(n==0) return 0;

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

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

    return memo[n]=min(dp(n-1)+abs(num[n-1]-num[n]), dp(n-2)+abs(num[n-2]-num[n]));

}



int main(){

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

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

    ios_base::sync_with_stdio(0);

    cin.tie(0);

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

    int n;

    cin>>n;

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

    cout<<dp(n-1)<<"\n";

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3803217932174252x):
#include <bits/stdc++.h>

#include <algorithm>

#include <climits>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <iostream>

#include <math.h>

using namespace __gnu_pbds;

using namespace std;

#define ll long long

typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> pbds;

#define vll vector<ll> 

#define nl cout << endl

#define fr(a,b) for(auto i=a;i<b;i++)

#define pll pair<ll,ll>

#define f first

#define s second

#define vpll vector<pll>

#define ump unordered_map

#define YES cout << "YES" << endl

#define NO cout << "NO" << endl

#define IN(a,b) ((a).find(b)!=(a).end())

#define pb push_back

#define mkp make_pair

#define PI M_PI

#define INF 1e18

#define deb(x) if(ON) cout << #x <<" : " << x << endl

#define fast ios_base::sync_with_stdio(false)

#define io cin.tie(NULL),cout.tie(NULL)

#define ON !false

ll dx[]={0,-1,0,1};

ll dy[]={1,0,-1,0};

const ll mod=1e9+7;

ll i,j;

ll add(ll a,ll b){

    return  ((a%mod)+(b%mod)+mod)%mod;

}

ll mul(ll a,ll b){

    return ((a%mod)*(b%mod)+mod)%mod;

}

void pause(ll id){

    cout << "Waiting " << id << endl;

    cin >> id;

}



const int mxn=1e5+10;

ll n,t;





signed main(){

    fast;

    io;

    if(!true){

        freopen("in.txt","r",stdin);

        freopen("out.txt","w",stdout);

    }



    cin >> n;

    ll h[n];

    vll ans(n,0);

    ll i,j;

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

        cin >> h[i];

    }

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

        ll mn=abs(h[i]-h[i-1])+ans[i-1];

        if(i>=2){

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

        }

        ans[i]+=mn;

    }

    cout << ans[n-1] << endl;

   

    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()

{



    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: 1.3661560982839294x):
#include <bits/stdc++.h>

#include <algorithm>

#include <climits>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <iostream>

#include <math.h>

using namespace __gnu_pbds;

using namespace std;

#define ll long long

typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> pbds;

#define vll vector<ll> 

#define nl cout << endl

#define fr(a,b) for(auto i=a;i<b;i++)

#define pll pair<ll,ll>

#define f first

#define s second

#define vpll vector<pll>

#define ump unordered_map

#define YES cout << "YES" << endl

#define NO cout << "NO" << endl

#define IN(a,b) ((a).find(b)!=(a).end())

#define pb push_back

#define mkp make_pair

#define PI M_PI

#define INF 1e18

#define deb(x) if(ON) cout << #x <<" : " << x << endl

#define fast ios_base::sync_with_stdio(false)

#define io cin.tie(NULL),cout.tie(NULL)

#define ON !false

ll dx[]={0,-1,0,1};

ll dy[]={1,0,-1,0};

const ll mod=1e9+7;

ll i,j;

ll add(ll a,ll b){

    return  ((a%mod)+(b%mod)+mod)%mod;

}

ll mul(ll a,ll b){

    return ((a%mod)*(b%mod)+mod)%mod;

}

void pause(ll id){

    cout << "Waiting " << id << endl;

    cin >> id;

}



const int mxn=1e5+10;

ll n,t;





signed main(){

    fast;

    io;

    if(!true){

        freopen("in.txt","r",stdin);

        freopen("out.txt","w",stdout);

    }



    cin >> n;

    ll h[n];

    vll ans(n,0);

    ll i,j;

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

        cin >> h[i];

    }

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

        ll mn=abs(h[i]-h[i-1])+ans[i-1];

        if(i>=2){

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

        }

        ans[i]+=mn;

    }

    cout << ans[n-1] << endl;

   

    return 0;

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



using namespace std;



const int N = 1e5 + 9;



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



int solve(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]) + solve(i + 1), abs(h[i] - h[i + 2]) + solve(i + 2));

}



int main(){



    cin >> n;

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

        cin >> h[i];



    memset(dp, -1, sizeof dp);



    cout << solve(0);

}


// Faster C++ solution for the same problem (measured speedup: 1.3691337907754526x):
#include <bits/stdc++.h>

#include <algorithm>

#include <climits>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <iostream>

#include <math.h>

using namespace __gnu_pbds;

using namespace std;

#define ll long long

typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> pbds;

#define vll vector<ll> 

#define nl cout << endl

#define fr(a,b) for(auto i=a;i<b;i++)

#define pll pair<ll,ll>

#define f first

#define s second

#define vpll vector<pll>

#define ump unordered_map

#define YES cout << "YES" << endl

#define NO cout << "NO" << endl

#define IN(a,b) ((a).find(b)!=(a).end())

#define pb push_back

#define mkp make_pair

#define PI M_PI

#define INF 1e18

#define deb(x) if(ON) cout << #x <<" : " << x << endl

#define fast ios_base::sync_with_stdio(false)

#define io cin.tie(NULL),cout.tie(NULL)

#define ON !false

ll dx[]={0,-1,0,1};

ll dy[]={1,0,-1,0};

const ll mod=1e9+7;

ll i,j;

ll add(ll a,ll b){

    return  ((a%mod)+(b%mod)+mod)%mod;

}

ll mul(ll a,ll b){

    return ((a%mod)*(b%mod)+mod)%mod;

}

void pause(ll id){

    cout << "Waiting " << id << endl;

    cin >> id;

}



const int mxn=1e5+10;

ll n,t;





signed main(){

    fast;

    io;

    if(!true){

        freopen("in.txt","r",stdin);

        freopen("out.txt","w",stdout);

    }



    cin >> n;

    ll h[n];

    vll ans(n,0);

    ll i,j;

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

        cin >> h[i];

    }

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

        ll mn=abs(h[i]-h[i-1])+ans[i-1];

        if(i>=2){

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

        }

        ans[i]+=mn;

    }

    cout << ans[n-1] << endl;

   

    return 0;

}
// 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: 1.7885763660089802x):
#include<bits/stdc++.h>

using namespace std;

int n; vector<int>v;

long long dp[100005];

long long solve(int i) {

	if (i >= n)return 1e18;

	if (i == n-2)return abs(v[i]-v[n-1]);

	if (i == n-3)return abs(v[i]-v[n-1]);

	long long& ret = dp[i];

	if (ret!=2e9)return ret;

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

	return ret;

}

int main() {

	cin >> n; v.resize(n); 

	for (auto& t : v)cin >> t;

	fill(dp, dp + n, 2e9);

	cout << solve(0);

}
// 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: 1.3842869015356627x):
#include<bits/stdc++.h>

using namespace std;

int n; vector<int>v;

long long dp[100005];

long long solve(int i) {

	if (i >= n)return 1e18;

	if (i == n-2)return abs(v[i]-v[n-1]);

	if (i == n-3)return abs(v[i]-v[n-1]);

	long long& ret = dp[i];

	if (ret!=2e9)return ret;

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

	return ret;

}

int main() {

	cin >> n; v.resize(n); 

	for (auto& t : v)cin >> t;

	fill(dp, dp + n, 2e9);

	cout << solve(0);

}
// Slower C++ solution:


/**

    And in the end, the love you take is equal to the love you make.



        I LOVE YOU LAURA.

**/



#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <ext/rope>



#define _CRT_SECURE_NO_DEPRECATE

#define _CRT_DISABLE_PERFCRIT_LOCKS



#define findOrder find_by_order

#define orderKey order_of_key



#define isOn(mask, i) (mask & (1 << i))

#define turn(mask, i) (mask |= (1 << i))

#define switch(mask, i) (mask &= ~(1 << i))

#define change(mask, i) (mask ^= (1 << i))

#define lowbit(mask) (mask & (-mask))

#define turnAll(mask, n) (mask = (1 << n) - 1)

#define switchLastBit(mask) (mask & (mask - 1))

#define turnLastZero(mask) (mask | (mask + 1))

#define modulo(x, mod) ((x) & (mod - 1))

#define log2(x) (31 - __builtin_clz(x))



#define mkp make_pair

#define pb push_back

#define pf push_front

#define eb emplace_back

#define ef emplace_front

#define ins insert

#define rsz resize

#define f first

#define s second

#define lb lower_bound

#define ub upper_bound



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

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

#define rall(x) rbegin(x), rend(x)



#define debug(x) (cerr << x << " ")



using namespace std;

using namespace __gnu_pbds;

using namespace __gnu_cxx;



typedef double db;

typedef long long ll;

typedef long double ld;

typedef unsigned long long ull;



typedef pair<int, int> ii;

typedef pair<ll,ll> pll;

typedef pair<ld,ld> pdb;

typedef pair<int, ii> tri;

typedef complex<ld> cd;



typedef vector<int> vi;

typedef vector<ii> vii;

typedef vector<ld> vdb;

typedef vector<cd> vcd;

typedef vector<ll> vll;

typedef vector<string> vs;

typedef vector<vll> matrix;



typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_tree;



const ll mod = 1e9+7;

const int maxn = 2e5+2;

const int pi = 2 * acos(0.0);

const db eps = 1e-9;



ll arr[maxn], dp[maxn];



int main()

{

    ios_base::sync_with_stdio(0);

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



    int n; cin >> n;



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

        cin >> arr[i];

    }



    memset(dp, 63, sizeof dp);



    dp[1] = 0;

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

        if (i <= 2){

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

        }

        else{

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

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

        }

    }



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



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.622940016033936x):
#include<bits/stdc++.h>

using namespace std;

int n; vector<int>v;

long long dp[100005];

long long solve(int i) {

	if (i >= n)return 1e18;

	if (i == n-2)return abs(v[i]-v[n-1]);

	if (i == n-3)return abs(v[i]-v[n-1]);

	long long& ret = dp[i];

	if (ret!=2e9)return ret;

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

	return ret;

}

int main() {

	cin >> n; v.resize(n); 

	for (auto& t : v)cin >> t;

	fill(dp, dp + n, 2e9);

	cout << solve(0);

}