stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 1.3823254877244129x):
#include<bits/stdc++.h>



#define endl '\n'

#define int long long int



using namespace std;

     

int gcd(int a, int b)

{

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

}

int lcm(int a, int b)

{

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

}

int ncr(int n, int k)

{

        int res = 1;

        if ( k > n - k )

            k = n - k;

     

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

        {

            res *= (n - i);

            res /= (i + 1);

        }

        return res;

}



map<int, int> dpArr;



int dp(int arr[],int n)

{

        if(n<=1)

        {

            if( n == 0) return dpArr[0];

            return dpArr[1];

        }

        if(dpArr.count(n)){

            return dpArr[n];

        }

        dpArr[n] =  dp(arr,n-1) + abs(arr[n-1] - arr[n]);

        if(n > 1) dpArr[n] = min(dpArr[n], dp(arr, n-2)+ abs(arr[n-2] - arr[n]));

        return dpArr[n];

}

     

signed main()

{

        int n;

        cin >> n;

        int arr[n];

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

            cin >> arr[i];

        }

        dpArr[0] = 0;

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

        cout << dp(arr,n-1);

     

        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: 1.7920614000392832x):
#include<bits/stdc++.h>

using namespace std;



int n, dp[101000], h[101000];



int main()

{

    cin>>n;

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

    {

        cin>>h[i];

    }

    dp[1]=0;

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

    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];

}
// 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: 1.3869841791309823x):
#include<bits/stdc++.h>

using namespace std;



int n, dp[101000], h[101000];



int main()

{

    cin>>n;

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

    {

        cin>>h[i];

    }

    dp[1]=0;

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

    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];

}
// 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: 2.6280508043620396x):
#include<bits/stdc++.h>

using namespace std;



int n, dp[101000], h[101000];



int main()

{

    cin>>n;

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

    {

        cin>>h[i];

    }

    dp[1]=0;

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

    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];

}
// 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: 9.421550881044823x):
#include<bits/stdc++.h>

using namespace std;



int n, dp[101000], h[101000];



int main()

{

    cin>>n;

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

    {

        cin>>h[i];

    }

    dp[1]=0;

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

    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];

}
// 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: 1.2515031022243939x):
#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;

}
// 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: 6.579629556852292x):
#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;

}
// 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: 1.7684738123665777x):
#include<bits/stdc++.h>

#define ll long long

#define MAXN 100010

#define INF 1000000010



using namespace std;



ll n, dp[MAXN], h[MAXN];



void init(){

    for(int i=1; i<=n; i++) dp[i]=INF;

    dp[1]=0;

    h[0]=INF;

}



int main(){



    ios_base::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);



    cin>>n;



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



    init();



    for(int i=2; 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];



return 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: 1.368728325327541x):
#include<bits/stdc++.h>

#define ll long long

#define MAXN 100010

#define INF 1000000010



using namespace std;



ll n, dp[MAXN], h[MAXN];



void init(){

    for(int i=1; i<=n; i++) dp[i]=INF;

    dp[1]=0;

    h[0]=INF;

}



int main(){



    ios_base::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);



    cin>>n;



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



    init();



    for(int i=2; 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];



return 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: 2.5934597023189654x):
#include<bits/stdc++.h>

#define ll long long

#define MAXN 100010

#define INF 1000000010



using namespace std;



ll n, dp[MAXN], h[MAXN];



void init(){

    for(int i=1; i<=n; i++) dp[i]=INF;

    dp[1]=0;

    h[0]=INF;

}



int main(){



    ios_base::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);



    cin>>n;



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



    init();



    for(int i=2; 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];



return 0;

}
// 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: 9.297541928330018x):
#include<bits/stdc++.h>

#define ll long long

#define MAXN 100010

#define INF 1000000010



using namespace std;



ll n, dp[MAXN], h[MAXN];



void init(){

    for(int i=1; i<=n; i++) dp[i]=INF;

    dp[1]=0;

    h[0]=INF;

}



int main(){



    ios_base::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);



    cin>>n;



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



    init();



    for(int i=2; 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];



return 0;

}