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

using namespace std;

int n , a[1000000] , frog , memo[1000000];



int f(int n){

	if(n == 1)return 0;

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

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

	memo[n] = min(f(n-1) + abs(a[n] - a[n-1]) , f(n-2) + abs(a[n] - a[n-2]));

	return memo[n];

}

int main(){

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

	scanf("%d" , &n);

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

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

	}

	printf("%d" , f(n));

}

// Faster C++ solution for the same problem (measured speedup: 4.94909007560155x):
#include<bits/stdc++.h>



using namespace std;

int main(){

	long long n,a[100000],dp[100000];

	scanf("%lld", &n);

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

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

	}

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

	dp[2]=abs(a[0]-a[2]);

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

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

	printf("%lld", dp[n-1]);

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

using namespace std;

const int mx = 100005;

int dp[mx];

int A[mx];

void memo(int i,int n)

{

	if(i>n) return;

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

	memo(i+1,n);

}

int main()

{

     int n;cin>>n;

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

     	cin>>A[i];

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

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

     memo(3,n);

     cout<<dp[n]<<endl;

     return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3808622549337852x):
#include <bits/stdc++.h>

 

#define ll long long

#define X first

#define Y second

#define MP make_pair

#define ld long double

 

using namespace std;

 

const int N = 5e5 + 123;

const ll mod = 1e9 + 7;



ll a[N], dp[N];

int n;



int main (){

	ios::sync_with_stdio(0);

	cin.tie(0);

	cout.tie(0);



	cin >> n;

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

		cin >> a[i];

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

		int x = i - 1, y = max(1, i - 2);

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

	}

	cout << dp[n];

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

using namespace std;



#define INF 1e18;

#define MAX_N 100005

long long N, h[MAX_N], memo[MAX_N];



long long dp(int pos) {

  if (pos < 1) {

  	return INF;

  }

  if (pos == 1) {

  	return 0;

  }

  if (pos == 2) {

  	return abs(h[2] - h[1]);

  }

  if (memo[pos] != -1) {

  	return memo[pos];

  }

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

}



int main() {

  cin >> N;

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

  	cin >> h[i];

  }

  /*

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

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

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

  }

  */

  /*

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

  	cout << i << ' ' << memo[i] << '\n';

  }

  */

  // cout << memo[N] << '\n';

  memset(memo, -1, sizeof memo);

  cout << dp(N) << '\n';

}

// Faster C++ solution for the same problem (measured speedup: 1.7750609110356788x):
#include <bits/stdc++.h>

 

#define ll long long

#define X first

#define Y second

#define MP make_pair

#define ld long double

 

using namespace std;

 

const int N = 5e5 + 123;

const ll mod = 1e9 + 7;



ll a[N], dp[N];

int n;



int main (){

	ios::sync_with_stdio(0);

	cin.tie(0);

	cout.tie(0);



	cin >> n;

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

		cin >> a[i];

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

		int x = i - 1, y = max(1, i - 2);

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

	}

	cout << dp[n];

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

using namespace std;

int n , a[1000000] , frog , memo[1000000];



int f(int n){

	if(n == 1)return 0;

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

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

	memo[n] = min(f(n-1) + abs(a[n] - a[n-1]) , f(n-2) + abs(a[n] - a[n-2]));

	return memo[n];

}

int main(){

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

	scanf("%d" , &n);

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

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

	}

	printf("%d" , f(n));

}

// Faster C++ solution for the same problem (measured speedup: 5.001792304725045x):
#include <bits/stdc++.h>

 

#define ll long long

#define X first

#define Y second

#define MP make_pair

#define ld long double

 

using namespace std;

 

const int N = 5e5 + 123;

const ll mod = 1e9 + 7;



ll a[N], dp[N];

int n;



int main (){

	ios::sync_with_stdio(0);

	cin.tie(0);

	cout.tie(0);



	cin >> n;

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

		cin >> a[i];

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

		int x = i - 1, y = max(1, i - 2);

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

	}

	cout << dp[n];

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

using namespace std;

const int mx = 100005;

int dp[mx];

int A[mx];

void memo(int i,int n)

{

	if(i>n) return;

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

	memo(i+1,n);

}

int main()

{

     int n;cin>>n;

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

     	cin>>A[i];

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

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

     memo(3,n);

     cout<<dp[n]<<endl;

     return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4027368207412305x):
#include <bits/stdc++.h>                                           

 

//#define int long long

//#pragma GCC optimize("Ofast")

//#pragma comment(linker, "/stack:200000000")

//#pragma GCC target("sse,sse2,sse3,ssse3,sse4")

 

 

#define file(s) freopen(s".in","r",stdin); freopen(s".out","w",stdout);

#define forev(i, b, a) for(int i = (b); i >= (a); --i)

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

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

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

#define pb push_back

#define ppb pop_back

#define mp make_pair

#define s second

#define f first

 

 

using namespace std;

 

 

typedef pair < long long, long long > pll;    

typedef pair < int, int > pii;

typedef unsigned long long ull;         

typedef vector < pii > vpii;

typedef vector < int > vi;

typedef long double ldb; 

typedef long long ll;  

typedef double db;                         

 

 

const int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1}, block = 555;

const pii base = mp(1171, 3307), Mod = mp(1e9 + 7, 1e9 + 9);

const int inf = 1e9, N = 2e5 + 48, mod = 1e9 + 7;

const db eps = 1e-12, pi = 3.14159265359;

const ll INF = 1e18;





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



main () {

	cin >> n;

	forn (i, 1, n)

		cin >> h[i];	

	forn (i, 2, n)

		dp[i] = min(dp[i - 1] + abs(h[i] - h[i - 1]), i < 3 ? inf : dp[i - 2] + abs(h[i] - h[i - 2]));		



	cout << dp[n];

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

using namespace std;



#define INF 1e18;

#define MAX_N 100005

long long N, h[MAX_N], memo[MAX_N];



long long dp(int pos) {

  if (pos < 1) {

  	return INF;

  }

  if (pos == 1) {

  	return 0;

  }

  if (pos == 2) {

  	return abs(h[2] - h[1]);

  }

  if (memo[pos] != -1) {

  	return memo[pos];

  }

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

}



int main() {

  cin >> N;

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

  	cin >> h[i];

  }

  /*

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

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

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

  }

  */

  /*

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

  	cout << i << ' ' << memo[i] << '\n';

  }

  */

  // cout << memo[N] << '\n';

  memset(memo, -1, sizeof memo);

  cout << dp(N) << '\n';

}

// Faster C++ solution for the same problem (measured speedup: 1.803180071054674x):
#include <bits/stdc++.h>                                           

 

//#define int long long

//#pragma GCC optimize("Ofast")

//#pragma comment(linker, "/stack:200000000")

//#pragma GCC target("sse,sse2,sse3,ssse3,sse4")

 

 

#define file(s) freopen(s".in","r",stdin); freopen(s".out","w",stdout);

#define forev(i, b, a) for(int i = (b); i >= (a); --i)

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

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

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

#define pb push_back

#define ppb pop_back

#define mp make_pair

#define s second

#define f first

 

 

using namespace std;

 

 

typedef pair < long long, long long > pll;    

typedef pair < int, int > pii;

typedef unsigned long long ull;         

typedef vector < pii > vpii;

typedef vector < int > vi;

typedef long double ldb; 

typedef long long ll;  

typedef double db;                         

 

 

const int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1}, block = 555;

const pii base = mp(1171, 3307), Mod = mp(1e9 + 7, 1e9 + 9);

const int inf = 1e9, N = 2e5 + 48, mod = 1e9 + 7;

const db eps = 1e-12, pi = 3.14159265359;

const ll INF = 1e18;





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



main () {

	cin >> n;

	forn (i, 1, n)

		cin >> h[i];	

	forn (i, 2, n)

		dp[i] = min(dp[i - 1] + abs(h[i] - h[i - 1]), i < 3 ? inf : dp[i - 2] + abs(h[i] - h[i - 2]));		



	cout << dp[n];

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

using namespace std;

int n , a[1000000] , frog , memo[1000000];



int f(int n){

	if(n == 1)return 0;

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

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

	memo[n] = min(f(n-1) + abs(a[n] - a[n-1]) , f(n-2) + abs(a[n] - a[n-2]));

	return memo[n];

}

int main(){

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

	scanf("%d" , &n);

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

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

	}

	printf("%d" , f(n));

}

// Faster C++ solution for the same problem (measured speedup: 5.0810268804649175x):
#include <bits/stdc++.h>                                           

 

//#define int long long

//#pragma GCC optimize("Ofast")

//#pragma comment(linker, "/stack:200000000")

//#pragma GCC target("sse,sse2,sse3,ssse3,sse4")

 

 

#define file(s) freopen(s".in","r",stdin); freopen(s".out","w",stdout);

#define forev(i, b, a) for(int i = (b); i >= (a); --i)

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

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

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

#define pb push_back

#define ppb pop_back

#define mp make_pair

#define s second

#define f first

 

 

using namespace std;

 

 

typedef pair < long long, long long > pll;    

typedef pair < int, int > pii;

typedef unsigned long long ull;         

typedef vector < pii > vpii;

typedef vector < int > vi;

typedef long double ldb; 

typedef long long ll;  

typedef double db;                         

 

 

const int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1}, block = 555;

const pii base = mp(1171, 3307), Mod = mp(1e9 + 7, 1e9 + 9);

const int inf = 1e9, N = 2e5 + 48, mod = 1e9 + 7;

const db eps = 1e-12, pi = 3.14159265359;

const ll INF = 1e18;





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



main () {

	cin >> n;

	forn (i, 1, n)

		cin >> h[i];	

	forn (i, 2, n)

		dp[i] = min(dp[i - 1] + abs(h[i] - h[i - 1]), i < 3 ? inf : dp[i - 2] + abs(h[i] - h[i - 2]));		



	cout << dp[n];

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

using namespace std;

const int mx = 100005;

int dp[mx];

int A[mx];

void memo(int i,int n)

{

	if(i>n) return;

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

	memo(i+1,n);

}

int main()

{

     int n;cin>>n;

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

     	cin>>A[i];

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

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

     memo(3,n);

     cout<<dp[n]<<endl;

     return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4075083072803432x):
#include<iostream>

#include<cstdio>

#include<cmath>

using namespace std;

const int maxn=100010;

const int INF=0x7fffffff;

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

void init(){

	scanf("%d",&n);

	for(int i=1;i<=n;i++) scanf("%d",&h[i]);

	for(int i=2;i<=n;i++) dp[i]=INF;

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

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

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

	}

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

}

int main(){

	init();

	return 0;

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

using namespace std;



#define INF 1e18;

#define MAX_N 100005

long long N, h[MAX_N], memo[MAX_N];



long long dp(int pos) {

  if (pos < 1) {

  	return INF;

  }

  if (pos == 1) {

  	return 0;

  }

  if (pos == 2) {

  	return abs(h[2] - h[1]);

  }

  if (memo[pos] != -1) {

  	return memo[pos];

  }

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

}



int main() {

  cin >> N;

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

  	cin >> h[i];

  }

  /*

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

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

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

  }

  */

  /*

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

  	cout << i << ' ' << memo[i] << '\n';

  }

  */

  // cout << memo[N] << '\n';

  memset(memo, -1, sizeof memo);

  cout << dp(N) << '\n';

}

// Faster C++ solution for the same problem (measured speedup: 1.8093136873606088x):
#include<iostream>

#include<cstdio>

#include<cmath>

using namespace std;

const int maxn=100010;

const int INF=0x7fffffff;

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

void init(){

	scanf("%d",&n);

	for(int i=1;i<=n;i++) scanf("%d",&h[i]);

	for(int i=2;i<=n;i++) dp[i]=INF;

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

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

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

	}

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

}

int main(){

	init();

	return 0;

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

using namespace std;

int n , a[1000000] , frog , memo[1000000];



int f(int n){

	if(n == 1)return 0;

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

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

	memo[n] = min(f(n-1) + abs(a[n] - a[n-1]) , f(n-2) + abs(a[n] - a[n-2]));

	return memo[n];

}

int main(){

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

	scanf("%d" , &n);

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

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

	}

	printf("%d" , f(n));

}

// Faster C++ solution for the same problem (measured speedup: 5.098310273191571x):
#include<iostream>

#include<cstdio>

#include<cmath>

using namespace std;

const int maxn=100010;

const int INF=0x7fffffff;

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

void init(){

	scanf("%d",&n);

	for(int i=1;i<=n;i++) scanf("%d",&h[i]);

	for(int i=2;i<=n;i++) dp[i]=INF;

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

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

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

	}

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

}

int main(){

	init();

	return 0;

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

using namespace std;

const int mx = 100005;

int dp[mx];

int A[mx];

void memo(int i,int n)

{

	if(i>n) return;

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

	memo(i+1,n);

}

int main()

{

     int n;cin>>n;

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

     	cin>>A[i];

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

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

     memo(3,n);

     cout<<dp[n]<<endl;

     return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3921032800726763x):
#include<bits/stdc++.h>

using namespace std;

int a[1000001],n;

int main()

{

	cin>>n;

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

	int f[100001];

	f[1]=0;

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

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

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

	cout<<f[n];

	return 0;

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

using namespace std;



#define INF 1e18;

#define MAX_N 100005

long long N, h[MAX_N], memo[MAX_N];



long long dp(int pos) {

  if (pos < 1) {

  	return INF;

  }

  if (pos == 1) {

  	return 0;

  }

  if (pos == 2) {

  	return abs(h[2] - h[1]);

  }

  if (memo[pos] != -1) {

  	return memo[pos];

  }

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

}



int main() {

  cin >> N;

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

  	cin >> h[i];

  }

  /*

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

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

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

  }

  */

  /*

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

  	cout << i << ' ' << memo[i] << '\n';

  }

  */

  // cout << memo[N] << '\n';

  memset(memo, -1, sizeof memo);

  cout << dp(N) << '\n';

}

// Faster C++ solution for the same problem (measured speedup: 1.7895109434358847x):
#include<bits/stdc++.h>

using namespace std;

int a[1000001],n;

int main()

{

	cin>>n;

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

	int f[100001];

	f[1]=0;

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

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

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

	cout<<f[n];

	return 0;

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

using namespace std;

int n , a[1000000] , frog , memo[1000000];



int f(int n){

	if(n == 1)return 0;

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

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

	memo[n] = min(f(n-1) + abs(a[n] - a[n-1]) , f(n-2) + abs(a[n] - a[n-2]));

	return memo[n];

}

int main(){

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

	scanf("%d" , &n);

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

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

	}

	printf("%d" , f(n));

}

// Faster C++ solution for the same problem (measured speedup: 5.042509814987952x):
#include<bits/stdc++.h>

using namespace std;

int a[1000001],n;

int main()

{

	cin>>n;

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

	int f[100001];

	f[1]=0;

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

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

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

	cout<<f[n];

	return 0;

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

using namespace std;

const int mx = 100005;

int dp[mx];

int A[mx];

void memo(int i,int n)

{

	if(i>n) return;

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

	memo(i+1,n);

}

int main()

{

     int n;cin>>n;

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

     	cin>>A[i];

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

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

     memo(3,n);

     cout<<dp[n]<<endl;

     return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4184529877363314x):
#include<cstdio>

#include<iostream>

#include<cmath>

using namespace std;

const int MAXN = 100010;

int n, a[MAXN], f[MAXN];

int main()

{

	scanf("%d",&n);

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

		scanf("%d",&a[i]),f[i]=0x3f3f3f3f;

	f[0] = f[1] = 0; a[0] = a[1];

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

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

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

	return 0;

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

using namespace std;



#define INF 1e18;

#define MAX_N 100005

long long N, h[MAX_N], memo[MAX_N];



long long dp(int pos) {

  if (pos < 1) {

  	return INF;

  }

  if (pos == 1) {

  	return 0;

  }

  if (pos == 2) {

  	return abs(h[2] - h[1]);

  }

  if (memo[pos] != -1) {

  	return memo[pos];

  }

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

}



int main() {

  cin >> N;

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

  	cin >> h[i];

  }

  /*

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

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

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

  }

  */

  /*

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

  	cout << i << ' ' << memo[i] << '\n';

  }

  */

  // cout << memo[N] << '\n';

  memset(memo, -1, sizeof memo);

  cout << dp(N) << '\n';

}

// Faster C++ solution for the same problem (measured speedup: 1.8233827767225543x):
#include<cstdio>

#include<iostream>

#include<cmath>

using namespace std;

const int MAXN = 100010;

int n, a[MAXN], f[MAXN];

int main()

{

	scanf("%d",&n);

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

		scanf("%d",&a[i]),f[i]=0x3f3f3f3f;

	f[0] = f[1] = 0; a[0] = a[1];

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

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

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

	return 0;

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

using namespace std;

int n , a[1000000] , frog , memo[1000000];



int f(int n){

	if(n == 1)return 0;

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

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

	memo[n] = min(f(n-1) + abs(a[n] - a[n-1]) , f(n-2) + abs(a[n] - a[n-2]));

	return memo[n];

}

int main(){

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

	scanf("%d" , &n);

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

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

	}

	printf("%d" , f(n));

}

// Faster C++ solution for the same problem (measured speedup: 5.137954356652352x):
#include<cstdio>

#include<iostream>

#include<cmath>

using namespace std;

const int MAXN = 100010;

int n, a[MAXN], f[MAXN];

int main()

{

	scanf("%d",&n);

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

		scanf("%d",&a[i]),f[i]=0x3f3f3f3f;

	f[0] = f[1] = 0; a[0] = a[1];

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

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

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

	return 0;

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

using namespace std;

const int mx = 100005;

int dp[mx];

int A[mx];

void memo(int i,int n)

{

	if(i>n) return;

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

	memo(i+1,n);

}

int main()

{

     int n;cin>>n;

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

     	cin>>A[i];

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

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

     memo(3,n);

     cout<<dp[n]<<endl;

     return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4004761866000714x):
#include <functional>

#include <algorithm>

#include <iostream>

#include <cstring>

#include <cstdlib>

#include <iomanip>

#include <vector>

#include <cstdio>

#include <stack>

#include <queue>

#include <deque>

#include <cmath>

#include <list>

#include <set>

#include <map>



using namespace std;



int a[1000010];

int f[1000010];



int main(void)

{

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

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

	int n;

	cin >> n;

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

	{

		cin >> a[i];

	}

	f[1] = 0;

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

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

	{

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

		//cout << f[i] << endl;

	}

	cout  << f[n] << endl;

	return 0;

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

using namespace std;



#define INF 1e18;

#define MAX_N 100005

long long N, h[MAX_N], memo[MAX_N];



long long dp(int pos) {

  if (pos < 1) {

  	return INF;

  }

  if (pos == 1) {

  	return 0;

  }

  if (pos == 2) {

  	return abs(h[2] - h[1]);

  }

  if (memo[pos] != -1) {

  	return memo[pos];

  }

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

}



int main() {

  cin >> N;

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

  	cin >> h[i];

  }

  /*

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

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

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

  }

  */

  /*

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

  	cout << i << ' ' << memo[i] << '\n';

  }

  */

  // cout << memo[N] << '\n';

  memset(memo, -1, sizeof memo);

  cout << dp(N) << '\n';

}

// Faster C++ solution for the same problem (measured speedup: 1.8002740872870775x):
#include <functional>

#include <algorithm>

#include <iostream>

#include <cstring>

#include <cstdlib>

#include <iomanip>

#include <vector>

#include <cstdio>

#include <stack>

#include <queue>

#include <deque>

#include <cmath>

#include <list>

#include <set>

#include <map>



using namespace std;



int a[1000010];

int f[1000010];



int main(void)

{

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

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

	int n;

	cin >> n;

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

	{

		cin >> a[i];

	}

	f[1] = 0;

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

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

	{

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

		//cout << f[i] << endl;

	}

	cout  << f[n] << endl;

	return 0;

}