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

using namespace std;

int n, arr[100005], best, memo[100005];



int dp(int x){

	if(x==1 || x == 0){

		return 0;

	}

	if (x == 2){

		return abs(arr[x] - arr[x-1]);

	}

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

		return memo[x];

	}

	best=min(dp(x-1)+abs(arr[x]-arr[x-1]), dp(x-2)+abs(arr[x-2]-arr[x]));

	memo[x] = best;

	return best;

}



int main(){

	ios_base::sync_with_stdio(0);

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

	cin.tie();

	cout.tie();

	cin >> n;

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

		cin >> arr[i];

	}

	cout << dp(n) << endl;

}

// Faster C++ solution for the same problem (measured speedup: 1.4222572127075095x):
#include<iostream>

#include<cstdio>

#include<algorithm>

#include<cstring>

using namespace std;

const int MAXN=100010;

int stones[MAXN];

int mins[MAXN]= {-1};

int main()

{

    int n;

    cin>>n;

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

    {

        cin>>stones[i];

    }

    for(int i=n-2; i>=0; --i)

    {

        if(i==n-2)

        {

            mins[n-2]=abs(stones[i]-stones[i+1]);

        }

        else if(i==n-3)

        {

            int one=abs(stones[i]-stones[i+1])+mins[i+1];

            int two=abs(stones[i]-stones[i+2]);

            mins[i]=min(one,two);

        }

        else

        {

            int one=abs(stones[i]-stones[i+1])+mins[i+1];

            int two=abs(stones[i]-stones[i+2])+mins[i+2];

            mins[i]=min(one,two);

        }

    }

    printf("%d\n",mins[0]);

    return 0;

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

using namespace std;

int n, arr[100005], best, memo[100005];



int dp(int x){

	if(x==0 or x==1){

		return 0;

	}

	else if(x==2){

		return dp(x-1)+abs(arr[x]-arr[x-1]);

	}

	else if(memo[x]!=-1){

		return memo[x];

	}

	best=min(dp(x-1)+abs(arr[x]-arr[x-1]), dp(x-2)+abs(arr[x-2]-arr[x]));

	memo[x]=best;

	return best;

}



int main(){

	ios_base::sync_with_stdio(0);

	cin.tie();

	cout.tie();

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

	cin >> n;

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

		cin >> arr[i];

	}

	cout << dp(n) << endl;

}


// Faster C++ solution for the same problem (measured speedup: 1.4222572127075095x):
#include<iostream>

#include<cstdio>

#include<algorithm>

#include<cstring>

using namespace std;

const int MAXN=100010;

int stones[MAXN];

int mins[MAXN]= {-1};

int main()

{

    int n;

    cin>>n;

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

    {

        cin>>stones[i];

    }

    for(int i=n-2; i>=0; --i)

    {

        if(i==n-2)

        {

            mins[n-2]=abs(stones[i]-stones[i+1]);

        }

        else if(i==n-3)

        {

            int one=abs(stones[i]-stones[i+1])+mins[i+1];

            int two=abs(stones[i]-stones[i+2]);

            mins[i]=min(one,two);

        }

        else

        {

            int one=abs(stones[i]-stones[i+1])+mins[i+1];

            int two=abs(stones[i]-stones[i+2])+mins[i+2];

            mins[i]=min(one,two);

        }

    }

    printf("%d\n",mins[0]);

    return 0;

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

using namespace std;

int dif(int a, int b)

{

    int ans=a-b;

    if(ans<0) ans*=-1;

    return ans;

}

int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    int N, h[100005],i,j;

    int f1,f2;

    cin>>N;

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

        cin>>h[i];

    int M[100005]={0};

    M[N-2]=dif(h[N-1],h[N-2]);

    for(i=N-3;i>=0;i--)

    {

        f1=dif(h[i+1],h[i])+M[i+1];

        f2=dif(h[i+2],h[i])+M[i+2];

        M[i]=min(f1,f2);

    }

    cout<<M[0]<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4388383517871075x):
#include <iostream>

#include <queue>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

const int maxn = 100010;

int H[maxn];

int main(){

    int n;

    scanf("%d",&n);

    int dp[n+1];

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

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

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

    }

    dp[1]=0;

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

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

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

    }

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

    return 0;



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

using namespace std;

int n;

int h[1000005], memo[1000005];

int f(int n) {

    if(n == 1) {                        // bisa juga base case nya n = 0, f(0) harus tinggi, misal 1e9;

        return 0;

    }

    if(n == 2) {

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

    } 

    else if(memo[n] != -1) {

        return memo[n];

    }

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

    return memo[n];

}



int main() {

    scanf("%d", &n);

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

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

    }

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

    f(n);

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

}

// Faster C++ solution for the same problem (measured speedup: 5.131146336094921x):
#include <iostream>

#include <queue>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

const int maxn = 100010;

int H[maxn];

int main(){

    int n;

    scanf("%d",&n);

    int dp[n+1];

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

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

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

    }

    dp[1]=0;

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

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

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

    }

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

    return 0;



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

using namespace std;



typedef signed long long ll;

const int MX = 1e5 + 5;

const int INF = 1e9 + 5;

int n, A[MX], dp[MX];



int solve(int pos) {

    if(pos > n) return INF;

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

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

    int ans = INF;

    if(pos + 1 < n) ans = min(ans, abs(A[pos] - A[pos + 1]) + solve(pos + 1));

    if(pos + 2 < n) ans = min(ans, abs(A[pos] - A[pos + 2]) + solve(pos + 2));

    return dp[pos] = ans;

}



int main() {

    scanf("%d", &n);



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

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

    }

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

    printf("%d\n", solve(0));

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.406045711848312x):
#include <iostream>

#include <queue>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

const int maxn = 100010;

int H[maxn];

int main(){

    int n;

    scanf("%d",&n);

    int dp[n+1];

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

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

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

    }

    dp[1]=0;

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

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

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

    }

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

    return 0;



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

#include<cmath>

#include<algorithm>

#include<cstring>

using namespace std;



int h[100055],dp[100055];

const int inf = 0x3f3f3f;

int main()

{

	int n;

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

	cin>>n;

	h[1]=inf;

	h[0]=inf;

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

		cin>>h[i];

	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]<<endl;

	return 0;

} 

//max(h[i],h[i-2])-min(h[i],h[i-2])

// Faster C++ solution for the same problem (measured speedup: 1.4218563768026435x):
#include <iostream>

#include <queue>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

const int maxn = 100010;

int H[maxn];

int main(){

    int n;

    scanf("%d",&n);

    int dp[n+1];

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

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

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

    }

    dp[1]=0;

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

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

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

    }

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

    return 0;



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

using namespace std;

#define IO ios::sync_with_stdio(false), cin.tie(0)

typedef long long ll;

const int maxn = 1e6 + 5;

const int inf = 0x3f3f3f3f;

ll a[maxn];

ll dp[maxn];



int main()

{

    IO;

    int n, k=2;

    cin >> n ;

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

        cin >> a[i];

    memset(dp, inf, sizeof(dp));

    dp[1] = 0;

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

    {

        for (int j = 1; j <= k; ++j)

        {

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

        }

    }

    cout << dp[n] << endl;

}


// Faster C++ solution for the same problem (measured speedup: 9.529340711207537x):
#include <iostream>

#include <queue>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

const int maxn = 100010;

int H[maxn];

int main(){

    int n;

    scanf("%d",&n);

    int dp[n+1];

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

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

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

    }

    dp[1]=0;

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

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

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

    }

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

    return 0;



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

using namespace std;

const int nax = 1e5+3;

int number;

int dyn[nax];

int a[nax];

int op1(int i){

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

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

	int res = abs(a[i]-a[i+1])+op1(i+1);

	if(i!=number-2) res = min(res, abs(a[i]-a[i+2])+op1(i+2));

	return dyn[i]=res;

}

int main(){

	cin>>number;

	for(int i=0; i<number; i++) cin>>a[i];

	memset(dyn, -1, sizeof dyn);

	cout<<op1(0)<<endl;

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4185120123340718x):
#include <iostream>

#include <queue>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

const int maxn = 100010;

int H[maxn];

int main(){

    int n;

    scanf("%d",&n);

    int dp[n+1];

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

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

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

    }

    dp[1]=0;

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

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

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

    }

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

    return 0;



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

using namespace std;

int n, arr[100005], best, memo[100005];



int dp(int x){

	if(x==1 || x == 0){

		return 0;

	}

	if (x == 2){

		return abs(arr[x] - arr[x-1]);

	}

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

		return memo[x];

	}

	best=min(dp(x-1)+abs(arr[x]-arr[x-1]), dp(x-2)+abs(arr[x-2]-arr[x]));

	memo[x] = best;

	return best;

}



int main(){

	ios_base::sync_with_stdio(0);

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

	cin.tie();

	cout.tie();

	cin >> n;

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

		cin >> arr[i];

	}

	cout << dp(n) << endl;

}

// Faster C++ solution for the same problem (measured speedup: 1.4346071188978227x):
#include <iostream>

#include <queue>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

const int maxn = 100010;

int H[maxn];

int main(){

    int n;

    scanf("%d",&n);

    int dp[n+1];

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

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

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

    }

    dp[1]=0;

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

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

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

    }

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

    return 0;



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

using namespace std;

int n, arr[100005], best, memo[100005];



int dp(int x){

	if(x==0 or x==1){

		return 0;

	}

	else if(x==2){

		return dp(x-1)+abs(arr[x]-arr[x-1]);

	}

	else if(memo[x]!=-1){

		return memo[x];

	}

	best=min(dp(x-1)+abs(arr[x]-arr[x-1]), dp(x-2)+abs(arr[x-2]-arr[x]));

	memo[x]=best;

	return best;

}



int main(){

	ios_base::sync_with_stdio(0);

	cin.tie();

	cout.tie();

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

	cin >> n;

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

		cin >> arr[i];

	}

	cout << dp(n) << endl;

}


// Faster C++ solution for the same problem (measured speedup: 1.4346071188978227x):
#include <iostream>

#include <queue>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

const int maxn = 100010;

int H[maxn];

int main(){

    int n;

    scanf("%d",&n);

    int dp[n+1];

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

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

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

    }

    dp[1]=0;

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

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

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

    }

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

    return 0;



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

using namespace std;

int n;

int h[1000005], memo[1000005];

int f(int n) {

    if(n == 1) {                        // bisa juga base case nya n = 0, f(0) harus tinggi, misal 1e9;

        return 0;

    }

    if(n == 2) {

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

    } 

    else if(memo[n] != -1) {

        return memo[n];

    }

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

    return memo[n];

}



int main() {

    scanf("%d", &n);

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

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

    }

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

    f(n);

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

}

// Faster C++ solution for the same problem (measured speedup: 3.629728924396555x):
#include <bits/stdc++.h>

using namespace std;

int arr[100010],n;

int dp[100010];

int f(int pos){

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

	int &ret=dp[pos];

	if(ret^-1)return ret;

	if(pos+1==n-1)return ret=abs(arr[pos+1]-arr[pos]);

	return ret=min(f(pos+1)+abs(arr[pos]-arr[pos+1]),f(pos+2)+abs(arr[pos]-arr[pos+2]));

}



int main(int argc, char const *argv[])

{

	memset(dp,-1,sizeof dp);

	cin>>n;

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

		cout<<f(0)<<endl;

	return 0;

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

using namespace std;

#define IO ios::sync_with_stdio(false), cin.tie(0)

typedef long long ll;

const int maxn = 1e6 + 5;

const int inf = 0x3f3f3f3f;

ll a[maxn];

ll dp[maxn];



int main()

{

    IO;

    int n, k=2;

    cin >> n ;

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

        cin >> a[i];

    memset(dp, inf, sizeof(dp));

    dp[1] = 0;

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

    {

        for (int j = 1; j <= k; ++j)

        {

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

        }

    }

    cout << dp[n] << endl;

}


// Faster C++ solution for the same problem (measured speedup: 6.740973915825536x):
#include <bits/stdc++.h>

using namespace std;

int arr[100010],n;

int dp[100010];

int f(int pos){

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

	int &ret=dp[pos];

	if(ret^-1)return ret;

	if(pos+1==n-1)return ret=abs(arr[pos+1]-arr[pos]);

	return ret=min(f(pos+1)+abs(arr[pos]-arr[pos+1]),f(pos+2)+abs(arr[pos]-arr[pos+2]));

}



int main(int argc, char const *argv[])

{

	memset(dp,-1,sizeof dp);

	cin>>n;

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

		cout<<f(0)<<endl;

	return 0;

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

using namespace std;

int n;

int h[1000005], memo[1000005];

int f(int n) {

    if(n == 1) {                        // bisa juga base case nya n = 0, f(0) harus tinggi, misal 1e9;

        return 0;

    }

    if(n == 2) {

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

    } 

    else if(memo[n] != -1) {

        return memo[n];

    }

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

    return memo[n];

}



int main() {

    scanf("%d", &n);

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

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

    }

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

    f(n);

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

}

// Faster C++ solution for the same problem (measured speedup: 3.6173248644715765x):
#include<bits/stdc++.h>

#define ll long long

using namespace std;

const int maxn = 1e5+5;

int a[maxn], dp[maxn];



int main(){

	int n; cin >> n;

	memset(dp, 0x3f, sizeof dp);

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

	dp[1] = 0;

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

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

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

	}

	cout << dp[n] << endl;

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

using namespace std;

#define IO ios::sync_with_stdio(false), cin.tie(0)

typedef long long ll;

const int maxn = 1e6 + 5;

const int inf = 0x3f3f3f3f;

ll a[maxn];

ll dp[maxn];



int main()

{

    IO;

    int n, k=2;

    cin >> n ;

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

        cin >> a[i];

    memset(dp, inf, sizeof(dp));

    dp[1] = 0;

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

    {

        for (int j = 1; j <= k; ++j)

        {

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

        }

    }

    cout << dp[n] << endl;

}


// Faster C++ solution for the same problem (measured speedup: 6.717937637870284x):
#include<bits/stdc++.h>

#define ll long long

using namespace std;

const int maxn = 1e5+5;

int a[maxn], dp[maxn];



int main(){

	int n; cin >> n;

	memset(dp, 0x3f, sizeof dp);

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

	dp[1] = 0;

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

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

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

	}

	cout << dp[n] << endl;

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

using namespace std;

int dif(int a, int b)

{

    int ans=a-b;

    if(ans<0) ans*=-1;

    return ans;

}

int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    int N, h[100005],i,j;

    int f1,f2;

    cin>>N;

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

        cin>>h[i];

    int M[100005]={0};

    M[N-2]=dif(h[N-1],h[N-2]);

    for(i=N-3;i>=0;i--)

    {

        f1=dif(h[i+1],h[i])+M[i+1];

        f2=dif(h[i+2],h[i])+M[i+2];

        M[i]=min(f1,f2);

    }

    cout<<M[0]<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.402439136984015x):
#include <bits/stdc++.h>

using ll = long long;

using namespace std;



int n, k = 2;

int arr[100100], dp[100100];

bool vis[100100];



int func( int pos ) {

  if (pos == n) return 0;

  if (vis[pos]) return dp[pos];

  vis[pos] = 1;

  int res = INT_MAX;

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

    if (pos + i > n) break;

    res = min(res, func(pos + i) + abs(arr[pos] - arr[pos + i]));

  }

  return dp[pos] = res;

}



int main() {

  ios::sync_with_stdio(false);

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



  cin >> n;

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

  cout << func(1) << endl;

  return 0;

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

using namespace std;

int n;

int h[1000005], memo[1000005];

int f(int n) {

    if(n == 1) {                        // bisa juga base case nya n = 0, f(0) harus tinggi, misal 1e9;

        return 0;

    }

    if(n == 2) {

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

    } 

    else if(memo[n] != -1) {

        return memo[n];

    }

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

    return memo[n];

}



int main() {

    scanf("%d", &n);

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

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

    }

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

    f(n);

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

}

// Faster C++ solution for the same problem (measured speedup: 5.001340442721532x):
#include <bits/stdc++.h>

using ll = long long;

using namespace std;



int n, k = 2;

int arr[100100], dp[100100];

bool vis[100100];



int func( int pos ) {

  if (pos == n) return 0;

  if (vis[pos]) return dp[pos];

  vis[pos] = 1;

  int res = INT_MAX;

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

    if (pos + i > n) break;

    res = min(res, func(pos + i) + abs(arr[pos] - arr[pos + i]));

  }

  return dp[pos] = res;

}



int main() {

  ios::sync_with_stdio(false);

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



  cin >> n;

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

  cout << func(1) << endl;

  return 0;

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

using namespace std;



typedef signed long long ll;

const int MX = 1e5 + 5;

const int INF = 1e9 + 5;

int n, A[MX], dp[MX];



int solve(int pos) {

    if(pos > n) return INF;

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

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

    int ans = INF;

    if(pos + 1 < n) ans = min(ans, abs(A[pos] - A[pos + 1]) + solve(pos + 1));

    if(pos + 2 < n) ans = min(ans, abs(A[pos] - A[pos + 2]) + solve(pos + 2));

    return dp[pos] = ans;

}



int main() {

    scanf("%d", &n);



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

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

    }

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

    printf("%d\n", solve(0));

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3704760734487191x):
#include <bits/stdc++.h>

using ll = long long;

using namespace std;



int n, k = 2;

int arr[100100], dp[100100];

bool vis[100100];



int func( int pos ) {

  if (pos == n) return 0;

  if (vis[pos]) return dp[pos];

  vis[pos] = 1;

  int res = INT_MAX;

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

    if (pos + i > n) break;

    res = min(res, func(pos + i) + abs(arr[pos] - arr[pos + i]));

  }

  return dp[pos] = res;

}



int main() {

  ios::sync_with_stdio(false);

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



  cin >> n;

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

  cout << func(1) << endl;

  return 0;

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

#include<cmath>

#include<algorithm>

#include<cstring>

using namespace std;



int h[100055],dp[100055];

const int inf = 0x3f3f3f;

int main()

{

	int n;

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

	cin>>n;

	h[1]=inf;

	h[0]=inf;

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

		cin>>h[i];

	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]<<endl;

	return 0;

} 

//max(h[i],h[i-2])-min(h[i],h[i-2])

// Faster C++ solution for the same problem (measured speedup: 1.385886765891102x):
#include <bits/stdc++.h>

using ll = long long;

using namespace std;



int n, k = 2;

int arr[100100], dp[100100];

bool vis[100100];



int func( int pos ) {

  if (pos == n) return 0;

  if (vis[pos]) return dp[pos];

  vis[pos] = 1;

  int res = INT_MAX;

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

    if (pos + i > n) break;

    res = min(res, func(pos + i) + abs(arr[pos] - arr[pos + i]));

  }

  return dp[pos] = res;

}



int main() {

  ios::sync_with_stdio(false);

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



  cin >> n;

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

  cout << func(1) << endl;

  return 0;

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

using namespace std;

#define IO ios::sync_with_stdio(false), cin.tie(0)

typedef long long ll;

const int maxn = 1e6 + 5;

const int inf = 0x3f3f3f3f;

ll a[maxn];

ll dp[maxn];



int main()

{

    IO;

    int n, k=2;

    cin >> n ;

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

        cin >> a[i];

    memset(dp, inf, sizeof(dp));

    dp[1] = 0;

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

    {

        for (int j = 1; j <= k; ++j)

        {

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

        }

    }

    cout << dp[n] << endl;

}


// Faster C++ solution for the same problem (measured speedup: 9.288270879389195x):
#include <bits/stdc++.h>

using ll = long long;

using namespace std;



int n, k = 2;

int arr[100100], dp[100100];

bool vis[100100];



int func( int pos ) {

  if (pos == n) return 0;

  if (vis[pos]) return dp[pos];

  vis[pos] = 1;

  int res = INT_MAX;

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

    if (pos + i > n) break;

    res = min(res, func(pos + i) + abs(arr[pos] - arr[pos + i]));

  }

  return dp[pos] = res;

}



int main() {

  ios::sync_with_stdio(false);

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



  cin >> n;

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

  cout << func(1) << endl;

  return 0;

}