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: 1.3575453872870975x):
#include <bits/stdc++.h>

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

#define ii          pair<int,int>

#define F           first

#define S           second

#define pb          push_back

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

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

#define fore(i,a,b) for(int i = a;i < b; i+= 1)

#define forr(i,a)   for(int i = a; i >= 0; i--)

#define fori(i,m)   for(auto i = m.begin(); i != m.end(); i++) 

#define w(t)        while(t--)

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

#define cls(a,car)  memset(a,car,sizeof (a))

#define db(x)       cerr << #x << " is " << x << '\n'

#define FL          freopen("input.txt", "r", stdin),freopen("output.txt", "w", stdout);

#define EjecuteTime cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n"

#define angle(x)    double(x * acos(-1) / 180.0)

using namespace std;

typedef long long   ll;

typedef vector<int> vi;

typedef vector<ii>  vii;

const int N = 1 * 1e5 + 5;

const ll mod = 1e9 + 7;

const double E = 1e-9;

//int dx[] = {1, 0}, dy[] = {0, 1};

//int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};

//int dx[] = {2, 1, -1, -2, -2, -1, 1, 2}, dy[] = {1, 2, 2, 1, -1, -2, -2, -1};

//int dx[] = {1, 1, 0, -1, -1, -1, 0, 1}, dy[] = {0 ,1, 1, 1, 0, -1, -1, -1};

// fflush(stdout)

// cout << flush

// a + b = a ^ b + 2 * (a & b)

int v[N];

int dp[N];

int n, k = 2;

int f(int i){

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

    int &ans = dp[i];

    if(ans == -1){

        ans = 1e9;

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

            if(i + j < n)

                ans = min(ans, f(i + j) + abs(v[i] - v[i + j]));

            else

                break;

    }

    return ans;

}

int main(){

    //FL;

    FIN;

    cin >> n;

    fore(i, 0, n)cin >> v[i];

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

        dp[i] = 1e9;

    for(int i = n - 1; i >= 0; i--){

        int &ans = dp[i];

        if(i == n - 1){

            ans = 0;

            continue;

        }

        ans = 1e9;

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

            if(i + j < n)

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

            else

                break;

        }

    }

    

    cout << dp[0] << '\n';

    EjecuteTime;

    return 0;

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

#include<stdio.h>

#include<cstring>

#include<cmath>

#include<algorithm>

using namespace std;

#define maxn 100005

int dp[maxn];

int h[maxn];

int n;

int main(){

    scanf("%d",&n);

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

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

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

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

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

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

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

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

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

    }

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.7349191403404547x):
#include <bits/stdc++.h>

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

#define ii          pair<int,int>

#define F           first

#define S           second

#define pb          push_back

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

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

#define fore(i,a,b) for(int i = a;i < b; i+= 1)

#define forr(i,a)   for(int i = a; i >= 0; i--)

#define fori(i,m)   for(auto i = m.begin(); i != m.end(); i++) 

#define w(t)        while(t--)

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

#define cls(a,car)  memset(a,car,sizeof (a))

#define db(x)       cerr << #x << " is " << x << '\n'

#define FL          freopen("input.txt", "r", stdin),freopen("output.txt", "w", stdout);

#define EjecuteTime cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n"

#define angle(x)    double(x * acos(-1) / 180.0)

using namespace std;

typedef long long   ll;

typedef vector<int> vi;

typedef vector<ii>  vii;

const int N = 1 * 1e5 + 5;

const ll mod = 1e9 + 7;

const double E = 1e-9;

//int dx[] = {1, 0}, dy[] = {0, 1};

//int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};

//int dx[] = {2, 1, -1, -2, -2, -1, 1, 2}, dy[] = {1, 2, 2, 1, -1, -2, -2, -1};

//int dx[] = {1, 1, 0, -1, -1, -1, 0, 1}, dy[] = {0 ,1, 1, 1, 0, -1, -1, -1};

// fflush(stdout)

// cout << flush

// a + b = a ^ b + 2 * (a & b)

int v[N];

int dp[N];

int n, k = 2;

int f(int i){

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

    int &ans = dp[i];

    if(ans == -1){

        ans = 1e9;

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

            if(i + j < n)

                ans = min(ans, f(i + j) + abs(v[i] - v[i + j]));

            else

                break;

    }

    return ans;

}

int main(){

    //FL;

    FIN;

    cin >> n;

    fore(i, 0, n)cin >> v[i];

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

        dp[i] = 1e9;

    for(int i = n - 1; i >= 0; i--){

        int &ans = dp[i];

        if(i == n - 1){

            ans = 0;

            continue;

        }

        ans = 1e9;

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

            if(i + j < n)

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

            else

                break;

        }

    }

    

    cout << dp[0] << '\n';

    EjecuteTime;

    return 0;

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



using namespace std;

int v[100010];

int memo[100010];

int c;



int f(int i)

{



    if(memo[i]!=-1)

        return memo[i];



    if(i == 0)

        return memo[0] = 0;



    if(i == 1)

        return memo[1] = abs(v[0] - v[1]);



    return memo[i] = min(abs(v[i] - v[i-1]) + f(i-1), abs(v[i] - v[i-2]) + f(i-2));

}



int main()

{

    ios_base::sync_with_stdio(false);

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

    cin>>c;



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

    {

        cin>>v[i];

    }



    cout<<f(c-1)<<endl;



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3731225805662304x):
#include <bits/stdc++.h>

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

#define ii          pair<int,int>

#define F           first

#define S           second

#define pb          push_back

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

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

#define fore(i,a,b) for(int i = a;i < b; i+= 1)

#define forr(i,a)   for(int i = a; i >= 0; i--)

#define fori(i,m)   for(auto i = m.begin(); i != m.end(); i++) 

#define w(t)        while(t--)

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

#define cls(a,car)  memset(a,car,sizeof (a))

#define db(x)       cerr << #x << " is " << x << '\n'

#define FL          freopen("input.txt", "r", stdin),freopen("output.txt", "w", stdout);

#define EjecuteTime cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n"

#define angle(x)    double(x * acos(-1) / 180.0)

using namespace std;

typedef long long   ll;

typedef vector<int> vi;

typedef vector<ii>  vii;

const int N = 1 * 1e5 + 5;

const ll mod = 1e9 + 7;

const double E = 1e-9;

//int dx[] = {1, 0}, dy[] = {0, 1};

//int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};

//int dx[] = {2, 1, -1, -2, -2, -1, 1, 2}, dy[] = {1, 2, 2, 1, -1, -2, -2, -1};

//int dx[] = {1, 1, 0, -1, -1, -1, 0, 1}, dy[] = {0 ,1, 1, 1, 0, -1, -1, -1};

// fflush(stdout)

// cout << flush

// a + b = a ^ b + 2 * (a & b)

int v[N];

int dp[N];

int n, k = 2;

int f(int i){

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

    int &ans = dp[i];

    if(ans == -1){

        ans = 1e9;

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

            if(i + j < n)

                ans = min(ans, f(i + j) + abs(v[i] - v[i + j]));

            else

                break;

    }

    return ans;

}

int main(){

    //FL;

    FIN;

    cin >> n;

    fore(i, 0, n)cin >> v[i];

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

        dp[i] = 1e9;

    for(int i = n - 1; i >= 0; i--){

        int &ans = dp[i];

        if(i == n - 1){

            ans = 0;

            continue;

        }

        ans = 1e9;

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

            if(i + j < n)

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

            else

                break;

        }

    }

    

    cout << dp[0] << '\n';

    EjecuteTime;

    return 0;

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

using namespace std;

#define Me3za() ios_base::sync_with_stdio(0);cout.tie(0);cin.tie(0);srand(time(NULL))

#define endl "\n"

int n;

int arr[100005];

int mem[100005];

long long solve(int i,int cur){

if(i>=n)return 1e9;

//cout<<i<<" "<<cur<<endl;

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

long long ch1=solve(i+1,arr[i+1])+abs(cur-arr[i+1]);

long long ch2=0;

if(i+2<=n)ch2=solve(i+2,arr[i+2])+abs(cur-arr[i+2]);

return mem[i]=min(ch1,ch2);

}

int main() {

Me3za();

cin>>n;

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

    cin>>arr[i];

}

memset(mem,-1,sizeof mem);

cout<<solve(0,arr[0]);

}




// Faster C++ solution for the same problem (measured speedup: 1.368518738313693x):
#include <bits/stdc++.h>

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

#define ii          pair<int,int>

#define F           first

#define S           second

#define pb          push_back

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

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

#define fore(i,a,b) for(int i = a;i < b; i+= 1)

#define forr(i,a)   for(int i = a; i >= 0; i--)

#define fori(i,m)   for(auto i = m.begin(); i != m.end(); i++) 

#define w(t)        while(t--)

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

#define cls(a,car)  memset(a,car,sizeof (a))

#define db(x)       cerr << #x << " is " << x << '\n'

#define FL          freopen("input.txt", "r", stdin),freopen("output.txt", "w", stdout);

#define EjecuteTime cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n"

#define angle(x)    double(x * acos(-1) / 180.0)

using namespace std;

typedef long long   ll;

typedef vector<int> vi;

typedef vector<ii>  vii;

const int N = 1 * 1e5 + 5;

const ll mod = 1e9 + 7;

const double E = 1e-9;

//int dx[] = {1, 0}, dy[] = {0, 1};

//int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};

//int dx[] = {2, 1, -1, -2, -2, -1, 1, 2}, dy[] = {1, 2, 2, 1, -1, -2, -2, -1};

//int dx[] = {1, 1, 0, -1, -1, -1, 0, 1}, dy[] = {0 ,1, 1, 1, 0, -1, -1, -1};

// fflush(stdout)

// cout << flush

// a + b = a ^ b + 2 * (a & b)

int v[N];

int dp[N];

int n, k = 2;

int f(int i){

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

    int &ans = dp[i];

    if(ans == -1){

        ans = 1e9;

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

            if(i + j < n)

                ans = min(ans, f(i + j) + abs(v[i] - v[i + j]));

            else

                break;

    }

    return ans;

}

int main(){

    //FL;

    FIN;

    cin >> n;

    fore(i, 0, n)cin >> v[i];

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

        dp[i] = 1e9;

    for(int i = n - 1; i >= 0; i--){

        int &ans = dp[i];

        if(i == n - 1){

            ans = 0;

            continue;

        }

        ans = 1e9;

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

            if(i + j < n)

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

            else

                break;

        }

    }

    

    cout << dp[0] << '\n';

    EjecuteTime;

    return 0;

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

using namespace std;



int const N=1e5+5, M=1e9+7, OO = 0x3f3f3f3f;



int n;

int arr[N];



int mem[N];

int solve(int i)

{

    if(i==n-1)

        return 0;

    if(mem[i]!=-1)

        return mem[i];

    int op1=solve(i+1)+abs(arr[i]-arr[i+1]);

    int op2=(i<=n-3) ? solve(i+2)+abs(arr[i]-arr[i+2]) : OO;

    return mem[i]=min(op1,op2);

}





int main()

{

//    freopen("input.txt","rt",stdin);

//    freopen("output.txt","wt",stdout);

    scanf("%d",&n);

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

        scanf("%d",arr+i);

    memset(mem, -1, sizeof mem);

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



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3454024130485864x):
#include <bits/stdc++.h>

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

#define ii          pair<int,int>

#define F           first

#define S           second

#define pb          push_back

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

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

#define fore(i,a,b) for(int i = a;i < b; i+= 1)

#define forr(i,a)   for(int i = a; i >= 0; i--)

#define fori(i,m)   for(auto i = m.begin(); i != m.end(); i++) 

#define w(t)        while(t--)

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

#define cls(a,car)  memset(a,car,sizeof (a))

#define db(x)       cerr << #x << " is " << x << '\n'

#define FL          freopen("input.txt", "r", stdin),freopen("output.txt", "w", stdout);

#define EjecuteTime cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n"

#define angle(x)    double(x * acos(-1) / 180.0)

using namespace std;

typedef long long   ll;

typedef vector<int> vi;

typedef vector<ii>  vii;

const int N = 1 * 1e5 + 5;

const ll mod = 1e9 + 7;

const double E = 1e-9;

//int dx[] = {1, 0}, dy[] = {0, 1};

//int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};

//int dx[] = {2, 1, -1, -2, -2, -1, 1, 2}, dy[] = {1, 2, 2, 1, -1, -2, -2, -1};

//int dx[] = {1, 1, 0, -1, -1, -1, 0, 1}, dy[] = {0 ,1, 1, 1, 0, -1, -1, -1};

// fflush(stdout)

// cout << flush

// a + b = a ^ b + 2 * (a & b)

int v[N];

int dp[N];

int n, k = 2;

int f(int i){

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

    int &ans = dp[i];

    if(ans == -1){

        ans = 1e9;

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

            if(i + j < n)

                ans = min(ans, f(i + j) + abs(v[i] - v[i + j]));

            else

                break;

    }

    return ans;

}

int main(){

    //FL;

    FIN;

    cin >> n;

    fore(i, 0, n)cin >> v[i];

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

        dp[i] = 1e9;

    for(int i = n - 1; i >= 0; i--){

        int &ans = dp[i];

        if(i == n - 1){

            ans = 0;

            continue;

        }

        ans = 1e9;

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

            if(i + j < n)

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

            else

                break;

        }

    }

    

    cout << dp[0] << '\n';

    EjecuteTime;

    return 0;

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



using namespace std;

typedef long long ll;

typedef double dd;

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

#define endl "\n"

#define clr(n, r) memset(n,r,sizeof(n));



void fast() {

    cin.tie(0);

    cin.sync_with_stdio(0);

}



const int N = 1e5 + 9;

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



int sol(int i){

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

    if(i == n)  return 1e9;



    int &ret = dp[i];

    if(~ret)    return ret;



    return ret = min(abs(h[i] - h[i + 1]) + sol(i + 1), abs(h[i] - h[i + 2]) + sol(i + 2));

}





int main() {



    cin >> n;



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



    memset(dp, -1, sizeof dp);



    cout << sol(0);

}

// Faster C++ solution for the same problem (measured speedup: 1.3557009191536165x):
#include <bits/stdc++.h>

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

#define ii          pair<int,int>

#define F           first

#define S           second

#define pb          push_back

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

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

#define fore(i,a,b) for(int i = a;i < b; i+= 1)

#define forr(i,a)   for(int i = a; i >= 0; i--)

#define fori(i,m)   for(auto i = m.begin(); i != m.end(); i++) 

#define w(t)        while(t--)

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

#define cls(a,car)  memset(a,car,sizeof (a))

#define db(x)       cerr << #x << " is " << x << '\n'

#define FL          freopen("input.txt", "r", stdin),freopen("output.txt", "w", stdout);

#define EjecuteTime cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n"

#define angle(x)    double(x * acos(-1) / 180.0)

using namespace std;

typedef long long   ll;

typedef vector<int> vi;

typedef vector<ii>  vii;

const int N = 1 * 1e5 + 5;

const ll mod = 1e9 + 7;

const double E = 1e-9;

//int dx[] = {1, 0}, dy[] = {0, 1};

//int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};

//int dx[] = {2, 1, -1, -2, -2, -1, 1, 2}, dy[] = {1, 2, 2, 1, -1, -2, -2, -1};

//int dx[] = {1, 1, 0, -1, -1, -1, 0, 1}, dy[] = {0 ,1, 1, 1, 0, -1, -1, -1};

// fflush(stdout)

// cout << flush

// a + b = a ^ b + 2 * (a & b)

int v[N];

int dp[N];

int n, k = 2;

int f(int i){

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

    int &ans = dp[i];

    if(ans == -1){

        ans = 1e9;

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

            if(i + j < n)

                ans = min(ans, f(i + j) + abs(v[i] - v[i + j]));

            else

                break;

    }

    return ans;

}

int main(){

    //FL;

    FIN;

    cin >> n;

    fore(i, 0, n)cin >> v[i];

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

        dp[i] = 1e9;

    for(int i = n - 1; i >= 0; i--){

        int &ans = dp[i];

        if(i == n - 1){

            ans = 0;

            continue;

        }

        ans = 1e9;

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

            if(i + j < n)

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

            else

                break;

        }

    }

    

    cout << dp[0] << '\n';

    EjecuteTime;

    return 0;

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



using namespace std;



const int N=1e5;

int n;

int arr[N];

int mem[N];



int solve(int i){

    if(i==n-1)

        return 0;

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

    int op1=solve(i+1)+abs(arr[i]-arr[i+1]);

    int op2=i+2<=n-1?solve(i+2)+abs(arr[i]-arr[i+2]):1000000000;

    return mem[i]=min(op1,op2);

}



int main() {

    scanf("%d",&n);

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

        scanf("%d",arr+i);

    memset(mem,-1,sizeof mem);

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3450712066964536x):
#include <bits/stdc++.h>

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

#define ii          pair<int,int>

#define F           first

#define S           second

#define pb          push_back

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

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

#define fore(i,a,b) for(int i = a;i < b; i+= 1)

#define forr(i,a)   for(int i = a; i >= 0; i--)

#define fori(i,m)   for(auto i = m.begin(); i != m.end(); i++) 

#define w(t)        while(t--)

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

#define cls(a,car)  memset(a,car,sizeof (a))

#define db(x)       cerr << #x << " is " << x << '\n'

#define FL          freopen("input.txt", "r", stdin),freopen("output.txt", "w", stdout);

#define EjecuteTime cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n"

#define angle(x)    double(x * acos(-1) / 180.0)

using namespace std;

typedef long long   ll;

typedef vector<int> vi;

typedef vector<ii>  vii;

const int N = 1 * 1e5 + 5;

const ll mod = 1e9 + 7;

const double E = 1e-9;

//int dx[] = {1, 0}, dy[] = {0, 1};

//int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};

//int dx[] = {2, 1, -1, -2, -2, -1, 1, 2}, dy[] = {1, 2, 2, 1, -1, -2, -2, -1};

//int dx[] = {1, 1, 0, -1, -1, -1, 0, 1}, dy[] = {0 ,1, 1, 1, 0, -1, -1, -1};

// fflush(stdout)

// cout << flush

// a + b = a ^ b + 2 * (a & b)

int v[N];

int dp[N];

int n, k = 2;

int f(int i){

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

    int &ans = dp[i];

    if(ans == -1){

        ans = 1e9;

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

            if(i + j < n)

                ans = min(ans, f(i + j) + abs(v[i] - v[i + j]));

            else

                break;

    }

    return ans;

}

int main(){

    //FL;

    FIN;

    cin >> n;

    fore(i, 0, n)cin >> v[i];

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

        dp[i] = 1e9;

    for(int i = n - 1; i >= 0; i--){

        int &ans = dp[i];

        if(i == n - 1){

            ans = 0;

            continue;

        }

        ans = 1e9;

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

            if(i + j < n)

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

            else

                break;

        }

    }

    

    cout << dp[0] << '\n';

    EjecuteTime;

    return 0;

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

#include <bitset>

using namespace std;

#define ll long long

#define mod 1000000009



// يلعن ابو الي يعقدها على نفسه تاني



int main()

{

    ios_base::sync_with_stdio(false);    cin.tie(NULL);



    ll n; cin >> n;

    ll arr[100005];

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

        cin >> arr[i];

    }

    if(n == 2){

        cout << abs(arr[1]-arr[0]);

    }else{

        ll sol[100005] = {};

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

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

            sol[i] = min(sol[i-1]+abs(arr[i]-arr[i-1]),sol[i-2]+abs(arr[i]-arr[i-2]));

        }









        cout << sol[n-1];

    }















    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.5574610586248654x):
#include <bits/stdc++.h>

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

#define ii          pair<int,int>

#define F           first

#define S           second

#define pb          push_back

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

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

#define fore(i,a,b) for(int i = a;i < b; i+= 1)

#define forr(i,a)   for(int i = a; i >= 0; i--)

#define fori(i,m)   for(auto i = m.begin(); i != m.end(); i++) 

#define w(t)        while(t--)

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

#define cls(a,car)  memset(a,car,sizeof (a))

#define db(x)       cerr << #x << " is " << x << '\n'

#define FL          freopen("input.txt", "r", stdin),freopen("output.txt", "w", stdout);

#define EjecuteTime cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n"

#define angle(x)    double(x * acos(-1) / 180.0)

using namespace std;

typedef long long   ll;

typedef vector<int> vi;

typedef vector<ii>  vii;

const int N = 1 * 1e5 + 5;

const ll mod = 1e9 + 7;

const double E = 1e-9;

//int dx[] = {1, 0}, dy[] = {0, 1};

//int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};

//int dx[] = {2, 1, -1, -2, -2, -1, 1, 2}, dy[] = {1, 2, 2, 1, -1, -2, -2, -1};

//int dx[] = {1, 1, 0, -1, -1, -1, 0, 1}, dy[] = {0 ,1, 1, 1, 0, -1, -1, -1};

// fflush(stdout)

// cout << flush

// a + b = a ^ b + 2 * (a & b)

int v[N];

int dp[N];

int n, k = 2;

int f(int i){

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

    int &ans = dp[i];

    if(ans == -1){

        ans = 1e9;

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

            if(i + j < n)

                ans = min(ans, f(i + j) + abs(v[i] - v[i + j]));

            else

                break;

    }

    return ans;

}

int main(){

    //FL;

    FIN;

    cin >> n;

    fore(i, 0, n)cin >> v[i];

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

        dp[i] = 1e9;

    for(int i = n - 1; i >= 0; i--){

        int &ans = dp[i];

        if(i == n - 1){

            ans = 0;

            continue;

        }

        ans = 1e9;

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

            if(i + j < n)

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

            else

                break;

        }

    }

    

    cout << dp[0] << '\n';

    EjecuteTime;

    return 0;

}