#include using namespace std; typedef long long ll; #define vi vector #define vb vector #define pub push_back #define emb emplace_back #define rep(i,a,b) for(long long i = a; i < b; i++) #define rrep(i,a,b) for(long long i = a; i >= b; i--) #define iterm(key,val,ok) for(auto const& [key, val]:ok) #define all(x) x.begin(),x.end() #define py cout<<"YES"<; template istream& operator>>(istream& is, vector& v) { for (auto& i : v) is >> i; return is; } template ostream& operator<<(ostream& os, const vector& v) { for (auto& i : v) os << i << " "; return os; } template istream& operator>>(istream& is, pair& v) { is >> v.first >> v.second; return is; } template ostream& operator<<(ostream& os, const pair& v) { os << v.first << " " << v.second; return os; } /* problems with mod or small constraints are DP (2d,3d) or combo ascii: 1=49 a=97 A=65 check edge cases and constraints CHECK FOR JUST ONE ELEMENT OR SAME NUMBERS ETC always think about reversing a process see if theres any dumb tricks if youre not getting it: like only having to check small values or sth (a-b)modm = (a%m+(m-b%m))%m mark visited after adding to queue think about transforming representations of graphs, if some graph is too big to be represented, think about bipartite representation, if its too small maybe you can expand according to problem to make it easier you can also bfs on pairs of nodes, different representations of nodes etc etc lower bound can give beyond end also (check segfaults etc if using) many smol vectors are slowwww, use std:array instead to search for and remove particular things fastly just use multiset, like u can use prefix multiset and find closest value of something also like so many applications in just logn time */ void solve(){ ll n;cin>>n; vector a(n); vector b(n); vector x(n); ll d1,d2,d3; cin>>a>>b; ll dp[n][3]; ll ans=0; rep(j,0,3){ if(j==0){ dp[0][0]=dp[0][1]=dp[0][2]=a[0]; } else if(j==1){ dp[0][0]=dp[0][1]=dp[0][2]=b[0]; } else{ dp[0][0]=dp[0][1]=dp[0][2]=(a[0]+b[0])/2; } rep(i,1,n){ vector arr1; x[i]=(a[i]+b[i])/2; if(a[i]>=dp[i-1][1] && a[i]<=dp[i-1][2]){ dp[i][0]=a[i]; dp[i][1]=dp[i-1][1]; dp[i][2]=dp[i-1][2]; continue; } else if(b[i]>=dp[i-1][1] && b[i]<=dp[i-1][2]){ dp[i][0]=b[i]; dp[i][1]=dp[i-1][1]; dp[i][2]=dp[i-1][2]; continue; } else if(x[i]>=dp[i-1][1] && x[i]<=dp[i-1][2]){ dp[i][0]=x[i]; dp[i][1]=dp[i-1][1]; dp[i][2]=dp[i-1][2]; continue; } else{ d1=min(abs(a[i]-dp[i-1][1]),abs(a[i]-dp[i-1][2])); d2=min(abs(b[i]-dp[i-1][1]),abs(b[i]-dp[i-1][2])); d3=min(abs(x[i]-dp[i-1][1]),abs(x[i]-dp[i-1][2])); if(d1<=d2 && d1<=d3){ dp[i][0]=a[i]; dp[i][1]=min(dp[i-1][1],a[i]); dp[i][2]=max(dp[i-1][2],a[i]); } else if(d2<=d1 && d2<=d3){ dp[i][0]=b[i]; dp[i][1]=min(dp[i-1][1],b[i]); dp[i][2]=max(dp[i-1][2],b[i]); } else{ dp[i][0]=x[i]; dp[i][1]=min(dp[i-1][1],x[i]); dp[i][2]=max(dp[i-1][2],x[i]); } } } if(j==0){ ans=dp[n-1][2]-dp[n-1][1]; } else{ ans=min(ans,dp[n-1][2]-dp[n-1][1]); } } pa; } signed main() { ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int t=1; //cin>>t; while(t--){ solve(); } }