[Codeforces] Round #333 (Div. 2) B. Approximating a Constant Range

Codeforces Round #333 (Div. 2) B. Approximating a Constant Range

  • Time : O(n)
  • Space : O(1)
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#include <bits/stdc++.h>

#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("Ofast")//Comment optimisations for interactive problems (use endl)
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")

using namespace std;

struct PairHash {inline std::size_t operator()(const std::pair<int, int> &v) const { return v.first * 31 + v.second; }};

// speed
#define Code ios_base::sync_with_stdio(false);
#define By ios::sync_with_stdio(0);
#define Sumfi cout.tie(NULL);

// alias
using ll = long long;
using ld = long double;

// constants
const ld PI = 3.14159265358979323846; /* pi */
const ll INF = 1e18;
const ld EPS = 1e-9;
const ll MAX_N = 505050;
const ll mod = 1e9 + 7;

// typedef
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef array<ll,3> all3;
typedef array<ll,5> all5;
typedef vector<all3> vall3;
typedef vector<all5> vall5;
typedef vector<ld> vld;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<vs> vvs;
typedef unordered_set<ll> usll;
typedef unordered_set<pll, PairHash> uspll;
typedef unordered_map<ll, ll> umll;
typedef unordered_map<pll, ll, PairHash> umpll;

// macros
#define rep(i,m,n) for(ll i=m;i<n;i++)
#define rrep(i,m,n) for(ll i=n;i>=m;i--)
#define all(a) begin(a), end(a)
#define rall(a) rbegin(a), rend(a)
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define sz(x) ll((x).size())
#define pyes cout<<"YES\n";
#define pno cout<<"NO\n";
#define pneg1 cout<<"-1\n";
#define CASE(x) cout<<"Case #"<<x<<": ";

// utility functions
template <typename T>
void print(T &&t) { cout << t << "\n"; }
template<typename T>
void printv(vector<T>v){ll n=v.size();rep(i,0,n)cout<<v[i]<<" ";cout<<"\n";}
//void readf() {freopen("", "rt", stdin);}

ll __gcd(ll x, ll y) { return !y ? x : __gcd(y, x % y); }
ll __lcm(ll x, ll y) { return x * y / __gcd(x,y); }
ll modpow(ll n, ll x, ll MOD = mod) { if(!x) return 1; ll res = modpow(n,x>>1,MOD); res = (res * res) % MOD; if(x&1) res = (res * n) % MOD; return res; }

ll solve(vll& A) {
ll res = 0, l = 0, r = 0, n = sz(A);
ll mi = A[0], ma = A[0];
umll freq;
while(r < n) {
while(r < n and abs(ma-mi) <= 1) {
freq[A[r]]++;
ma = max(ma, A[r]);
mi = min(mi, A[r]);
r++;
if(abs(ma - mi) <= 1)
res = max(res, r - l);
}
while(abs(ma-mi) > 1) {
if(--freq[A[l]] == 0) {
freq.erase(A[l]);
if(A[l] == ma) {
ma = -INF;
for(auto [n,_] : freq) ma = max(ma,n);
}
if(A[l] == mi) {
mi = INF;
for(auto [n,_] : freq) mi = min(mi,n);
}
}
l++;
}
}
return res;
}

int main() {
Code By Sumfi

cout.precision(12);

ll tc = 1;
//cin>>tc;
for (ll i = 1; i <= tc; i++) {
ll n;
cin>>n;
vll A(n);
rep(i,0,n) cin>>A[i];
print(solve(A));
}
return 0;
}
Author: Song Hayoung
Link: https://songhayoung.github.io/2022/07/19/PS/Codeforces/div2-333-b/
Copyright Notice: All articles in this blog are licensed under CC BY-NC-SA 4.0 unless stating additionally.