Codeforces Round #622 (Div. 2) Problem C1

Problem: https://codeforces.com/contest/1313/problem/C1
Hard Version : https://codeforces.com/contest/1313/problem/C2

What’s the intution behind it and how to approach towards the solution for hard version? Any solution/logic with a brief explanation would help. Thank you.

The basic logic behind it is that in the final answer, one element must be a peak (greater than or equal to previous element and also greater than or equal to next element) and elements before it should be in non-descending order and elements after it should be in non-ascending order.
For the easy version, we can just check by taking every element as a peak and then forming the solution. The answer is the solution with the largest total sum.
I have no idea how to solve the hard version.

2 Likes

Yeah the greedy approach worked. Thanks.

I’ll try the harder version now.

for hard version during contest I overkilled with segment tree+binary search and got tle cz of high constant realize after contest a stack would suffice and there is no need of seg tree

Can you explain a bit more about the stack approach?

heres my code

#include <iostream>
#include <bits/stdc++.h>
#include <stdio.h>
 
using namespace std;
typedef   long long int llo;
#define pb push_back
llo inf=100000000000;
#define mp make_pair
llo it[500001];
 

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	llo n;
	cin>>n;
	for(llo i=0;i<n;i++){
		cin>>it[i];
	}
	
	//build(0,0,n-1);
	
	vector<pair<int,int>> stack;
	
	int ll[n];
	stack.pb(mp(it[0],0));
	for(int i=1;i<n;i++){
		while(stack.size()>0){
			if(stack[stack.size()-1].first<=it[i]){
				ll[i]=stack[stack.size()-1].second;
				break;
			}
			else{
				stack.pop_back();
			}
		}
		stack.pb(mp(it[i],i));
	}
	int rr[n];
	stack.clear();
	stack.pb(mp(it[n-1],n-1));
	for(int i=n-2;i>=0;i--){
		while(stack.size()>0){
			if(stack[stack.size()-1].first<=it[i]){
				rr[i]=stack[stack.size()-1].second;
				break;
			}
			else{
				stack.pop_back();
			}
		}
		stack.pb(mp(it[i],i));
	}
	llo l[n];
	llo r[n];
	l[0]=it[0];
	llo maa=it[0];
	for(llo i=1;i<n;i++){
		if(maa>it[i]){
			l[i]=(i+1)*it[i];
		}
		else{
			llo ind=ll[i];
			l[i]=l[ind]+(i-ind)*it[i];
		}
		maa=min(maa,it[i]);
	}
 
	r[n-1]=it[n-1];
	maa=it[n-1];
	for(llo i=n-2;i>=0;i--){
		if(maa>it[i]){
			r[i]=(n-i)*(it[i]);
		}
		else{
		
			llo ind=rr[i];
			
			r[i]=r[ind]+(ind-i)*it[i];
		}
		maa=min(maa,it[i]);
	}
	
	llo ans=0;
	llo ans2=0;
	for(llo i=0;i<n;i++){
		if(l[i]+r[i]-it[i]>ans){
			ans=l[i]+r[i]-it[i];
			ans2=i;
		}
	}
	llo arr[n];
	arr[ans2]=it[ans2];
	for(llo i=ans2-1;i>=0;i--){
		arr[i]=min(it[i],arr[i+1]);
	}
	for(llo i=ans2+1;i<n;i++){
		arr[i]=min(it[i],arr[i-1]);
	}
	for(llo i=0;i<n;i++){
		cout<<arr[i]<<" ";
 
	}
	cout<<endl;
 
	return 0;
}
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