Loser_'s blog

By Loser_, 7 months ago, In English

I just completed CSES Sorting and Searching section problems. And I didn't find any editorials for this. So I am writing my own approach for the problems.Please do correct me if I made any mistakes and do share your approach for the problems.

My solutions are here

complete editorial for CSES coding platform of all 27 problems in Sorting and Searching section.

1.Distinct Numbers

Editorial
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2.Apartments

Editorial
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3.Ferris Wheel

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4.Concert Tickets

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5.Restaurant Customers

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6.Movie Festival

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7.Sum of Two Values

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8.Maximum Subarray Sum

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9.Stick Lengths

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10.Playlist

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11.Towers

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12.Traffic Lights

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13.Room Allocation

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14.Factory Machines

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15.Tasks and Deadlines

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16.Reading Books

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17.Sum of Three Values

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18.Sum of Four Values

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19.Nearest Smaller Values

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20.Subarray Sums I

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21.Subarray Sums II

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22.Subarray Divisibility

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23.Array Division

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24.Sliding Median

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25.Sliding Cost

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26.Movie Festival II

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27.Maximum Subarray Sum II

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7 months ago, # |
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Thank You so much

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7 months ago, # |
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thanks

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7 months ago, # |
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.

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7 months ago, # |
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Thanks a lot,man

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6 months ago, # |
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too helpful, thanks

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6 months ago, # |
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how is this working in 25th problem sliding cost

abs(p-a[i+m])-abs(P-a[i]) and if m is even we decrease the extra value p-P

can someone explain why the common elements in the the windows are not considered for new cost ? Please explain the math behind it

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    6 months ago, # ^ |
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    In this problem,everytime we check for a window of k elements.

    First we check for first k elements and store it's mid value in P(capital). Then with each iteration we erase the first value from window and add the next value from the array.

    See,if the initial set is 2 3 4 after iteration it's 3 4 5.

    After the iteration mid value is p(small).Now after the new value added to set the cost is abs(p-a[i+m]) and the previous cost is abs(P-a[i]).So the change of total cost d is simply the difference between these two.

    Now when it's come to even value ,Like this one-

    8 4
    2 4 3 5 8 1 2 1
    

    Here the previous window is 2 3 4 5 and after 1st iteration 3 4 8 5. P(capital)=3, p(small)=4; Unlike the odd k ,even k has two mid value. So either 1st mid value gives you the min cost or the 2nd mid value. If you notice then P(capital) is 1st mid value, p(small) is 2nd mid value. That's the reason we simply erase the extra value(it's either add to total cost or decreases).

    I hope it makes sense.

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      6 months ago, # ^ |
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      If you notice then P(capital) is 1st mid value, p(small) is 2nd mid value this statement is not always true.

      You just explained the code, I want some kind of mathematical proof of why this works ? Why changing median doesn't effect cumulative cost of the common elements is the windows ?

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      2 months ago, # ^ |
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      Yes a mathematical proof would be helpful.

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6 months ago, # |
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Thanks a lot!

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5 months ago, # |
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I guess for question number 10 sliding window concept will be much more intuitive. Here is my ac solution,

#include<bits/stdc++.h>
using namespace std;
 
int main()
{
	int n;
	cin>>n;
	vector<int>k(n);
	for(int i=0;i<n;i++)
	cin>>k[i];
	set<int>st;
	int ans = 0,j=0;
	for(int i=0;i<n;i++)
	{
		if(st.find(k[i])!=st.end())
		{
			while(!st.empty() && k[j]!=k[i])
			{
			  st.erase(k[j]);
			  j++;	
		    }
		    st.erase(k[j]);
		    j++;
		}
		st.insert(k[i]);
		ans = max(ans,i-j+1);
	}
	cout<<ans;
	return 0;
}

Can you share intuition of updating j variable in your solution?

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    3 months ago, # ^ |
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    Hi, can you please share with me the expected time complexity of your code and how exactly could you determine if your code could pass the given test cases? It seems to me that complexity would be O(n^2) since there is a while loop inside the outer for loop?

    Yet your code seems to pass all the test cases within the accepted time limit, HOW?

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      2 months ago, # ^ |
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      the time complexity is $$$\mathcal O(N\log N)$$$ since it uses sliding window and a set. Sliding window uses two loops and takes $$$\mathcal O(N)$$$; notice that $$$j$$$ does not reset to $$$0$$$ and retains it's value every time $$$i$$$ increases.

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3 months ago, # |
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For 10. Playlist I used two-pointer technique with sliding window concept, It is much more intuitive.

Keep on increasing window length, while we have subarray a with unique numbers. As soon as we find a duplicate, we delete from the array till the subarray does not contain duplicate

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3 months ago, # |
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Can someone please help.me to solve Josephus problem 2

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    3 months ago, # ^ |
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    Main concept is — erase every $$$k$$$th element from the circular array(in I,$$$k=1$$$).Remember erase function is $$$O(n)$$$ so it definitely causes TLE.Use ordered set instead whose erase function complexity is $$$log(n)$$$.

            int n,k;cin>>n>>k;
    	ordered_set josep; 
    	for(int i=1;i<=n;++i)josep.insert(i);
    	int pos=0;
    	while(josep.size()>1)
    	{
    		pos=(pos+k)%(int)josep.size();
    		cout<<*(josep.find_by_order(pos))<<' ';
    		josep.erase(*(josep.find_by_order(pos)));
    		pos%=(int)josep.size();
    	}
    	cout<<*(josep.find_by_order(0))<<endl;
    
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3 months ago, # |
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There are 35 problems in this section .

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    3 months ago, # ^ |
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    They are newly added problems which I didn't solve entirely. If someone wants to contribute happy to add them all

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2 months ago, # |
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In problem no 6 why it is necessary to sort elements based on ending time. Why can't we sort elements based on starting time?

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    2 months ago, # ^ |
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    suppose there is a case 1 20 2 3 3 4 4 5 in this case when you select the movie that start early you will not be able to select another movie therefore it is better to sort movie that finish early so that we can select another movie if available.

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2 months ago, # |
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Loser_ I tried everthing but test case 5 in ques 6 (Movie Festival) is giving run time error

LINK TO MY CODE

i have given link to my code. Can anyone help?

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    2 months ago, # ^ |
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    comparator should return false in equal case. Else it gives error. in sortbysecinc function (return a.second<b.second); use this.

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2 months ago, # |
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Um.. Converting code to English words can't be called an editorial, but good job. Atleast people have some reference point where they can look for solutions.

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5 weeks ago, # |
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Can someone explain the proof for 15-> Tasks and Deadlines ?

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5 weeks ago, # |
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Can anyone please provide me a Multiset solution for the Apartments question?

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2 weeks ago, # |
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Thank You so much. Really helpful