### Loser_'s blog

By Loser_, 20 months ago, 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

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2.Apartments

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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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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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idea Comments (49)
 » Thank You so much
•  » » Welcome
 » thanks
•  » » welcome
 » 19 months ago, # | ← Rev. 2 →   .
 » Thanks a lot,man
 » how is this working in 25th problem sliding costabs(p-a[i+m])-abs(P-a[i]) and if m is even we decrease the extra value p-Pcan someone explain why the common elements in the the windows are not considered for new cost ? Please explain the math behind it
•  » » 18 months ago, # ^ | ← Rev. 2 →   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.
•  » » » 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 ?
•  » » » 15 months ago, # ^ | ← Rev. 2 →   Yes a mathematical proof would be helpful.
 » Thanks a lot!
 » I guess for question number 10 sliding window concept will be much more intuitive. Here is my ac solution, #include using namespace std; int main() { int n; cin>>n; vectork(n); for(int i=0;i>k[i]; setst; int ans = 0,j=0; for(int i=0;i
•  » » 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?
•  » » » 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.
 » 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
•  » » 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))<
 » There are 35 problems in this section .
•  » » They are newly added problems which I didn't solve entirely. If someone wants to contribute happy to add them all
 » 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?
•  » » 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.
 » 15 months ago, # | ← Rev. 2 →   Loser_ I tried everthing but test case 5 in ques 6 (Movie Festival) is giving run time errorLINK TO MY CODEi have given link to my code. Can anyone help?
•  » » comparator should return false in equal case. Else it gives error. in sortbysecinc function (return a.second
•  » » » Thank you so much for the reply. it worked.
 » 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.
 » Can someone explain the proof for 15-> Tasks and Deadlines ?
•  » » Let's take the testcase given in the problem and write out all possible combinations: (10-6) + (12-11) + (15-19) => reward = 1 (10-6) + (15-14) + (12-19) => reward = -2 (15-8) + (10-14) + (12-19) => reward = -4 (15-8) + (12-13) + (10-19) => reward = -3 (12-5) + (10-11) + (15-19) => reward = 2 (12-5) + (15-13) + (10-19) => reward = 0 Observations: We can observe that the deadlines are always positive while calculating the solution. The other term(consisting of the duration) is the sum of individual durations. For example, the case with reward = 1 can be written as (10-6) + (12-(6+5)) + (15-(6+5+8)). Notice that the i-th duration we choose, gets repeated in calculating all the further terms. So it makes sense to choose the durations sorted in a non-decreasing order.
 » 14 months ago, # | ← Rev. 3 →   Can anyone please provide me a Multiset solution for the Apartments question?
•  » » Maybe it's too late but here it is: https://cses.fi/paste/74d273db0c74039e225cdc/
•  » » » Thanks a lot, I appreciate your effort.
 » Thank You so much. Really helpful
 » missing coin sum gfg linkhow it come in O(n) ? start with answer = 1 while(check next sum are possible or not)code : in c++ cin >> n; int count = 0; vi arr = i_vector(n); // input given array sortall(arr); int res = 1 ; for( int i = 0 ; i < n && arr[i] <= res ; i++ ) { res += arr[i ] ; } cout << res ;
 » for the Playlist question why it is getting AC for Map and TLE for Unordered_map.
•  » » Bcs in worst case scenario map perform better than unordered_map .why this happens?. For this you need to know internal working of map.
•  » » » Oh, now I get it. Thanks-
 » 10 months ago, # | ← Rev. 2 →   $O(n^2 \log n)$ solution for Sum of Four Values problem. Spoiler#include using namespace std; typedef long long ll; typedef vector vint; typedef vector vvint; typedef vector vbool; typedef vector vvbool; typedef pair pll; #define ff first #define ss second ll n, x; vint a; vector> twoSums; int main(){ cin >> n >> x; a.resize(n); for(int i = 0; i < n; i++){ cin >> a[i]; } for(int i = 0; i < n; i++){ for(int j = i + 1; j < n; j++) twoSums.push_back({a[i] + a[j], {i, j}}); } sort(twoSums.begin(), twoSums.end()); int l = 0, r = twoSums.size() - 1; while(l < r){ if(twoSums[l].ff + twoSums[r].ff < x){ l++; } else if(twoSums[l].ff + twoSums[r].ff > x){ r--; } else{ ll a = twoSums[l].ss.ff, b = twoSums[l].ss.ss, c = twoSums[r].ss.ff, d = twoSums[r].ss.ss; if(a == c or a == d or b == c or b == d){ if(l < r - 1 and twoSums[r - 1].ff == twoSums[r].ff) r--; else l++; } else{ cout << a + 1 << " " << b + 1 << " " << c + 1 << " " << d + 1; return 0; } } } cout << "IMPOSSIBLE"; return 0; } Explanation: We first find the sums which can be formed by two values in $x$ and store the sums in $twoSums$, in $O(n^2)$ time. Now, the problem boils down to "find if $x$ can be attainable using two values of $twoSums$ (note that these values must contain distinct indices)". So, we first sort the $twoSums$ array, which takes $O(n^2 \log n)$ time. And then, we run the two-pointer algo (takes $O(n^2)$ time) on $twoSums$ to check if $x$ is attainable using two values of $twoSums$.
 » 8 months ago, # | ← Rev. 3 →   I am having WA on Subarray Sums I. Could anyone please tell me what am I missing here?Code Link Thanks in advance
•  » » 8 months ago, # ^ | ← Rev. 2 →   Use long long for the currSum variable. The sum of the array values can go beyond INT_MAX.
•  » » » Thanks Brother ...
 » Your solution of Subarray sum is very nice. I was using sliding window but your solution works for negative numbers too!
 » 6 months ago, # | ← Rev. 2 →   What exactly you did for creating ordered_multiset in 24th(Sliding Median) ? I created an ordered_set of pair {key,val} & stored some random number(but distinct) as val. Is there any better method to create ordered_multiset ? Btw, thanx for this blog!
•  » » You need to change less to less_euqal Codetemplate using Multiset = tree, rb_tree_tag, tree_order_statistics_node_update>; 
 » Thanks for the post. orz. Meme •  » »
 » 25th Sliding cost seems wrong, I am getting wrong O/P with your code for case: 10 5 4 6 10 4 3 8 8 10 8 9 Correct O/P: 9 11 11 11 7 3 Your O/P: 11 9 9 9 5 -1
 » THank YOu SO MUch...
 » How 18 — "Sum of 4 values" is working?, cause I think its complexity is n^3, and according to constraints that should not work. Can anyone explain?
•  » » For the O(N^2) solution ,you can refer this Sum Of 4 Values