[Community Puzzle] Rectangle Partition


#42

1

This no longer looks like an easy puzzle. :thinking:

Рекурсия наше усё …


#43

It is harder than some other “easy” ones but as reflected by peer coders, it seems to be a good challenge to encourage them think and try out different algorithms to enhance performance, and finally helped enhancing programming skill. Review the many tips in this thread for assistance if you may need it.


#44

I invite everybody who is struggling to pass the high density tests to really spend time understanding what is the quickest way to compare values in a loop. I am quite new to Python and only used very common functions but it worked!


#45

I had an interesting case where I passed all the tests, but failed on submission item #3. I was using C++; changing some assignment items to do bounds checking (e.g. using vec.at(x) rather than vec[x] showed problems in the given test items.

(And yes, I was accessing memory out of bounds – the height can be bigger than the width. I was a little surprised it didn’t fail more issues).

Fixing the bounds issue solved the problem, although I also avoided it entirely moving to a map – I just find the syntax around maps for C++ (e.g. incrementing a value in one) so tedious I had avoided it in my first pass.


#46

Greate puzzle. With this hint and a variation that does not uses list i was able to pass every test, thanks.


#47
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Contributed by @Niako. This is a “Super Hi-density” testcase. This test might be too harsh for some already submitted solutions. Challenge yourself to improve with this optional test.


#48

Thank you. I did this with Groovy. And got 42042751 with green light :slight_smile:


#50

42042751 is indeed the right answer to that case (just confirming as it wasn’t mentioned).


#51

I also did this “Super Hi-density” with Python3. I timed my solver in GC’s own IDE and got solution in 0.0527 seconds.


#52

Hm. I think this puzzle should be listed as Medium difficulty. I may of course have overlooked something, but my first, straightforward approach took too long, and I needed to apply some brain and experience to find a faster solution. And still get only 77% on the new solution, presumably because it is a bit more complex (harder to debug).

Update: I did solve it, though. And there were even simpler and faster solutions than mine. But I still think it should perhaps be Medium. I struggled more with it (and ended up with longer code) than with “The Resistance”, which is supposed to be Very hard.


#53

I don’t understand … I pass all test excepted the additional validator 3 …
More information about this validator ? :frowning:


#54

I agree that it should be listed under Medium difficulty. It took me a while to come up with a solution for the first testcases, but than I got stuck on testcase 7, 8 and 9. I had to rethink my solution and worry about performance. Good practice though and at the end solved it, but it was difficult.


#55

Hi, I’m having problem solving hi density.
What I did was getting all the possible edges of each height and width and store it in array

Sample 1:
widthLength = [2,3,5,8,5,10]
heightLength = [3, 2, 5]
and compare those two array. (2x2, 3x3, 5x5,5x5) = 4.

My problem I think is that my algorithm for storing the lengths is not optimize. Can you suggest an optimize way to do that?


#56

Really unfair :smiley:, I coded the same code from C++ to Python 3 and I passed Hi Density. That’s why most people here said they solved it using Python.


#57

Continuing the discussion from [Community Puzzle] Rectangle Partition:

What did you finally do to solve this. After reading the comments made, i realise i have the same code but it doesn’t work except the 3 first tests.

here is my code :

import java.util.;
import java.io.
;
import java.math.*;

/**

  • Auto-generated code below aims at helping you parse

  • the standard input according to the problem statement.
    **/
    class Solution {

    public static void main(String args[]) {

     Scanner in = new Scanner(System.in);
     int w = in.nextInt();
     int h = in.nextInt();
     int countX = in.nextInt();
     int countY = in.nextInt();
     
     List<Integer> listeX = new ArrayList<Integer>();
     List<Integer> listeY = new ArrayList<Integer>();
    
     
     for (int i = 0; i < countX; i++) {
         int x = in.nextInt();
         listeX.add(x);
     }
         
     for (int i = 0; i < countY; i++) {
         int y = in.nextInt();
         listeY.add(y);
      }
      
     listeX.add(w);
     listeY.add(h);
    
    
     for(int i = 1; i <= countX ; i++){
         for(int j = 0; j < i; j++){
                 listeX.add(listeX.get(i) - listeX.get(j));
         }
     }
    
     
     for(int i = 1; i <= countY; i++){
         for(int j = 0; j < i; j++){
                 listeY.add(listeY.get(i) - listeY.get(j));
         }
     }
     
     
    
     int nbCarres = 0;
     
     for(int i = 0; i < listeX.size(); i++){
         for(int j = 0; j < listeY.size(); j++){
             if(listeX.get(i) == listeY.get(j)){
                 nbCarres = nbCarres +1;
             }
         }
     }
    
     // Write an action using System.out.println()
     // To debug: System.err.println("Debug messages...");
    
     System.out.println(nbCarres);
    

    }
    }


#58

Your solution is way too slow, you have too many loops to count the squares.
Hints for faster solutions have been given earlier in this thread, you should try and read it.


#59

The expected complexity is O(n^2) where n = max(number of x-axis measurements, number of y-axis measurements). However, the test cases don’t cover the extreme case where there are 500 measurements along each axis, allowing O(n^4) solutions to pass.


#60

Hello guys,
I have nearly completed this puzzles, but i can’t find out why i cant’ pass 3 tests out of 9 (Bigger-2, Hi-density 1, Imbalance). It makes me crazy !:exploding_head: Here is my Javascript code :

var inputs = readline().split(" ");

const w = parseInt(inputs[0]);
const h = parseInt(inputs[1]);
const countX = parseInt(inputs[2]);
const countY = parseInt(inputs[3]);
const wSlices = [...readline().split(" "), w].map(el => +el);
const hSlices = [...readline().split(" "), h].map(el => +el);
let nombreDeCarres = 0;

let wCombination = [...wSlices];
let hCombination = [...hSlices];

calculateCombination(countX, wSlices, wCombination);
calculateCombination(countX, hSlices, hCombination);

function calculateCombination(count, slices, resultArray) {
  for (let i = 0; i < count; i++) {
    const nb = slices.shift();
    slices.forEach(el => {
      resultArray.push(el - nb);
    });
  }
}

for(let i=0;i<wCombination.length;i++){
    for(let j=0;j<hCombination.length;j++){
        if(hCombination[j]===wCombination[i])nombreDeCarres++
    }
}

console.log(nombreDeCarres);

If anyone could give me some help, i would appreciate, it’s so frustrating ^^

Thanks !


#61

This is a very challenging problem. I agree it should be in medium difficulty. I made a first solution very quickly but then I realize it’s so slow for the Hi-density 1 and Hi-density 2 test cases. After that I tried to optimize it but I continue to time out. I’m using c++ to solve it. The algorithm I use it’s very simple store in a vector all the possible heights and all the possible widths for the sub-rectantes and then compare the 2 vectors and count how many are equals.I also has an extra vector to save the count of each possible width or height value because there can be repeated. I’m still thinking how can I improve the performance…


#62

Read some earlier posts in this thread. There are some tips answering this exact question.