kandi X-RAY | Codility Summary
kandi X-RAY | Codility Summary
Top functions reviewed by kandi - BETA
- Calculate the solution of the solution
- Compares two integers .
- Returns a list of fibs that are at most once .
- Recursively traverse the tree .
- Gets gcd .
- Adds a new block to the channel .
- Checks if the wall should grow .
Codility Key Features
Codility Examples and Code Snippets
Trending Discussions on Codility
There's this demo task on Codility that I'm getting low-end score for, even though it's really simple:
Given the array
$a, which contains (non-unique) integers in range
1,000,000, find lowest integer that does not exist in array.
I've tried first using array_filter on original array to filter out only positive integers. Another idea to speed it up is only extract unique values. From there only a basic for loop that stops at
$i that's not in array.
ANSWERAnswered 2022-Apr-11 at 21:02
There are a lot of unnecessary and expensive steps (full-array iterations) in what you do.
- Instead of
array_unique, you can
array_flipfor an array with unique keys.
- If you have a large array,
in_arrayat each iteration is expensive.
- However, checking
isseton an array key is a cheap operation.
- No need to filter negative integers, or sort any of it:
- ... just loop from 1 up++ and check matches; first "no-match" is the lowest missing integer.
I'm trying to understand how the XOR operation managed to solve the problem PermMissingElem
I know one of the properties of XOR is that:...
ANSWERAnswered 2022-Mar-15 at 14:01
The XOR operation is commutative, the order of application does not matter.
Two XORs with the same value cancel each other.
So, starting from 0 and applying any sequence of XORs, followed by the same sequence that includes an extra element, only the effect of the extra element remains.
E.g. 0 ⊕2⊕3⊕1⊕5 ⊕1⊕2⊕3⊕4⊕5 has the same effect as 0⊕1⊕1⊕2⊕2⊕3⊕3⊕4⊕5⊕5, or 0⊕4.
I am trying to use recursion to solve the OddOccurrencesInArray Problem in Codility, in which
- we are given an array with N elements, N is always odd
- all of the elements of the array except for one has a total even number of occurrences
- we need to write code that returns the one unpaired value
For example, if the array given is [9, 3, 9, 3, 7, 9, 9], the code must return 7, because that is the only element in the array which is unpaired.
My solution pseudocode/thought process was:
- sort the array
- if the first two elements are equal to each other, remove them and run the solution algorithm again recursively on the array minus the first two elements (after sorting) i.e. if the unpaired element is not found, we keep reducing the size of the array
- if the first two elements are NOT equal to each other, the first element of the array must be the unpaired item
My implementation was:...
ANSWERAnswered 2022-Jan-05 at 18:47
You don't return anything from your recursive call, which means you are returning
I'm solving the MinPerimeterRectangle problem from Codility:
An integer N is given, representing the area of some rectangle.
The area of a rectangle whose sides are of length A and B is A * B, and the perimeter is 2 * (A + B).
The goal is to find the minimal perimeter of any rectangle whose area equals N. The sides of this rectangle should be only integers.
For example, given integer N = 30, rectangles of area 30 are:
(1, 30), with a perimeter of 62, (2, 15), with a perimeter of 34, (3, 10), with a perimeter of 26, (5, 6), with a perimeter of 22. Write a function:
that, given an integer N, returns the minimal perimeter of any rectangle whose area is exactly equal to N.
For example, given an integer N = 30, the function should return 22, as explained above.
Write an efficient algorithm for the following assumptions:
N is an integer within the range [1..1,000,000,000].
I initially used this code:
''' import math...
ANSWERAnswered 2022-Jan-10 at 15:17
The answer to your question is pretty simple. I assume that you are aware that the minimum of perimeter occurs when the length is equal to breadth meaning that if there was no constraint that l and b have to be natural numbers, you could have twice the square root of your area as your perimeter, but because you have the constraint, both the length and breadth should be as close to each other as possible.
ex: For an area of 30, the minimum perimeter is at (5,6) and for 20, it is at (4,5) and for 12, it is at (3,4).......so on.
Translating this to nerdy computer language would mean that you should get the pair of dimensions in square root of are number of iterations i.e O(n^0.5) which you very well implemented. Now the problem with prime numbers is that they have no other factors apart from one and themselves......meaning that even you started from the square root of the area, you have to iterate all the way up until n or all the way down until 1. For large numbers, iterating back to 1 is easier than iterating up to n (remember that this is n^0.5 and not n/2 and hence the twist). To better explain this you can think of it this way, iterating from 5 to 1 is easier than 5 to 25 and iterating from 8 to 1 is faster than 8 to 64 :).
Hence you were able to find the minimal pair sooner when you iterated down the ladder rather than up. Hope this answers your question.
I have already read through the answers on this question - Codility NailingPlanks. This is not a duplicate, as I'm trying to solve this problem using a different approach - instead of running a binary search on the planks that a given nail can cover, I'm trying to run it on the total number of nails required to cover all the planks.
This is my code:...
ANSWERAnswered 2022-Jan-02 at 15:26
The problem with your
if condition can be seen with this sample input:
I'm trying to solve the Codility FibFrog problem and I came up with the following solution:...
ANSWERAnswered 2021-Dec-31 at 18:49
Please refer to the first point in my previous answer.
If len(A) = 100000, you are calculating 100003 fibonacci numbers, while we only need fibonacci numbers which are less than 100k, which would be <30 of them.
fibonacci function is still returning
N fibonacci numbers instead of just returning fibonacci numbers which are less than
N=100k, it should be just 25 numbers instead of over 100k.
Please update your
fibonacci function to this -
I'm trying to solve the FibFrog Codility problem and I came up with the following approach:
len(A)is 0 we know we can reach the other side in one jump.
len(A) + 1is a fibonacci number, we can also reach it in one jump.
- Else, we loop through A, and for the positions we can reach, we check if we can either reach them directly from -1 using a fibonacci number
(idx + 1 in fibonaccis)or if we can reach them by first jumping to another position (
reachables) and then jumping to the current position. In either case, we also check if we can go from the current position to the end of the river - if we can, then we can return because we found the minimum number of steps required.
- Finally, if
Trueonce this loop completes, this means we can't reach any position using a Fibonacci number, so we return -1.
I'm getting 83% correctness and 0% performance with this approach.
I understand the solution is
O(n^2), assuming the array consists of only
1, the nested loop
for v in reachables: would run
n times - however I'm not sure how else I can compute this, since for each of the positions I need to check whether we can reach it from the start of the array, or from any previous positions using a fibonacci number.
ANSWERAnswered 2021-Dec-31 at 10:09
Some suggestions for improving performance of your algorithm -
len(A) = 100000, you are calculating 100003 fibonacci numbers, while we only need fibonacci numbers which are less than 100k, which would be <30 of them.
- Your solution is
O(n^4), since each
X in reachablesor
Y in fibonaccisoperation is
len(A). (and length of
Nbecause of above issue)
- Since you are doing a lot of
item in listoperations on
reachables, consider making it a
- Even with the above changes, the algorithm would be
O(N^2)because of nested looping across
reachables, so you need to come up with a better approach.
With your existing implementation, you need to traverse through all the paths and then in the end you will get the smallest number of jumps.
Instead of this approach, if you start at
0, and then keep a count of the number of jumps you have taken so far, and maintain how far(and to which numbers) you can reach after each jump then you can easily find the minimum jumps required to reach the end. (this will also save on redundant work you would have to do in case you have all
1s in A.
Recently, I'm practicing code exercises in Codility. Here you can find the problem, it is in the Exercises 6 - SQL section. Just start a test to see the problem description! SqlEventsDelta
I wrote this solution to the SqlEventDelta Question in SQLite. It works fine in local tool But, It was not working in web tool.
Can anyone give any advice on how can I solve this problem?
※ I searched this problem in Stackoverflow and I know a better code then my own way. But, If possible, I wanna use my own SQLite code logic and function....
ANSWERAnswered 2021-Oct-25 at 13:58
I tried to use a somehow naive approach. I'm aware that it is very bad for performance due to many subqueries but the catch here is the "DISTINCT ON" of PostgreSQL, however I got 100% 😃
Hope you like it!
I've written the below algorithm as a solution to Codility Flags. This passes the correctness checks, however it times out on most of the performance checks.
The complexity of this should be
m is the number of peaks in
n is the length of
A. However, the
while potentialK > maxFlags loop should only execute until a suitable number of flags is found which satisfies the criteria. I'm not sure how to further optimise this for time complexity.
ANSWERAnswered 2021-Dec-28 at 23:28
I managed to optimize it as follows:
Since the distance between the individual flags has to be >= the number of flags, we know the maximum number of flags will be the root of the last element of peaks - the first element of peaks:
sqrt(peaks[-1] - peaks)
I was then able to update the initial value of potentialK to
I'm practicing on Codility. There is an Easy Level question, yet I'm stuck on performance. The test result analysis marks this code as O(N**2), but obviously there are not any nested loops. Can anyone tell me why my code is O(N**2)?...
ANSWERAnswered 2021-Dec-17 at 16:37
The other loop is hidden in the remove method of
ArrayList. The remove method of
ArrayList is O(N) because it has to shift the elements to fill the gap.
No vulnerabilities reported
You can use Codility like any standard Java library. Please include the the jar files in your classpath. You can also use any IDE and you can run and debug the Codility component as you would do with any other Java program. Best practice is to use a build tool that supports dependency management such as Maven or Gradle. For Maven installation, please refer maven.apache.org. For Gradle installation, please refer gradle.org .
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