codesignal | My solutions to codeSignals problems in Python | SQL Database library
kandi X-RAY | codesignal Summary
kandi X-RAY | codesignal Summary
My solutions to codeSignals problems in Python.
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Top functions reviewed by kandi - BETA
- Reverse in parentheses .
- Return the encoding of a string .
- Returns True if a sequence is almost increasing .
- Returns the number of votes that voted in the vote .
- Count the common character between two strings .
- Checks if inputString is aBeautiful string .
- Blurb a box .
- Line up a string .
- Return the number of squares in a matrix .
- Return a list of spiral numbers .
codesignal Key Features
codesignal Examples and Code Snippets
Community Discussions
Trending Discussions on codesignal
QUESTION
A sequence (e.g. c(1,2,3,4)) is almost increasing when we can remove exactly one element from the sequence and get a strictly increasing sequence (i.e. a0 < a1 < ... < an). I'm trying to find a way to check whether a sequence is almost increasing. If it is, I want to return TRUE; if it isn't I want to output FALSE. I've got this far:
...ANSWER
Answered 2022-Mar-21 at 13:37Here is a solution which works for me. The idea is that diff(x) has negative elements for every downwards step in x. For example, min(diff(x)) is positive, if x is strictly increasing. If diff(x)[i] <= 0 for exactly one index i, we have to check whether either removing x[i] or removing x[i+1] makes the sequence strictly increasing. The following function passed all tests I tried:
QUESTION
I have a problem with the following task from the platform Codesignal:
After some investigation, you've created a database containing a foreignCompetitors table, which has the following structure:
competitor: the name of the competitor;
country: the country in which the competitor is operating.
In your report, you need to include the number of competitors per country and an additional row at the bottom that contains a summary: ("Total:", total_number_of_competitors)
Given the foreignCompetitors table, compose the resulting table with two columns: country and competitors. The first column should contain the country name, and the second column should contain the number of competitors in this country. The table should be sorted by the country names in ascending order. In addition, it should have an extra row at the bottom with the summary, as described above.
Example
For the following table foreignCompetitors
my solution:
...ANSWER
Answered 2022-Mar-09 at 08:23You want a GROUP BY WITH ROLLUP here:
QUESTION
I couldn't figure out why I passed 22/23 on this challenge and not able to solve the last test case since it was hidden.. The feedback from the CodeSignal is
Tests passed: 22/23. Execution time limit exceeded: Program exceeded the execution time limit. Make sure that it completes execution in a few seconds for any possible input.
Challenge Given an array a that contains only numbers in the range from 1 to a.length, find the first duplicate number for which the second occurrence has the minimal index. In other words, if there are more than 1 duplicated numbers, return the number for which the second occurrence has a smaller index than the second occurrence of the other number does. If there are no such elements, return -1.
Example
For a = [2, 1, 3, 5, 3, 2], the output should be solution(a) = 3.
There are 2 duplicates: numbers 2 and 3. The second occurrence of 3 has a smaller index than the second occurrence of 2 does, so the answer is 3.
For a = [2, 2], the output should be solution(a) = 2; For a = [2, 4, 3, 5, 1], the output should be solution(a) = -1.
Input/Output
[execution time limit] 4 seconds (js)
[input] array.integer a
Guaranteed constraints: 1 ≤ a.length ≤ 105, 1 ≤ a[i] ≤ a.length.
[output] integer
The element in a that occurs in the array more than once and has the minimal index for its second occurrence. If there are no such elements, return -1.
My code
...ANSWER
Answered 2021-Dec-05 at 01:19In bad cases, you're iterating over the whole array for every element in it - O(n ^ 2). The inner while(pointer < a.length) results in the argument taking too much time.
Instead, make a Set of elements found so far, and return when the first duplicate element is found (which will be the minimal second index).
QUESTION
I am trying to call/print the following linked list:
...ANSWER
Answered 2021-Jun-23 at 15:09Looking at the comments you made, there are some misunderstandings to resolve here:
How to create a listIt seems you expected one of these to create a list:
QUESTION
I'm asked to check whether a given string can be a palindrome after rearranging and then return true if it can be a palindrome or false if it cannot be a palindrome.
I'm getting a runtime error: Segmentation fault while running tests.
Here's my code:
...ANSWER
Answered 2021-Jun-08 at 19:38"Why am I getting “runtime error: Segmentation fault” for random test cases?"
Aside from your function not being logically viable for testing palindromes, it will fail on random occasions due to undefined behavior:
QUESTION
so Im working on a codesignal problem, and these are the instructions:
...ANSWER
Answered 2021-May-20 at 03:40Add a break statement inside the if condition.
QUESTION
I'm having this issue where my .map won't update any of my data when it rerenders. I have a simple player that is moving to the next element in the playlist array and after its updated or every time I click next or prev in my Playerbar it supposed to show Artist of the selected song {artist.name} and the Selected song {song.name} but for some reason nothing changes. I think I have missed something obvious but can't see it Can someone help me out please? Thanks.
ANSWER
Answered 2021-Mar-09 at 13:38const PlayerBar = (props) => {
const [playing, setPlaying] = React.useState(false);
function clickPlayButton() {
setPlaying(!playing)
}
function renderPlayButton() {
let buttonUrl = '';
if (playing) {
buttonUrl = 'https://codesignal.s3.amazonaws.com/uploads/1557137524244/rounded-pause-button.svg';
} else {
buttonUrl = 'https://codesignal.s3.amazonaws.com/uploads/1557136695174/play-right-arrow-circular-button.svg';
}
return (
clickPlayButton()}
/>
);
}
function renderNextButton() {
let buttonUrl = 'https://codesignal.s3.amazonaws.com/uploads/1557137539567/next-button.svg';
return (
props.nextButton()}
/>
);
};
function renderPrevButton() {
let buttonUrl = 'https://codesignal.s3.amazonaws.com/uploads/1557138446191/previous-button.svg';
return (
props.prevButton()}
/>
);
};
return (
{renderPlayButton()}
{renderPrevButton()}
{renderNextButton()}
{props.listArtist} // here ...
{props.listName} // here ...
);
}
const App = () => {
const playlist = [
{
id: 1,
name: 'Yesterday',
artist: 'Beatles'
},
{
id: 2,
name: 'Nothing else matters',
artist: 'Metallica'
},
{
id: 3,
name: 'Always',
artist: 'Bon Jovi'
},
{
id: 4,
name: 'Waka Waka',
artist: 'Shakira'
}
];
const [curItemIndex, setCurItemIndex] = React.useState(0);
const [list, setPlaylist] = React.useState(playlist)
function getSongClass(index) {
let className = 'list-group-song song row';
if (index === curItemIndex) {
className += ' selected';
}
return className;
}
function renderItems() {
return list.map((song, index) => {
return (
{song.artist}
{song.name}
);
});
}
function clickNextButton() {
if(curItemIndex === list.length - 1) // here ...
return;
setCurItemIndex(curItemIndex => curItemIndex + 1)
}
function clickPrevButton() {
if(curItemIndex === 0)
return;
setCurItemIndex(curItemIndex => curItemIndex - 1)
}
return (
{renderItems()}
);
}
ReactDOM.render(
,
document.getElementById('app')
);
QUESTION
I am working through some problems on CodeSignal. I came across this one, arrayMaxConsecutiveSum. I got it to pass almost all tests, but it is timing out on the last one. If I move the test into custom tests, it passes there, so I'm not sure what to do. How do I code it better so that it doesn't time out?
Problem:
...Given array of integers, find the maximal possible sum of some of its k consecutive elements.
Example:
For inputArray = [2, 3, 5, 1, 6] and k = 2, the output should be arrayMaxConsecutiveSum(inputArray, k) = 8. All possible sums of 2 consecutive elements are:
2 + 3 = 5;
3 + 5 = 8;
5 + 1 = 6;
1 + 6 = 7.
Thus, the answer is 8.
ANSWER
Answered 2021-Feb-15 at 03:18Your current algorithm is O(n ^ 2) because it requires a nested loop.
You can make it O(n) by using a rolling sum instead. Start with the sum of elements 0 to k, then on each iteration, subtract the earliest element that makes up the sum and add the next element not included in the sum yet.
For example, with a k of 2:
- start out with the sum of elements [0] and [1]
- subtract [0], add [2], compare the new sum
- subtract [1], add [3], compare the new sum
and so on.
QUESTION
EDIT: So I changed my code to the following:
...ANSWER
Answered 2020-Nov-09 at 15:20You are given two arrays of integers a and b of the same length. The length is the same so we need to iterate only once improving it from O(n^2) to O(n). You still need to check every element, so that is the best possible complexity for this problem.
The if statement checking for duplicates is as unneeded as the variable pairs.
You can use a Set that will check for duplicates and in the end, return its length instead of counting the pairs manually.
I'm attaching the examplary solution below:
QUESTION
I took a practice CodeSignal exam and was able to pass 14/16 test cases for this problem. You are given a vector as input (list of ints) and the solution will be long long.
Originally I simply used a brute-force solution of two for loops and adding the current a[i] concat a[j] to a running total. However, I tried to optimize this by using memoization. I used a unordered_map of pairs to check if I already computed the (i,j) pair and if so, simply return the cached result. Even with my optimization, I still don't pass any additional test cases and receive a 14/16 result. What insight or optimizations am I missing?
I have found similar online problems, however their insight doesn't seem to be applicable to this specific problem.
Ex: Similar Problem
Question:
Given an array of positive integers a, your task is to calculate the sum of every possible concat(a[i], a[j]), where concat(a[i],a[j]) is the concatenation of the string representations of a[I] and a[j] respectively.
Ex:
...ANSWER
Answered 2020-Jun-15 at 21:26Let's calculate the contribution a_i integer to answer in all pairs. There are two cases. The first case when number a_i is low part. When total sum is n * a_i to answer (n is total number integers). The second case is high part. Then let's find all offsets in decimal notation. Denote by k_j as total number integers length j (length in decimal notation). Then high part add to answer k_j * a_i * 10^j for all value j (1 <= j <= 7). Knowing k_j we can calculate the answer for all numbers a_i in linear time.
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