kandi X-RAY | CodeSignal Summary
kandi X-RAY | CodeSignal Summary
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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.
QUESTION
Given two strings pattern and s. The first string pattern contains only the
symbols 0 and 1, and the second string s contains only lowercase English
letters.
Let's say that pattern matches a substring s[l..r] of s if the following 3
conditions are met:
- they have equal length;
- for each 0 in pattern the corresponding letter in the substring is a vowel;
- for each 1 in pattern the corresponding letter is a consonant.
the task is to calculate the number of substrings of
sthat matchpattern.
Note: In this we define the vowels as a,e,i,o,u, and y. All other
letters are consonants.
I am not challenging anyone here, I have tried different ways but could not achieve. This question was asked in codesignal test assessment recently.
...ANSWER
Answered 2020-Jun-12 at 17:58You could take check for length first and then check the test with a regular expression for consonants against the pattern and count.
QUESTION
I've been solving some challenges at codesignal.com using C-Lisp to learn it and I've been avoiding using loops to make lisp style code.
In this challenge called alternatingSums (which gives you an int array a that can be very large and ask you to return an array/list {sumOfEvenIndexedElements, sumOfOddIndexedElements}) i have been receiving stack overflow error with this code:
...ANSWER
Answered 2020-Jun-19 at 07:32Common Lisp compilers are not required to optimize tail calls. Many do, but not all implementations compile your code by default; you have to compile the file using compile-file, or else the function individually with (compile 'alternatingsums).
CLISP contains both an interpreter, which processes the nested-list representation of Lisp source code, and a byte code compiler. The compiler supports tail recursion, whereas the interpreter doesn't:
QUESTION
I have a types of labels as 'Commercial', 'Organization', 'network', 'Information' which corresponds to domains .com, .org, .net, .info respectively. Here, for the given list of domains I am trying to return the list of labels.
Eg: If input is ["en.wiki.org", "codesignal.com", "happy.net", "code.info"]
then output should be as ["organization", "commercial", "network", "information"]
I have written the code as below in Python:
...ANSWER
Answered 2020-Apr-27 at 16:43return immediately stops the function and returns whatever you tell it to. You need to gather a list of results instead. E.g.:
QUESTION
i'm currently doing the inversePermutation challenge on codesignal. Here a brief resume of the task:
Given a permutation, produce its inverse permutation.
Example
For permutation = [1, 3, 4, 2], the output should be inversePermutation(permutation) = [1, 4, 2, 3].
In the solutions given by other users, I see this:
...ANSWER
Answered 2020-Apr-26 at 08:53From an online definition: "An inverse permutation is a permutation which you will get by inserting position of an element at the position specified by the element value in the array"
I have the feeling the input expected is a sequence of n numbers with n in range(1, n).
In your example you are including 0 and numbers exceeding list.len()
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