interview-prep | Data Structures / Algorithms for Coding Interviews | Dataset library

I'm doing this question on hackerrank: https://www.hackerrank.com/challenges/ctci-bubble-sort/problem?h_l=interview&playlist_slugs%5B%5D=interview-preparation-kit&playlist_slugs%5B%5D=sorting

I wrote the solution in intellij, and it gives me the correct output there, but when I copied it over to the hackerrank ide, it gave me an error.

This is the code I'm talking about:

I have the following solution:

I am trying to practise a question on hackerrank. Problem link

I am trying to read the variable q

Given a numeric vector, I'd like to find the smallest absolute difference in combinations of size 2. However, the point of friction comes with the use of combn to create the matrix holding the pairs. How would one handle issues when a matrix/vector is too large?

When the number of resulting pairs (number of columns) using combn is too large, I get the following error:

Error in matrix(r, nrow = len.r, ncol = count) : invalid 'ncol' value (too large or NA)

This post states that the size limit of a matrix is roughly one billion rows and two columns.

Here is the code I've used. Apologies for the use of cat in my function output -- I'm solving the Minimum Absolute Difference in an Array Greedy Algorithm problem in HackerRank and R outputs are only counted as correct if they're given using cat:

Can someone explain what is wrong with my code? I am getting "Abort called" in 7 test cases. Rest are successful.

I have a 2d dp array with of size: n+1 x m+1 were n and m are sizes of a and b respectively. So, the row represent string a and columns represent string b.

First, I set dp[0][0] to 1 since it is possible to turn empty string into empty.

So, initially, i am checking if I can turn any substring of a into the empty string (in the first single for-loop). This is true for all substrings of a without any capital letters. As soon as there is a capital letter, the rest of the substrings cannot be converted.

Then (in the next double for-loop), I am examining all the cases.

Case 1: a[i-1] == b[i-1] -> if both the letter are the exact same, then dp[i][j] = dp[i-1][j-1]

Case 2: a[i-1] is lower case (this has 2 sub cases):

Case 2.1: a[i-1] and b[j-1] are the same letter (but not the same case) -> then we can either change a[i-1] or delete it . So: dp[i][j] = dp[i-1][j-1] || dp[i-1][j].
Case 2.2: a[i-1] and b[j-1] are not the same -> in this case, we can only delete a[i-1] since it is lower case . So: dp[i][j] = dp[i-1][j]

Link to problem: https://www.hackerrank.com/challenges/abbr/problem?h_l=interview&playlist_slugs%5B%5D=interview-preparation-kit&playlist_slugs%5B%5D=dynamic-programming

P.S. The main logic of the program is just inside the abbreviation() function.

Code (EDITTED):

So I've been solving this HackerRank problem and I can't really see why my answer isn't working. I've now seen other people's answers and they make sense but it's a completely different approach and I really want to understand why mine doesn't work. It passes two test cases but not the third one. If you all think it's just mathematically or logically off and can explain why that would be awesome. Thanks!

This is the link to the question: https://www.hackerrank.com/challenges/new-year-chaos/problem?h_l=interview&playlist_slugs%5B%5D=interview-preparation-kit&playlist_slugs%5B%5D=arrays

This is my code:

public class Solution {

I am solving this hackerrank problem for counting the number of 'a's in a given string.

My solution was to store the string in a pattern variable. While the length of the pattern is less than n, it will just add the string to itself. Then I would loop over the pattern and add the number of 'a's in the string.

This solution works fine when n < 1000000. But add one more 0 and when n = 10000000, I get a RangeError for my string in hackerrank because it's too damn long.

Is there a way to get around this RangeError problem? I know there are other ways to solve this problem, but I just want to know how I could edit my code to make it pass the hackerrank test.

We have a method in which an array and number of rotations are passed as input and we have to return the array after left rotations. Here is my solution.

Question Desc: Sherlock considers a string to be valid if all characters of the string appear the same number of times. It is also valid if he can remove just 1 character at 1 index in the string, and the remaining characters will occur the same number of times. Given a string s, determine if it is valid. If so, return YES, otherwise return NO.