Triplet sum divisor hackerearth solution
Please Login in order to post a comment. Can you please explain the logic here of how did you deduce the total number of pairs modding to k?
The idea is that you separate elements into buckets depending on their mod k. Because you don't need to print the actual pairs, you need to store only the number of elements in each list.
It is an optimal solution, but keep in mind: Reading the input sizes is really important in a competition. Small input size: You don't have to think that hard and you can implementend a simpler solution, which needs less time to write the code for and debug.
Yes, but this second solution is a better learning tool for modular arithmetics. The other solution is trivial. There is always a tradeoff between space and time complexity. Otherwise we might be in a better world!! The space complexity can be reduced to O min n,k if a hashtable is used instead of an array.
So space complexity and time complexity both scale linearly. But in the problem it is given as we have to print the number of i,j pairs and the given condition is i. Thanks usinha02 for the algo, and petreeftime for the explanation.
Here's an implementation in Python. Some bits are a little clunky. Feedback Please. So it will always be 1 value bigger than i. Is this possible in O n time. Where are you comparing the values of ar[i] and ar[j], which is the necessary condition in the given problem statement.
Cause the second number has to be at a later position, there's never a pair for the last position. Thank man!. I've just spent half an hour looking through this and the parent comments above which are given some credit for the idea - it's a really clever solution to the problem, and if you read petreeftime's description above, it covers it pretty well. I found a key to understanding it was to alter the code so that you see the contents of all the relevant variables at each step, and then it became clear how it was working I'm using Python 3.
The first step works out the mod value of the current array element - that's really the only thing that matters - a 3 and a 6 will have the same effect as a 3 and a 3 for a k of 3. The third step adds the current array element to the nums list, which will allow that to be counted towards future possibilities in other steps. Great - just spent half an hour trying to work out how this worked, and now I have I feel like I've achieved something today!
Thanks for the code. Thanks for the explanation petreeftime! Thank you usinha02 for the elegant solution, and thank you petreeftime for an awesome explanation! Although I understood most of it, and this maybe most naive doubt that I have. I wasn't able to understand the following condition:.Q1 FizzBuzz Write a program that prints the numbers in the given range.
Print a new line after each string or number. Next line will have T integers, denoted by N. But follow the rules given in the problem statement. N is an integer. Please read the below instructions carefully. Q2 Sum of Two Numbers. You are given two numbers a and b. Your task is to find the sum of two numbers. Input The first line contains an integer t, denoting the number of test cases.
Next t lines contain two integers, a and b separated by a space. Output For every number a and b, output the sum of two numbers a and b. HackerEarth Questions Asked on 24th June Q Monk and the Magical Candy Bags Our Monk loves candy!
He picks up a bag, eats all the candies in it and drops it on the ground. But as soon as he drops the bag, the number of candies in the bag increases magically! Amazed by the magical spell, Monk can now have a lot more candies! Output : Print the answer to each test case in a new line.
Input Constraint. Each digit is made of a certain number of dashes, as shown in the image below. For instance 1 is made of 2 dashes, 8 is made of 7 dashes and so on. You have to write a function that takes this string message as an input and returns a corresponding value in terms of a number. This number is the count of dashes in the string message. Today is her birthday. Chandu wants to give her some sorted arrays on her birthday. But the shop has only unsorted arrays. But, he doesn't have much time to sort the arrays manually as he is getting late for the birthday party.
Help him, or his girlfriend will kill him. He took three hours to solve it. The teacher got agitated at Chandu and asked you the same question. Can you solve it? Answer Chandu is very fond of strings. Or so he thinks! But, he does not like strings which have same consecutive letters. No one has any idea why it is so. Now, the problem is quite simple. You simply need to perform one operation - if there are two same consecutive letters, delete one of them.
Output: For each test case, print the answer to the given problem. Subscribe to: Posts Atom.Skip to content. Instantly share code, notes, and snippets. Code Revisions 12 Forks 1. Embed What would you like to do? Embed Embed this gist in your website.
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Hence, today I kept all those suggestions in mind while solving a new problem question. The question is such: Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
The sum that is closest to the target is 2. It would be great to hear back if I need to make further improvements. Any help will be highly appreciated. Set kMax to k. Define iMin to be i - 1. Start your for loop loop with index at iMin. At the start of each iteration, set k to kMax. When [i][j][k-1] is no longer too high, resume your uppy-downy scheme. Whenever you increment icheck if [i][k-1][k] is too low.
If so, increase i until [i][k-1][k] is no longer too low. Check if [i-1][k-1][k] gives a better match and store it as temp. AFter changing scope like this, check the opposite end of the scope. So if you just reduced kcheck if [i][k-1][k] is too low. Do this until scope is stable. So you can kill off any number that's belowprovided it's not the last number at or below Additionally, once we get toany number greater than that is not the last number at or above is not going to help.
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Judging by the complexity of your method Triplet. A way to accomplish this would be to sort the integers in the method hashCode before calculating the hash. In fact, Triplet. Regardless of the above, you are creating unnecessary List s in Triplet. Here, the explicit creation of an ArrayList from the contents of the List returned by the invocation of Arrays.
The same applies to the process of converting the elements of other to a List. You could even bypass the creation of an intermediate List completely by using a stream:. The easiest way to go about it is probably, just like in the method hashCodeto sort the integers from the triplets as a List and then compare the two List s for equality.
Actually, you could already sort the integer's in the Triplet 's constructor, which would ensure that you only ever have to sort the three integers of a Triplet once. True, this does not preserve the order of the integers, but then, this is not really necessary for the purpose of this program either. There is no need to add the result to a new ArrayList you could just use. You creating some unnecessary objects that still need to be cleaned up by the garbage collector.
Instead of:. The reason why you came up with two different solutions is: Your mind is preset to procedural solutions an premature optimization. The "sum of three" problem only differs in step 1. Yes, this solution might perform awfully for large input arrays but with the given example it is fast enough.
What do we need to change to adopt the previous solution to the "sum of two" solution to the "sum of three" problem? Now we have a similar solution for a similar problem. We could even go further and transform the factory method for the Pairs to be parameterized so that the same solution could be used for all "find combinations with matching sum" problems.
My approach is totally focused on the problem. No optimization or "clever tricks" have been used. That allowed me to easily extend the first approach to a more general version to handle similar problems. Sign up to join this community. The best answers are voted up and rise to the top.
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Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. Given an array A of N integers. So there are two triplets 1,2,2 and 2,2,1. Look at the ijk values which are indices. So, there are 12 possibilities of triplets of indices that can have array values as 221.
I tried this but i want to decrease the time complexity :. In this section what I did is ran 3 loops checked all the possibility of possible triplets with given constraints that should be divisible by only one of i, j and k.
All possible combinations of triplets won't be much.
Then iterated our previously created vector with 3 possible i, j, k. All the 3 i, j, k can't be equal because sum is divisible by only one and we have taken care of this in getpossiblenumbers function. But when two of i, j and k are equal take care of counts as I did in 3 cases above. As per the question, there can be duplicate as well as unique values in the array eg.
To get an optimized solution the algorithm is divided into two major parts- unique values, duplicate values. Main Idea Remove the duplicates and keep their counts. Numbers are less than Hence time complexity will reduce a lot.
Learn more. Good triplets sum and divisor Ask Question. Asked 6 months ago. Active 6 months ago. Viewed 3k times. Array values of a triplet i,j,k is A[i], A[j], A[k].
What have you tried so far?
Count Divisors of n in O(n^1/3)
What is the question? Show us what yozu have tried so far and where you got stuck. Used 3 nested for loops, then if else conditions. Your predicate is weird. Active Oldest Votes. Arjun Singh Arjun Singh 3 3 silver badges 8 8 bronze badges. Akhil Manchanda 53 4 4 bronze badges. Akash Swain Akash Swain 27 1 1 bronze badge. Expected output is wrong.
Can you please look into it once. Sign up or log in Sign up using Google. Sign up using Facebook.