HackerRank - Missing Numbers | Full solution with examples and visuals

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Missing Numbers is a programming challenge on HackerRank. You are given an array, and an artist is trying to transport those numbers to a different array. In the process, he misses some of those numbers. These are termed as missing numbers. We need to find these numbers in a sorted manner. Watch the video to understand the problem in a simplified manner. I then work along with you to solve it first using a Brute Force approach, and then an efficient approach. All along with visuals and explanations.
00:00 - Intro
01:04 - Problem Statement and Test Case
03:11 - Brute Force Solution
05:49 - Efficient Solution
09:54 - Dry-run of code
12:10 - Final thoughts
📚 Links I talk about in the video:
Actual problem on HackerRank: www.hackerrank.com/challenges...
Code on Github: github.com/nikoo28/java-solut...
Test cases on GitHub: github.com/nikoo28/java-solut...
📘 A text based explanation is available at: studyalgorithms.com/array/hac...
🔗 To see more videos like this, you can show your support on: www.buymeacoffee.com/studyalg...
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#hackerrank #programming #tutorial

Опубликовано:

15 июл 2024

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Комментарии : 30

@sikderabdurrahmanjony8741 3 года назад

NOBODY EXPLAINED IN SUCH DETAILS LIKE YOU.GO ON.

@nikoo28 3 года назад

Thank you so much for your support.

@shreysom2060 3 года назад

Great explanation brother

@vinoddiwan5792 3 года назад

Amazing, amazing, amazing...............

@sikderabdurrahmanjony8741 3 года назад

THANK YOU BROTHER.

@arundoss6785 3 года назад

Awesome 😊😎

@unemployedcse3514 Месяц назад

awesome ❤

@AjayMishra-ix3kj 3 года назад

Wow

@jritzeku 6 месяцев назад

I just finished your strings playlist in which you had similar problem. In that, you used the pointer approach. I used similar strategy: function missingNums(a1, a2) { let i = 0//pointer for a1 let result = [] for (let j = 0; j < a2.length; j++) { if (a2[j] === a1[i]) { i++ } else { result.push(a2[j]) } } //sort in ascending order return result.sort((a, b) => a - b) } Which approach would be considered better?

@nikoo28 6 месяцев назад

What is the time complexity of your approach?

@ambiliprasadr 9 месяцев назад

Hi Nikhil, Great Explanation. Can I ask you a doubt? I tried to work below code ,passed test cases except test case 1. Any suggestion would be appreciated def missingNumbers(arr, brr): brr_copy = brr.copy() # Create a copy of brr to avoid modifying the original list for i in arr: if i in brr_copy: brr_copy.remove(i) return sorted(brr_copy)

@nikoo28 8 месяцев назад

have you tried to debug with some test cases?

@kusumalakshman1133 10 месяцев назад

❤❤❤❤❤❤❤❤❤❤

@Robert-ob8bu 2 года назад

1 test case is failing by use of map.......i also did while loop to managee multiple frequencies ....even then its failing 1 test case

@nikoo28 2 года назад

Can you give me an example of the test case. The solution I submitted on hackerrank passes all cases.

@premanathareddydodda7565 2 года назад

can i use for this solution stream api in java?

@nikoo28 2 года назад

Yes, you should be able to

@shenth27 Год назад

I'm not a java guy but wonder time complexity of TreeMap to maintan the items in sorted order.

@nikoo28 Год назад

A TreeMap almost works in constant time.

@shreysom2060 3 года назад

in case of repeted freequency that is if 1 is more than 1 then the code is not right.

@nikoo28 3 года назад

The code is correct. We subtract frequency by "1" for every repeated number. Can you show me a test case for which the code fails?

@shreysom2060 3 года назад

@@nikoo28 i mean in case where 2 times 1 is there in IntegeerFreqmap after removing numbers from arr then in final loop how times 1 will be inserted. as result[i++]=IntegeerFreqmap.getKey() value in . We are not checking frequency to put in resultSet.

@nikoo28 3 года назад

Let us go over it step by step. Suppose you have two 1’s In the first loop you create the map and the frequency would be stored as “1 -> 2” In the next loop if you find one 1 in the other array you subtract a frequency so the map changes to “1 -> 1” In the final loop you look at the remaining elements in the frequency map. You find a frequency and hence the code works.

@shreysom2060 3 года назад

@@nikoo28 got it I was thinking some else thing only read the question again thankyou so much

@vilakshan.s 2 месяца назад

Still have similar doubt. What if in second array both the two 1s got missed from array 1 then in the MAP you will have like 1->2 then for such frequencies you will have to add the 1 entry 2 times in the final output

@AjayMishra-ix3kj 3 года назад

Is there another way to solve it???

@nikoo28 3 года назад

You can try to solve it using a frequency array where each index stores the count of each element in first array

@AjayMishra-ix3kj 3 года назад

@@nikoo28 okk