HackerRank - Missing Numbers | Full solution with examples and visuals

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

[@sikderabdurrahmanjony8741]

NOBODY EXPLAINED IN SUCH DETAILS LIKE YOU.GO ON.

[@nikoo28]

Thank you so much for your support.

[@shreysom2060]

Great explanation brother

[@sikderabdurrahmanjony8741]

THANK YOU BROTHER.

[@vinoddiwan5792]

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

[@unemployedcse3514]

awesome ❤

[@arundoss6785]

Awesome 😊😎

[@jritzeku]

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]

What is the time complexity of your approach?

[@ambiliprasadr]

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]

have you tried to debug with some test cases?

[@Robert-ob8bu]

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]

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

[@shenth27]

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

[@nikoo28]

A TreeMap almost works in constant time.

[@AjayMishra-ix3kj]

Wow

[@premanathareddydodda7565]

can i use for this solution stream api in java?

[@nikoo28]

Yes, you should be able to

[@kusumalakshman1133]

❤❤❤❤❤❤❤❤❤❤

[@shreysom2060]

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

[@nikoo28]

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]

@@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]

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]

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

[@vilakshan.s]

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]

Is there another way to solve it???

[@nikoo28]

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

[@AjayMishra-ix3kj]

@@nikoo28 okk