Given an array of integers, find the numbers that appears an odd number of times return an array of the numbers.

Question

input:
[1,1,1,2,2,3,3,3,2]

output:
[1,2,3]

Answer 1

For example, now in input you have number A
if you have A in array - delete A from array and continue
else add A to array

In this case you'll have in array only that numbers, that appears an odd number of times

Answer 2

notGreat I described an algorithm for your problem lol

Answer 3

notGreat and my algorithm is better than your solution, cause it is O(n) and needs less memory
Explanation: you have as input n numbers 1,1,1,2,...
And as a data structure for this task you should choose hashset
you add first number to hashmap - O(1)
from second number to last you you check is there a number in hashset - if no then add O(1), if yes then delete O(1)
Finally, you will have only numbers that appears an odd number of time

Answer 4

Post your attempt first..

Answer 5

https://code.sololearn.com/c9kWnK53xK88/?ref=app

Answer 6

notGreat
Pls tag the language also..

May you looking like this :

for(i in obj){
if(obj[i].length % 2 !== 0 && obj[i].length != 1) num.push(+i);
}

Answer 7

Jayakrishna🇮🇳

I don't completely know what your asking, but if that code is part of your solution to the asked problem then it's incorrect, you don't want to filter values with length not equal 1 you would want the first condition obj[I].length % 2 != 0 as 1 is an odd number, you can see that I had that idea on line 10 of my odd int problem

Answer 8

Maria Vasilyova the problem is to determine if there is a number(s) in an array that appears an odd number of time

you can view my solution on comment

Answer 9

Jayakrishna🇮🇳 I completed the challenge in JavaScript but you can complete in whatever language you please

Answer 10

notGreat
That is a about, dublicate numbers printing.. Can you give an example which is you not getting write? With expected output?