I think this solution is very readable compared with others, which is a huge plus.

Nice. This was most similar to my solution. Don't completely understand the syntax of the #1 solution for this problem. But in the meanwhile, this is great. Thanks!

ı like it

Nice solution. Gives me new idea to solve a problem.
Thanks all.

hey, a little note here,
you can still add arrow function at reduce

you probably forget arr.length - 1 instead of arr.length

parseInt really needs a radix passing to it: although base 10 is assumed it's good practice to ensure that parameter is always passed, eg. parseInt(x, 10);

Because arr[i] still needs to be counted - which would not happen when the If statement fails. This is especially important in this challenge where the value you want to return may only appear once.

Since the array is sorted first, incrementing num in the else statement is the equivalent of saying "This is the last time this number appears in the array, let's see if it showed up an odd number of times."

unsure why the else statement is incremeneting num?
if arr[i] !== arr[i+1], then why are we continuing to count num?

new road for me!thx