Day 1: Find All Duplicates in an Array (

[@OmoruyiOmofonmwan]

Gonna subscribe. You can do it my man

[@idkname7]

I'm gonna follow you , because I wanna do this too , currently studying dsa.

[@5foreign346]

this uses O(n) space not constant

[@SagarSingh-wj4pn]

correct! my guy mistook linear for constant space

[@codestreakwithleo]

Thanks for pointing it out, I missed the O(1) space complexity condition there. Had a further look at the problem again, and it looks like doing the nums[abs(nums[i]) - 1] < 0 approach should work

[@Micha-mx8ii]

What about this? Does this comply with the requirements? result = [] for k in range(len(nums)): if nums[k] in nums[:k]: result.append(nums[k]) return result

[@udaykiran2427]

No, will give a TLE as it's O(n^2)