Concatenation of Array - Leetcode 1929 - Python

[@thespicycabbage]

finally an easy problem to boost my self confidence T_T

[@thespicycabbage]

confidence is real low. Had to come back to see how this was done lol

[@the8426]

Another way is to create ans 2x len of nums with 0 as placeholder. Loop through nums and add the current value of nums to the corresponding index of ans and corresponding index ans + len of nums. So in the first iteration, we add the value at nums[0] to ans[0] and ans[0+len(nums)] which is ans[3] and so on. The time complexity improves to O(n) instead of O(n+n) lil difference.

[@软件实验室]

but the creation of the array ans with placeholders takes O(2n) time

[@ahmadmtera]

@@软件实验室 who cares... At the end we need to return an array of length 2n. Working with arrays is better and cheaper in terms of processing power compared to using an object like List. Storage is also cheap and nums is eventually discarded so in both cases we're left with a Space Complexity of O(2n) as requested.

[@piyusharyaprakash4365]

Return nums + nums Easy peasy But your solution works better if there is a 3rd argument Nice one 👍

[@akash-kumar737]

Use return nums*2 Easy peasy plus expandable.

[@TaoZeng-z2w]

How are you able to return nums + nums in Java?

[@ahmadmtera]

Python will screw you over the minute you step into any language that doesn't support such high level features. In problem solving, one should learn how to solve problems by learning how to type out the relatively low level instructions that deal with basic building blocks like arrays, and not by using a language specific feature that does that for them.

[@yfjsdgzjdnfn]

Simple Sol: def concatenation (nums,x): return x * nums;

[@business_central]

Please do "Last day where you can still cross" , even LC doesn't have an official solution yet on it

[@candylover2456]

since its a nested loop though, wouldn't this be O(n^2)? Confused here, can anyone please help?

[@软件实验室]

n is the input size, but the outer loop's iteration is independent with the inner loop, which in this case is still O(n)

[@lavanya_m01]

return ( nums + nums ) is very pythonic. Instead we can directly use the formula given in the qn 😌

[@doc9448]

I submitted 3 times: nested for loops, a while loop (with the same for loop inside), and then simply "return nums + nums". The return was the most efficient according to python. A true easy this was.

[@ingluissantana]

Greattttt to see some easy problems solved!!! 🍻🍻🍻🍻

[@dumbfailurekms]

ans[i] = nums[i % len(nums)] or ans.append and define the range of the for loop as 2 *len(nums)

[@dusvn1484]

Mby extend with same nums works good ?

[@qazyhn94]

is only append is O(1) but append in the middle of the begining in list will be O(n) operation?

[@dankr]

thanks

[@vishalroy7932]

Why did you gave up some of us were watching your videos daily

[@cyberrebel4623]

return nums*2

[@suspatrol9550]

nums.extend(nums) return nums

[@BennduR]

var getConcatenation = function(nums) { let ans = []; for(i = 0; i < nums.length; i++){ ans[i] = nums[i]; ans[i+nums.length] = nums[i]; } return ans; } 1ms runtime is this good or bad and why? Thx

[@turtlenekk4354]

basically just return nums + nums right?

[@aaditya_87]

yeah no idea why he made a video for this

[@gopro12378]

Just return the list multiplied by 2 that is return nums * 2

[@home390]

Return nums * 2 😜

[@novascotia2015]

dude i came from your tweet

[@satwiktatikonda764]

whts it?

[@pritishbhakat]

2444. Count Subarrays With Fixed Bounds plz solve this problem, I am stuck in it.