Shift 2D Grid - Leetcode 1260 - Python

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Problem Link: leetcode.com/problems/shift-2...
0:00 - Read the problem
1:40 - Drawing Explanation
6:37 - Coding Explanation
leetcode 1260
#facebook #interview #python
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[@GunnerJayx]

Wow I was just trying this LeetCode question of the day wishing there was a Neetcode video for it...and here we are

[@sivaganesh4489]

same here

[@udattam]

Loved your solution. I have previously seen problems where we convert a r,c to a single value but never got the intuition behind it. But it's all clear now. Thanks again for the clear explanation!

[@avedissimracing9628]

5:33 why do we mod by 2? We have to mod by 3 - the number of columns

[@NeetCode]

Yeah that's correct

[@studyaccount794]

Thanks, I was stuck in this problem. This is a tricky one for an easy problem I think.

[@NeetCode]

Yeah I agree!

[@sivaganesh4489]

Thanks for great explanation. expecting a gird series from you

[@hamdysaadpersonal]

love it and love the way you solve it

[@shantanukumar4081]

Great Explanation !!!

[@AKrieger94]

Wow! just in time :D I was literally gonna search for this problem in youtube, but it appeared in my youtube suggestions

[@NotKevinNguyen]

I was looking for your video earlier today haha

[@nehaa3778]

Clear and helpful!

[@Marcelo-yp9uz]

great explanation!

[@andrewgleason7588]

great use of helper functions, make it much more readable of a solution

[@spacex6997]

Would it be possible to convert the 2D list into 1D, do the scrolling and then for each 3 elements, add those into 3 lists for 9 items?

[@Raymond-Wu]

Glad we had the same approach! I immediately decided to flatten the arr when I saw this problem. Only difference is that I simulated the k shifting with popping and inserting at index 0. That way I didn't have to think about when things would go out of bounds

[@sprinklepancake]

thanks man

[@abheykalia3409]

(k/n)%m will tell us how many places do we need to shift the complete rows by. and do the k%n number of rotations on this new matrix. I did in this way

[@sarveshmaurya2168]

well explained

[@impostorX]

Thanks! But can we do it in place for O(1) space complexity?

[@hantinglu8050]

Check out the concept of 189. Rotate Array. Basically they're the same instead of this is a 2d problem.

[@genuinepriyanshu8685]

We can use inplace shifting using one temporary variable so that SC will be O(1)

[@shenyan9062]

nice solution

[@Faha_Kite]

Nice explanation. Can you share how you make your videos?

[@pranavsharma7479]

this shd be a medium problem it requires this 2d to 1d and 1d to 2d trick + shift circularly

[@vaibhavvashist238]

Hi How much time you spend in particular problem without looking for solution?

[@NeetCode]

I think 15-20min is a good rule of thumb.

[@timarleyfoster5417]

So is N the number of columns or values in a row, because he goes back to saying 'values in a row' in the code part?

[@linhnguyenduc641]

To recap: 1.Treat 2D array as 1D one 2.Shift items in 1D array because it is easy to find a new position after shifting 3.Convert a new position back to a new row and a new col in 2D array 4.Update value of new row and new col in result

[@artemis5507]

Flatten-> then used deque->again unflatten... faster but man to much space consumption 😮‍💨

[@lordsixth5944]

Damm never thought about this one. Tho how old are you?

[@Od253]

This easy is lowkey difficult

[@NeetCode]

Agreed

[@jerryjohnthomas4908]

Hi absoltely love your channel. Could you pleaseeeee do 979. Distribute Coins in Binary Tree.(or 968. Binary Tree Cameras ) i know its a lot to ask but you are awesome :')

[@ritikgoyal3622]

From where I can learn a python oops pls suggest me Any sources

[@jugsma6676]

I was putting all row and cols in 1D and back to grid cache = [] for i in range(len(grid)): for j in range(len(grid[i])): cache.append(grid[i][j]) tmp1 = cache[:-k] tmp2 = cache[-k:] cache = tmp2 + tmp1 p = 0 for i in range(len(grid)): for j in range(len(grid[i])): grid[i][j] = cache[p] p += 1 return grid

[@Marcelo-yp9uz]

5:50 Seven divided by three is also two*

[@cozywind2010]

I found this solution to be more intuitive and cleaner. the sum(grid,[]) is new to me though. class Solution: def shiftGrid(self, grid: List[List[int]], k: int) -> List[List[int]]: col=len(grid[0]) nums=sum(grid,[]) k = k % len(nums) move=len(nums)-k ans=grid count=0 nums = nums[move:] + nums[:move] for i in range(0,len(nums),col): ans[count]=nums[i:i+col] count+=1 return ans

[@numberonep5404]

fun pb

[@smokesmoker4301]

use c and use pointers.

[@BeLikeNexus]

I feel like this is not an easy lol

[@NeetCode]

Yeah agree

[@bhaveshverma8629]

Still not understand

[@dataman4503]

C++. vector shiftGrid(vector& grid, int k) { int rows = grid.size(); int cols = grid[0].size(); // -- Optimization -- (Edge case) // After rows*cols shifts, the result grid will be same as original input grid. if (k % (rows*cols) == 0) return grid; vector res(rows, vector(cols, 0)); for (int r=0; r

[@imohanreddy]

can someone help why this is not working ? Please class Solution: def shiftGrid(self, grid: List[List[int]], k: int) -> List[List[int]]: m, n = len(grid), len(grid[0]) ans = [[0]*n for i in range(m)] for r in range(m): for c in range(n): one=((r*n +c)+k)%(m*n) print(one) j,k=one//n, one%n ans[j][k]=grid[r][c] return ans

[@imohanreddy]

tried without functionand "one" doesn't seem to be calculating right , not able to figure it out

[@NeetCode]

Your issue is that 'k' is an input parameter, but your are reassigning it in your for loop.

[@imohanreddy]

@@NeetCode oh yes, thanks a lot

[@RajanPaudel-xh7xv]

hard

[@pphodaie]

I flattened the grid to an array, removed the last k % m * n elements of this array and added it to the start of the array. Finally I converted the shifted array back to a grid. func shiftGrid(_ grid: [[Int]], _ k: Int) -> [[Int]] { let m = grid.count let n = grid[0].count let array = grid.flatMap{$0} let size = array.count let tk = k % array.count let shifitedArray = Array(array[(size - tk)...] + array[..