Out of Boundary Paths - Leetcode 576 - Python 

That 3D DP solution is wild. I was able to get the memoization solution on my own today, but I still have trouble translating that into bottom up approaches. But honestly, I've come a long ways already for being able to solve a problem like this by myself with memo. Thanks Neet!

3D DP it is! I was thinking in terms of 2D and didn't know how to memoize it. Thanks NeetCode!

Your solutions are so real, nothing fancy, they are simple and easy to understand

Thanks for doing these daily problems. Very helpful

i actually got MLE on a bfs solution and a TLE on the dfs one xd

One thing I was confused about the recursive brute force solution is that why are we allowed to go back to the node we just came from? Would that not consitute a path?

Great explanation as always. Thank you .

You mentioned a BFS solution wouldn't work but in one of my approaches I considered it and it somehow worked. Gave TLE at 22/94 but it could possibly be optimized? # BFS solution MOD = 10**9 + 7 directions = [(0, -1), (0, 1), (-1, 0), (1, 0)] queue = [(startRow, startColumn, maxMove)] res = 0 while queue: node_i, node_j, curMoves = queue.pop(0) if curMoves > 0: curMoves -= 1 for delrow, delcol in directions: new_i, new_j = node_i + delrow, node_j + delcol if not (0

The second solution is less efficient in LC because we are calculating the possibilities for all positions in the grid. Different from the dfs solution were we calculate for the startRow/startCol, but in a case where we wanted to calculate all of them the DP is much more faster and memory efficient.

Well explained

Memoization solution is pretty straightforward, I dunno bout da tabu sol tho

Thank you so much

thanks, man

Thanks ❤

always on top

For a Brute Force Memoization I am getting TLE.

💪💪

day 26 Leetcode

Not to brag NeetCode, but I got the 15 view on this video.

for the if statments I mada a helper function: class Solution: def findPaths(self, m: int, n: int, maxMove: int, startRow: int, startColumn: int) -> int: ROWS,COLS = m,n MODULO = pow(10,9)+7 curGrid = [[0] * COLS for _ in range(ROWS)] prevGrid = [[0] * COLS for _ in range(ROWS)] def helper(r,c): if r < 0 or c < 0 or r == ROWS or c == COLS: return 1 return prevGrid[r][c] for _ in range(maxMove): for r in range(ROWS): for c in range(COLS): val = helper(r+1,c) val += helper(r-1,c) val += helper(r,c+1) val += helper(r,c-1) val %=MODULO curGrid[r][c] = val prevGrid, curGrid = curGrid,prevGrid return prevGrid[startRow][startColumn]

Great explanation as usual! A suggestion, is it not a bit more readible to remove the if/else checks with something similar to this? (I added this vcrsion to neetcode github if that's ok, instead of having all the if statements) class Solution { fun findPaths(m: Int, n: Int, maxMove: Int, startRow: Int, startColumn: Int): Int { val mod = 1_000_000_007 val dirs = intArrayOf(0, 1, 0, -1, 0) val dp = Array (m) { Array (n) { LongArray (maxMove + 1) } } for (k in 1..maxMove) { for (i in 0 until m) { for (j in 0 until n) { for (dir in 0..3) { val i2 = i + dirs[dir] val j2 = j + dirs[dir + 1] if (i2 < 0 || i2 == m || j2 < 0 || j2 == n) dp[i][j][k]++ else dp[i][j][k] = (dp[i][j][k] + dp[i2][j2][k - 1]) % mod } } } } return dp[startRow][startColumn][maxMove].toInt() } }