Pacific Atlantic Water Flow - Leetcode 417 - Graphs (Python)

[@GregHogg]

Master Data Structures & Algorithms For FREE at AlgoMap.io!

[@saideep7510]

I am currently doing NeetCode 150 and watch his videos regularly, but every once in a while when I feel like his solution is a little complex to understand I check if you have uploaded a video on the same question and if you have then I go through it cause coincidently most of the times your version of the solution is comparatively much easier to understand. Thanks!

[@ShivamSharma-t8c]

Real

[@anuradhasuresh3451]

Hi Greg, graph problems scared me at first, but after watching your videos and how wonderfully you break everything down into simple code, I'm getting more confident with them! Thanks a lot.

[@vivek2319]

I compared this solution to other channels' solution and I feel yours is the most easiest to read and understand. Thanks 😇

[@vikramkhurana1331]

Broke down a complex problem beautifully

[@auraSinhue]

Great explanation!

[@Anantlahamge07]

bro ngl till now the hardest problem and solution. but it's only because of Greg that a rookie like me could comprehend how to solve this Q 😭😭

[@vuanhkhoa9715]

Hi Greg ! Can I ask for your discord community server ? I saw it somewhere but now Im unable to find it

[@GregHogg]

Right here! discord.com/invite/g4ZnYzTB

[@codinggigs8367]

Here is my solution that I come up with after hearing your thought process. It might be simple to understand : from collections import deque class Solution: def pacificAtlantic(self, heights: List[List[int]]) -> List[List[int]]: rows , cols = len(heights), len(heights[0]) visited_p = [[0 for _ in range(cols)] for _ in range(rows)] visited_a = [[0 for _ in range(cols)] for _ in range(rows)] row_col_list = [(1,0),(-1,0),(0,1),(0, -1)] set_p = set() set_a = set() que_p = deque() que_a = deque() for r in range(rows): for c in range(cols): if r == 0 or c == 0: set_p.add((r,c)) que_p.append((r, c, heights[r][c])) visited_p[r][c] = -1 if r == rows - 1 or c == cols -1: set_a.add((r,c)) que_a.append((r, c, heights[r][c])) visited_a[r][c] = -1 while que_p: rqp, cqp, htqp = que_p.popleft() for rl, cl in row_col_list: rn = rqp + rl cn = cqp + cl if 0

[@bigk9000]

I still struggle on how you were able to come up with the solutions to most of these! @_@

[@hariharasudharsan3170]

Great work.... Can you make videos for DSA concepts too!

[@GregHogg]

Thanks! Yeah after this leetcode playlist is done I'll be doing a playlist on the generic theory

[@hariharasudharsan3170]

@@GregHogg fast as you can....🤝🏻

[@yygysgtyfugunvt]

Your discord server link is invalid

[@user-jm6gp2qc8x]

lol this seems to be the hardest solution in your series (and i have finished even the dp ones in yours)

[@dishandshah660]

thanks

[@_PulpoPaul]

This make me feel like a stupid

[@GarouNguyen]

Bro what is r, c = i+i_o, j+j_o, can you explain bro i don't understand 😢

[@vikramkhurana1331]

i_o and j_o are offsets.. so for [0,1], r ,c = i+0,j+1, so j is incremented, so it is square right to i,j. We go through the list of all 4 neighboring squares to check if water can flow from them to us