Minimum Time to Visit a Cell In a Grid | Leetcode 2577

[@sailendrachettri8521]

Leetcode is throwing hard level question frequently
Your explanation is always best :)

[@techdose4u]

Yeah, Leetcode is getting tougher!
As do the interviews :)

[@SantoshKumar-rt1it]

best explanation of solution

[@techdose4u]

Thanks! :)

[@PCCOERCoder]

please increase the font size when you are showing the code or the problem. It will help a lot for mobile users. Hope so you see this comment!

[@techdose4u]

yep.
i will try

[@PCCOERCoder]

Thank you sir

[@techdose4u]

welcome :)

[@PRANAVMAPPOLI]

Why we are not checking is there a better other option ro reach a cell or not.
Here we are just use a visited set, buy usually im dijakastra we will compare
if crntTime

[@PRANAVMAPPOLI]

class Solution:
def minimumTime(self, grid: List[List[int]]) -> int:
rows=len(grid)
cols=len(grid[0])
time=[[float('inf')]*cols for _ in range(rows)]
heap=[(grid[0][0],0,0)]
direc=[(0,1),(0,-1),(-1,0),(1,0)]
time[0][0]=grid[0][0]
if grid[0][1] > 1 and grid[1][0] > 1:
return -1

def isValid(i,j):
return 0

[@techdose4u]

Nope,
coz here the result may differ by value of 1 of you just push without calculating based on even odd diff. That I had explained in the video.
That calculation of even odd diff is the only difference. Rest everything is same.

[@prakhargarg4166]

Thankyou, sir

[@techdose4u]

welcome :)

[@saikishanrao4150]

nice

[@techdose4u]

welcome :)

[@21flame72]

Thankyou sir

[@techdose4u]

welcome :)