Largest Odd Number in String - Leetcode 1903 - Python

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Problem Link: leetcode.com/problems/largest...
0:00 - Read the problem
0:30 - Drawing Explanation
4:10 - Coding Explanation
leetcode 1903
#neetcode #leetcode #python

[@akshayiithyd]

r = len(num) - 1
res = ''
if len(num) == 0:
return res
while r >=0 and int(num[r])%2 == 0:
r -= 1
return res + num[:r+1]

[@shaiksoofi3741]

Thank you

[@34535fff]

Thanks, finally I did good solution before your video, yeah it's just an easy problem, but I am happy)

[@thebosslevelstudios995]

Can we use heap for the following?

[@amanchavhan4159]

If we start from 0->n(cpp)
It shows memory limit exceeded.
Why???

[@Vancha112]

Looks like you're out of bounds, try n-1 :)

[@dk4882]

Easy one

[@ajith4249]

def largestOddNumber(self, num):
for i in range(len(num)-1,-1,-1):
if int(num[i])%2==1:
return num[:i+1]
return ""

[@winwin2369]

Way too easy problem it won't be asked 😂

[@chrischika7026]

I thought list slicing was O(N) so why is it not O(N^2)

[@jessezigi2419]

You only do list slicing once

[@Azikkii]

What? Your answer is in your question.

[@Vancha112]

Yeah in the code there's only one slice, which makes it o(n) :)

[@leeroymlg4692]

it would only be N^2 if you were to do a slicing operation for every iteration of the loop of the input parameter. But in this case, it is only done once

[@RexSacriticulus]

A more elegant and pretentious way to check if a number is odd is n & 1.

[@itspurelypassionate]

yes it is lol