Insert Interval - Leetcode 57 - Python

0:00 - Read the problem
2:17 - Drawing Explanation
7:31 - Coding Explanation

leetcode 57

#sorted #array #python
Disclosure: Some of the links above may be affiliate links, from which I may earn a small commission.

[the_derpler]

This one kills me, I always get soooo close yet end up with a variety of edge cases. Thanks for your video.

[anthonyuccello]

At the start of my interviews, Im definitely going to use the line "Lets right some neetcode today"

[PippyPappyPatterson]

I come to your solutions first for new Leetcode problems. Sometimes I think, "wow there's got to be a more elegant way to do this." Then I look at the 🐍 solutions on Leetcode discussions and I realize how yours blows everything else out of the water.

The best part of following along with your solutions is I find myself thinking through algorithms similar to you. It's like being in the lectures of a great math professor and assimilating their thought processes into my repertoire.

[symbol]

Your solution is so compact and clean, mine had 2 separate while loops, then a for loop to get the answer, this is way cleaner and better. This also opened my mind to how to think about this problem more logically.
Thank you! Liked!

[JJ-qvco]

Dude you upload faster than I can solve each day

[karthik]

Preparing with your videos is really fun! Thanks!

[cw]

Deceptively tricky. Lots of edge cases to consider.

[licokr]

now it is a must after solving problems, watching your explanation. this is absoultely amazing. i didn't think of expanding the new interval array so i had to make a lot of cases and plus, conditions you wrote look amazingly easy. knowing well what exactly you have to do. thanks a lot. i learned a lot today too

[electric]

Clever solution using the for loop to basically ignore all the intervals that are in between newInterval.

[sandippatel]

Your explanation is very easy to understand and helpful. Thank you for such a nice video.

[mrfli]

Pure gold! Saved lots of time finding solution. Great explanation.

[edwardsolomon]

Just can't emphasize enough how good you explain a problem!

[hwang]

pretty smart solution, i feel like its difficult to come up with in an interview, heres the commented code if it helps anyone:

class Solution:
def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
res = []

for i in range(len(intervals)):
if newInterval[1] < intervals[i][0]: # Right bound of of new interval is to the left of current interval
res.append(newInterval)
return res + intervals[i:]
elif newInterval[0] > intervals[i][1]: # Left bound of new interval is to the right of current interval
res.append(intervals[i])
else: # Some part of the the new interval is overlapping with the current interval
newInterval = [min(newInterval[0], intervals[i][0]), max(newInterval[1], intervals[i][1])]

res.append(newInterval)

return res

[DavidDLee]

A more interesting question would be how to store these intervals such that insertion will take less than O(N), that we don't need to iterate.

[hey.............]

No doubt you are at Google. Really clean and simple code

[BobbyMully]

I did line sweep algorithm that was rather messy, but this way was much more elegant and efficient. An especially great solution for this one!

[wigglesort]

You explained the logic and code very easily. I code in C++ but easily understood python solution

[davidoh]

I think understanding the cycle update and doing it with an outer loop is something that people would more likely come up in the interview! I wonder what is the better way to solve this in actual interview

[sinnyman]

Hey Neet, could u make a video about ' Number of Closed Islands' ?

[fpnwrvd]

You could do this and avoid returning twice

res = []
i, n = 0, len(intervals)
while i < n:
if newInterval[1] < intervals[i][0]:
break
elif newInterval[0] > intervals[i][1]:
res.append(intervals[i])
else:
newInterval = [min(newInterval[0], intervals[i][0]), max(newInterval[1], intervals[i][1])]
i+=1
res.append(newInterval)
return res + intervals[i:]