Search Insert Position - Binary Search - Leetcode 35 - Python

0:00 - Concept
8:35 - Coding Solution

Leetcode 35
#python #leetcode #neetcode

[luckychitundu]

Nice One! I came here to find out why we are returning left. Thanks brother.

[CS_nb]

that breathing at the end freaked me out lol

[ernie]

Keep them coming! You’re doing a great job

[alphabee]

unlike others you had a clear understanding why to use binary search.

[pranavkul]

Thanks a lot. The video was really helpful and the way you explain it is so great

[duvincent]

Imagine array as people line up and facing a wall, whenever a person want to join the queue, we can either choose a person, push him back and sit in front of him, or join at the end of the queue.

The natural of array list cause that, we are searching right (I am bigger than you, then I join behind you so i + 1), if we are searching left (I am smaller than you, I don't have to move but push you behind as i).

That is why we can always choose the lower bound pointer. At first I am thinking why god create left and right bound unequally.

[Oyeah]

I am confused about when to use (less than)< vs <=(less than or equal to). What is your approach to knowing when to use which one?

[ruksharalam]

One o' the best LeetCode channels in YT

[lylelll]

like your videos! They are easy to understand.

[skms]

Who invents these problems !! Crazy stuff

[무광-rk]

Thank you sir. i am kotlin user but its really impressive thank you

[thinja]

wow that trick at the end was *chefs kiss*

[biomedd]

at 12:50, I am still not clear why left becomes 1 since the shift has not occured, so left should remains at 0. Any help with better explantion? thnks

[abdullahahmad]

I did something like this:

def binary(list1, key):
start = 0
end = len(list1) - 1
while start <= end:
mid = (start + end) // 2
if list1[mid] == key:
return mid
elif key > list1[mid]:
start = mid + 1
elif key < list1[mid]:
end = mid - 1
# print(mid)
if key > list1[mid]:
return mid + 1
elif key < list1[mid] and mid > 0:
return mid
else:
return 0

[kirich]

is there a recursive function solution?

[szymon]

one of the things I noticed when solving this problem is that if the target is less than nums[0] you should return 0, and if the target is greater than nums[len(nums)-1], to return len(nums). do you think this is efficient or just a waste of space/time?

[canr]

great video. I have one small question though. why do we add or subtract 1 from mid?

[srinathreddy]

Very well explained like why we should left, why not right??
Thanks a lot❤

[Famelhaut]

1 liner:
import bisect

class Solution:
def searchInsert(self, nums: List[int], target: int) -> int:
return bisect.bisect_left(nums, target)

[SeyyedMohammadLoghmanDastgheyb]

Hi 🙂 this solution doesn't work for 👉 nums=[1, 3, 5, 6], target=2 😵‍💫 What am I missing? 🧐