Basic Calculator II | Leet code 227 | Theory explained + Python code

[@symbol767]

This is the best explanation on the problem I've seen. After your explanation I was able to code it out on my own. Thank you so much.

[@jamesperalta35]

For 18:30, the reason it wasn't coming out correct was because the question says "The integer division should truncate toward zero." When you perform integer division, you get the floor which would result in rounding -0.333 to -1 when it should round to 0 based on the description. int() however rounds towards 0 which is why it works.

[@saianishmalla2646]

Thanks for explaining!

[@thechunwu]

In a situation like: "14-3/2" I find the int() rounds to -2 rather than -1 because in this case its rounding down. which is a greater negative number?

[@tzujuiyu]

@@thechunwu -1 is greater than -2. In your example, this expression int(-3/2) should evaluate to -1. I wonder if you added an extra ''/' by chance. int(-3//2) gives you -2

[@am_0x2a]

Great video! A further optimization would be to eliminate the stack altogether and keep a running sum. Instead of appending to the stack, just add. Instead of popping, just keep track of the most recent value and subtract it off. Then you can save yourself the extra loop to go through and sum all the values. Specifically, it will bring storage complexity from O(n) to O(1) and reduce time by a factor of 2 (though it will still be O(n)).

[@BreakthroughCreator]

this is one of the best coding teaching videos so far, nice voice, cadence, explanation, and visuals

[@anoops7974]

Thank you Sai Anish. Great video and explanation.

[@grovestreet9165]

if you're indian then your accent is realy too good

[@HarshitaKhazanchi]

this really cleared things up!

[@n.h.son1902]

Great visualization! It helps build my intuition a lot for this challenge. There're a couple of ways to handle the edge case about the operator == "/", one way we can think of is to take advantage of the built-in function math.floor() and math.ceil(). Cheers!

[@LARamones23]

Thanks for the nice explanation! Your videos are very helpful. Integer division for negative number is the issue for stack[-1] //= curr_num. For example -3//4 returns -1 in python but 3//4 returns 0. Following approach seems also to be working. if stack[-1] >=0 : stack[-1] //= curr_num else: stack[-1] = -(-stack[-1]//curr_num)

[@saianishmalla2646]

Glad the videos help! Thanks for explaining why int division didn't work it makes sense.

[@vidyutphagetra3804]

good explanation Anish

[@kevinzheng575]

Nice information. Thank you!

[@sugamff8825]

Big fan bro ❤️

[@saianishmalla2646]

ayyeee thank you!!!

[@ishayadav001]

Love your videos!!

[@saianishmalla2646]

Glad you like them!

[@linli7049]

Very good explanation!

[@sakshiramsinghani5284]

Thanks a lot! It's really helpful.

[@user-kg9lm4mp8j]

//是整除，int()是取整，题目要求向0取整，整除遇到负数会出错。

[@shrimpo6416]

Bc int div round towards zero. So it gets weird when rounding neg num

[@aminesfahani3563]

thanks i want to share this code class Solution: def myPow(self, x: float, n: int) -> float: if n==0: return 1 elif n

[@Chancemm]

Thank you!!

[@saianishmalla2646]

You're welcome !!

[@diegogarcia-pl1bt]

you just stole the answer from the discussion you clearly do not understand the code

[@PremPal-uy4nm]

It seems he didn't explain clearly how char & operator are working together to add values in stack to make correct calculation as we move through the string. In short char is leading in for loop and operator is lagging.

[@n.h.son1902]

@@PremPal-uy4nm if I remember it right, he did. His explanation in the visualization meant that the variable “operator” in the code is the previous operator in the illustration and the variable “char” means the current operator we are at.