Love that you completed all the vids for blind 75, but only use python for your codes. I have plenty of other people I could go to for C, so don't do any other language. Python only - that's your niche and why most people are here.
If you want to do the same thing in Python, you can use this code (and I'll explain it!): class Solution: def getSum(self, a: int, b: int) -> int: mask = 0xffffffff while b != 0: tmp = (a & b) mask // 2: return ~(a ^ mask) else: return a To explain, the hexadecimal number 0xffffffff is the same as the binary number 0b1111111111111111111111111111111, containing 32 1's. (It's just easier to type lol.) In order to make the code work like Java, we want to treat our numbers like they only have 32 bits. ANDing a number with the mask 0xffffffff, or 32 1's, basically turns all of a number's bits into 0's except for the rightmost 32. As a result, the number can be represented as if it only has 32 bits. We do what Neetcode describes in his video, using XOR for the sum and AND for the carry. We AND with the mask each time we set a and b in order to keep our numbers within 32 bits. After we exit the while loop, we have our answer a. If a is positive, then we can return it directly. However, in Python, negative numbers are represented in binary as having an unlimited number of leading 1's. The current answer would only have values in the rightmost 32 bits. Therefore, if the answer is negative, we need to convert it into Python's representation of negative numbers. First, we need to check if the answer is negative. We cannot just check to see if the answer is less than zero because our representation of the answer is not the same as Python's (since Python's have unlimited leading 1's). We are still treating our answer as if it only fits into 32 bits. A 32-bit signed integer is positive if the 32nd bit is a 0 and is negative if the 32nd bit is a 1. If we divide our mask (0xffffffff) by 2, we will get the binary number 0b0111111111111111111111111111111, which has 31 1's. This number is the greatest value we can have before the 32nd bit becomes a 1. Therefore, if our answer a > mask // 2, it is negative. Otherwise, it is positive and we can just return a itself. If the number is negative, we then need to convert it into Python's representation of negative numbers. To do so, we can XOR with the mask of 32 1's in order to flip the rightmost 32 bits, since XORing a bit with 1 flips the bit. We can then NOT the number in order to turn all of the leading 0's into 1's. For example, say that the answer is -3, and (....0000000) or (....1111111) denote leading 0's or 1's until the 32nd bit: Our representation of -3 in 32 bits: (...0000000)11111111111111111111111111111101 XOR with mask, aka flip rightmost 32 bits: (...0000000)00000000000000000000000000000010 NOT, aka flipping all bits: (...1111111)1111111111111111111111111111101 The result is Python's representation of -3, including an unlimited number of leading 1's. Overall, the code uses the same process as Neetcode's Java code, but with masking to get numbers into 32 bits and some manipulation to get those 32-bit numbers back into Python's representation before returning.
Man how tf do u know all this stuff. My code was failing to add -8 and -12, and here I was thinking what to do with -ve integers T_T. Having a non IT background surely makes it difficult
thank you man, i really appreciate the in depth explanation!!! this forced me to learn a little more about python today lol, i don't ever think about how integers are stored in languages and stuff so this was kind of cool to dive into.
@@lingzi6599 Unfortunately, since Python stores its integers in a different way than Java, all of this solution's parts are necessary. Python is as capable as any other language for handling binary operations, but in interviews (and beyond) it can be important to remember that Python represents its integers as having more than 32 bits. In many problems, remembering the bitmask of 0xffffffff can save you if you choose to use Python!
Thanks so much! You were the only one capable to clear in my mind these bitwise operations handled together. The use of the variable 'tmp', explaining its reason and its location on the code really clarified everything. Congrats for this awesome tutorial. God bless you. 🥰
Thanks for Explanation.....this code will work in Java but not worked same in C++ with negative number 🙃For C++ code-------- int getSum(int a, int b) { while(b!=0) { unsigned int temp=(a&b); a=a^b; b=(int)(temp
Hey man love your videos, your explanation is absolutely spot on: the default int val in python is too big. For people crying about python, mask to limit int size to 32bit (0xffffffff). Don't spread hate people :) cheers class Solution: def getSum(self, a: int, b: int) -> int: mask=0xffffffff while b & mask: a, b = (a ^ b), (a & b) 0 else a
Umm... I think I found out why Neet used Java. I converted it to Python but I got a time-limit exceed error. I guess Java automatically handles some of the edge cases (like when the input is a negative number ) :/ .
Lol I like seeing everyone complain about this being Java...People, I'm a C++/C# programmer and I watch his Python videos. It's not hard to convert one language to another...the point of these vids is to understand the algorithm and apply it!
I also code in java but I can easily understand c++/c and not to mention I'm here so I probably understand python too. It's not that difficult after some time you just start understanding them for some unknown reason (a good unknown reason).
Try implementing this code in c/c++ then come back . people are complaining because he chose the easy way out of this. and didn't wrote the code in c/c++.
@@dorondavid4698 ayyo never said cpp wasn't made for bit masking. Handling -ve number during bit operation is difficult in CPP. If u r so confident in ur coding skills , convert the code to CPP and run it for -ve number 👀.
Use whatever language you want we will still be here. Actually, I am a Java programmer. I didn't even know p of python but Neetcode is the only channel I watch for coding concepts. Most people don't understand programming languages are not important the purpose here is to understand the concept behind the questions. If you understand the concept you should be good enough to code it yourself if not then you don't know that programming language well enough. OMG, I want to write 2 pages regarding this but I think wise people will understand this.
You actually listened and coded in java🥺 Thanks man. Also I see so many requests to stick to python I hope you do what you feel right I am okay with any of the language because your explanation does the work for me💯
my solution in python, use log instead of binary: def getSum(self, a: int, b: int) -> int: # check one is zero and other is positive if min(a,b) == 0: return max(a,b) # if one is zero and other is negative if max(a,b) == 0: return min(a,b) # use property of ln, ln e^x = x x = math.e**a y = math.e**b z = x*y return int(math.log(z))
Coding in Java after coding in python for many days is too frustrating!!! So many semi-colon mistakes! Such a long method call for just print(debug purposes)..Kindly provide an alternative solution in Python too.
Doesn't work for negative integers, since left shift operator on negative integers is not standardized and would give different answers based on compiler.
LeetCode's official solution says that this is a popular question at Facebook and the creator of the Blind 75 list was a Facebook tech lead. That's probably one reason why it's on the list. It's also a tricky bit manipulation problem so it can help solidify your skills there. Since it's an annoying problem it's better to do it now than see it for the time in an interview.
You are NOT alone! I am pretty dependent on his channel too!!! I am a college student majoring at Stat with ZERO cs background so his videos means everything to me. Plus, I code in python only.
Can someone please explain why I can't use sum to solve this? class Solution: def missingNumber(self, nums: List[int]) -> int: #is 0 missing ? zero_missing = 0 in nums if not zero_missing: return 0 tot_sum = 0 for i in range(len(nums) + 1): tot_sum += i arr_sum = sum(nums) return tot_sum - arr_sum
People crying for python are going to be bad developers. It should be easy as f to port the solution in any language of your choice. I don't even understand python and I've been porting all solutions to C++ from this channel.
Normally I might agree with you but believe it or not sometimes the language does make a difference. This problem is quite a bit trickier in Python because integers have unlimited bit precision. It's not a simple port. To get the two's complement representation of a negative number you normally want to know how many bits your integer is.
just look at this overly complicated code in c++ by me.... just rubbish but it works perfectly fine in the constraints defined for a and b ...............................next line int getSum(int a, int b) { int ans = 0; int value = 1; if(a < 0 && b < 0) { a = abs(a); b = abs(b); value = -1; } if(a < 0 || b < 0) { if(a < 0) { if(abs(a) > b) { value = -1; } else { int t = b; b = a; a = t; } } else { if(abs(b) > a) { value = -1; int t = b; b = a; a = t; } } a = abs(a); b = abs(b); int c; int d; int i = 0; int e = 0; while(a != 0 || b != 0) { c = a&1; a >>= 1; d = b&1; b >>= 1; if(c + e - d == 0) { ans += 0; e = 0; } else if(c + e - d == 1) { ans += pow(2, i); } else if(c + e - d == -1) { ans += pow(2, i); e = -1; } i++; } } else { int c; int d; int e = 0; int i = 0; while(a != 0 || b != 0) { c = a&1; a >>= 1; d = b&1; b >>= 1; if(c + d + e == 1) { ans += pow(2, i); e = 0; } else if(c + d + e == 2) { if(a == 0 && b == 0) { ans += pow(2, i+1); } e = 1; } else if(c + d + e == 3) { if(a == 0 && b == 0) { ans += pow(2, i); ans += pow(2, i+1); } else { e = 1; ans += pow(2, i); } } i++; } } return ans*value; }
