Add Binary - Leetcode 67 - Python 

✏️ Have problem suggestions you would like me to solve? Feel free to comment them below :~)

an easy problem that can get boggled down in edge cases, really like some the tricks you used. great video as always!

Love the way you write the code. Concise. Love the loop of max of lengths. Great explanation as usual

how is this a easy problem :|

Can't we just use int(a[i]) to cast it to an integer?

Isn't the time complexity quadratic since the addition of the 'char' string to the 'res' string copies all the contents and creates a new string for every loop since strings in python are immutable?

Hi at 5:10, couldnt you use int() instead of ord() to cast the char?

"Let's go back to elementary school and add these together. What happens when we add 1 and 1?.....0?" 😂

thanks, you're a lifesaver when it comes to leetcode

You are a good teacher. Keep up the good work. Thank you!

For converting it to integer, can we not do int(a[i]) and int(b[i]) instead of ord(a[i])-ord("0") and ord(b[i])-ord("0")?

Thank you for explaining what happens when you add "11" and "11". I was completely lost but not anymore thanks to you!

This code is simply beautiful. Thanks.

def addBinary(self, a: str, b: str) -> str: a = int(a,2) b = int(b,2) res = a+b return bin(res)[2:]

What is the time complexity of the string concats in python(eg. char + res in the example above)

Amazing video and nice explanation!!!

we can use the same code for the question : leetcode 413(Add strings),just replace 2 by 10

Great Explanation! Might be cleaner to use the divmod() function instead of % and //. Eg. char, carry = divmod(total, 2) res = str(char) + res

The returned res should be reversed to get the original output. return res[::-1]

Why not just do bin(int(a, 2) + int(b, 2))[2:] ?

Wonderful Explanation 👍👍👍

Very well explained, but don't we need to return res[::-1]?

Convert the array into non non-decreasing array by replacing elements with its subarray sum if necessary. for example, input=[3, 8, 5, 2, 10], output=[3,8,17] input=[1,2,3,4,5],output=[1,2,3,4,5] input=[5,4,3,2,1],output=[5,10] input=[],output=[] Please solve this problem.

Youre an amazing teacher mr. neet!

Why are we usig "ord" instead of "int(str)" ?? @NeetCode

one thing I never knew before is that you can use f-strings to convert int to binary. def binary_conversion(num): return f'{num:b}' so binary_conversion(1000) will get a result of 1111101000

we could have appended 0* (a.size - b.size) to the shorter string, saving time in reversal of both the strings. Later reversing the final string

Why didn't you use int function on the values a[i] and b[i]?

thank u bro it is done and passed but I don't understand what I'm doing so what should I do to be better in Leetcode for everything I must find a tutorial so the tutorial killing me bro please help me out so more than a year I did a lot of HTML CSS javascript and 1000 of course and now python, why I can't, make a project and why I can't solve one easy problem in Leetcode or code war so I don't know bro and I fill I know everything but I can't code by my self must follow someone bro thx

is it normal to find "easy" questions hard? i mean i don't think most of the easy question are actually easy ngl :///

I havent run the code, but did you get a character short each time from your reversing method. a,b = reversed(a), reversed(b)

Nice explanation. Did you forget to reverse the result??

I was thinking that the interviewer would want the bit manipulation solution for this

Why use ord?? Since we know it can either be 0 or 1. Why not just a string comparison and assign. Like, if it's "0", then digitA is 0, else 1 Same for digitB Good implementation though

int(a[i]) worked succesfully. so why did u use ord?

i have a question for that carry = total//2 what if we add only 1 and 0 this still will give me a carry even though the result will be just 1 can anyone please help me with this?

should be medium

Why not just pad the shorter string with leading zeros?

Answer in 2 lines of code: c = bin(int(a,2) + int(b,2)) return c[2:]

I implemented same approach for Dart, it gave Time Limit Exceeded, but for java it worked :(

Python solution converted to JavaScript: var addBinary = function(a, b) { let totalSum = ''; let carry = 0; a = a.split('').reverse().join(''); b = b.split('').reverse().join(''); for (let i = 0; i < Math.max(a.length, b.length); i++){ const digitA = parseInt(i < a.length ? a[i] : 0); const digitB = parseInt(i < b.length ? b[i] : 0); const total = digitA + digitB + carry; const currentSum = total % 2; totalSum = currentSum + totalSum; carry = Math.floor(total / 2); } return carry ? 1 + totalSum : totalSum; };

can someone tell my error niothing is shown in the output class Solution { public: string solve(string a , string b){ int countfirst=0; int diff = a.size()-b.size(); // int countsecond=0; string s; int carry=0; for(int i=b.size()-1;countfirst > b.size();i--){ if (a[i + diff] + b[i] + carry == 0){ carry=0; s.push_back(0); } else if (a[i + diff] + b[i] + carry == 1){ carry=0; s.push_back(1); } else if (a[i + diff] + b[i] + carry == 2){ carry = 1; s.push_back(0); } else if (a[i + diff] + b[i] + carry == 3){ carry=1; s.push_back(1); } countfirst++; } for(int i=a.size()-b.size()-1;i>=0;i--){ if (a[i] + carry == 0){ carry=0; s.push_back(0); } else if (a[i] + carry == 1){ carry=0; s.push_back(1); } else if (a[i] + carry == 2){ carry=1; s.push_back(0); } } if (carry == 1){ s.push_back(1); } // reverse(s.begin() , s.end()); return s; } string addBinary(string a, string b) { if (a.size() >= b.size()){ string s = solve(a,b); return s; } string s=solve(b,a); return s; } };

idk if this counts as cheating but return str(bin(int(a,2)+int(b,2)))[2:]

U a binary God

Hello I hava a doubt can we use inbuild function like bin class Solution: def addBinary(self, a: str, b: str) -> str: return bin(int(a, 2) + int(b, 2))[2:] the code works pretty efficiently but is it the right way?

Why not bin to int -> add 2 ints-> int to bin and return

class Solution: def addBinary(self, a: str, b: str) -> str: a1=int(a,2) b1=int(b,2) sum1=a1+b1 return bin(sum1)[2:]

the second example showing wrong answer with this solution

How is this supposed to be easy 😭😭😭😭😭😭😭. Not understanding shit . Somebody tell me where to start ;

i was not even able to solve this one is this abnormal ?

Come on... this is really an "Easy" task?

class Solution: def addBinary(self, a: str, b: str) -> str: s1 = int(a, 2) s2= int(b, 2) s=s1+s2 return bin(s).replace("0b", "")

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