BASIC CALCULATOR III | LEETCODE # 772 | PYTHON SOLUTION

[@landocodes]

My Meta phone screen is tomorrow and I think I found a video for every Meta question I ran into (except for maybe 1 that I can't remember).. Just wanted to say thanks again for the hard work man. I'll be recommending this channel to everyone I see prepping for Meta and hopefully I'll pass to the onsite and get more utility out of it.

[@crackfaang]

Thanks for the kind words and best of luck with the phone screen. I'm sure you'll smash it!

[@landocodes]

Thanks man I passed!@@crackfaang Now it's time to check out your System Design playlist :)

[@aman4434]

how was it? Mine was supposed to be tomorrow and its now gotten rescheduled lol. I'm watching this channel too! But still nervous...

[@landocodes]

@@aman4434it went well! I heard that I passed the next day. If you’ve done the top meta questions, and get a good interviewer, then it should be a breeze! Just make sure to communicate your ideas and have a good understanding of the solution you’ve chosen.

[@landocodes]

It went well! I was a bit nervous for a second but my interviewer was amazing. She calmed me down and told me to just let my thoughts out. Once I did that I got back on track and everything went smoothly! Heard back the next morning that I passed! As long as you have the majority of the medium/easy top 100 meta questions complete then I'm sure you'll do fine! Good luck :)@@aman4434

[@ziyiyang4550]

Great solution! It's the most clean one I've seen. Implementing DFS using stack is definitely a pain lol

[@davidoh6342]

in line 34, if op=s[i], will it become ")" ???

[@def__init]

yes but it's fine because if the string still has more after ")", the following MUST be a new OP, like 2*(...)/3 from the examples. And when we process the "/" after, which will first call helper with op being ")", which will do nothing. Then business as usual, num back to 0 and op set to "/"

[@amadoufall6237]

Good job on your videos. Could you please do this one next? 1267. Count Servers that Communicate. I would really appreciate to hear your explanations

[@crackfaang]

Thanks for the kind words and your support. I’ll add that question to my work queue and make sure you subscribe so you don’t miss the upload

[@kn7419533]

Another solution to reuse Cracking FAANG's code for 227. Basic Calculator II without using stack, we can do class Solution { public: int calculate(string s) { int i = 0; return calculateHelper(s, i); } int calculateHelper(string& s, int& i) { int cur = 0, prev = 0, res = 0; char cur_operation = '+'; while (i < s.length()) { char cur_char = s[i]; if (isdigit(cur_char)) { cur = 0; // Extract the full number while (i < s.length() && isdigit(s[i])) { cur = cur * 10 + (s[i] - '0'); // Convert character to integer i++; } i--; // Adjust index to process the next character properly // Perform operation based on the previous operator if (cur_operation == '+') { res += cur; prev = cur; } else if (cur_operation == '-') { res -= cur; prev = -cur; } else if (cur_operation == '*') { res -= prev; prev = prev * cur; res += prev; } else if (cur_operation == '/') { res -= prev; prev = prev / cur; res += prev; } } else if (cur_char == '(') { i++; // Move past '(' cur = calculateHelper(s, i); // Recursively evaluate parentheses*** // Perform the operation (same as the previous block) if (cur_operation == '+') { res += cur; prev = cur; } else if (cur_operation == '-') { res -= cur; prev = -cur; } else if (cur_operation == '*') { res -= prev; prev = prev * cur; res += prev; } else if (cur_operation == '/') { res -= prev; prev = prev / cur; res += prev; } } else if (cur_char == ')') { // Return result if we encounter a closing parenthesis return res; } else if (cur_char != ' ') { // Store the operator for the next calculation cur_operation = cur_char; } // else --> ' ' (skip it) i++; } return res; }

[@garimagupta4208]

great explanation