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1235. Maximum Profit in Job Scheduling

Hi everyone, Today's post is little bit intriguing to me as it took me some time to get t the solution. I have been out of practice for a bit so I was not able to do it in the DP. I didn't want to copy the code. Though i understood the code after i checked its solution in the discussions. Anyways, i tried to do it with heap and with some thinking I was able to do it and with my surprise It performed real well. I would like to jump to the solution but first I would like to explain a little bit about heaps so you can get better understanding of the code. A heap is like a special type of list where the smallest (or largest) item is always at the front. Think of it like a priority line at a theme park, where the person with the highest priority goes to the front. There are two types of heaps: Min Heap: The smallest item is at the front. It's like standing in a line where the shortest person is always at the front. Max Heap: The largest item is at the front. It's like stand...

2125. Number of Laser Beams in a Bank

Hi Guys, Today is 3 January 2024 and below is the solution for problem of the day So, the question we have is Number of Laser Beams in a Bank .  In this we have to find total number of lasers. So just imagine it as we have to find number of lines connecting the dots.  so if there are 2 dots. we will have one line, if we have 3 dots we will get 3 lines, and so on... Now relate the above provided example, let's relate this to our problem statement.   We have 4 walls, first wall has 3 lasers, second wall has 0 lasers, third wall got 2 lasers and last wall has 1 laser. So as i suggested we need to connect dots so make sure to only join adjacent dots. So lasers from 1 and 3 wall  will connect to each other (6 lasers) and wall 3 and 4 will connect to each other (2 lasers). So total of 8 lasers, which is the required output anyways. class Solution : def numberOfBeams ( self , bank : List[ str ]) -> int : val = 0 # Initialize a variable `val` to store th...