The brute force solution is easier to understand because it directly checks how far each bar can expand. The optimized monotonic stack solution is more efficient because it avoids repeated scanning.

The main trick in this problem is realizing that we do not need to simulate the cars moving. We only need to calculate each car’s arrival time and process the cars from closest to farthest from the destination.

We need to return an array result, where result[i] tells us how many days we have to wait after day i to get a warmer temperature. The monotonic stack solution is more efficient because it keeps track of unresolved days and updates them as soon as a warmer temperature appears.

This problem is a great example of how useful stacks are when evaluating expressions. Since Reverse Polish Notation places operators after operands, a stack helps us easily keep track of the values needed for each operation.

The key insight isn’t about finding the minimum quickly—it’s about remembering the minimum as the stack evolves. By maintaining a second stack that mirrors the main stack and stores the minimum value at every level, we eliminate the need to recompute the minimum after every pop().

You are keeping the scores for a baseball game with a unique set of rules. At the beginning of the game, you start with an empty record. You're given a list of strings operations, where each element represents an operation to apply to the record.