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Baseball Game

Asked at:AmazonGoogle

You're keeping score for a baseball game with special operations. Each operation is either an integer score, '+' (sum of the last two scores), 'D' (double the last score), or 'C' (remove the last score). Return the total of all recorded scores.

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Problem

You are keeping the scores for a baseball game with strange rules. You are given a list of strings operations. Return the sum of all scores after applying all operations.

Input

An array of strings `ops` where each element is an integer (as a string), '+', 'D', or 'C'.

Output

The sum of all recorded scores after processing every operation.

Examples

Input: ops = ["5","2","C","D","+"]

Output: 30

"5" → [5]. "2" → [5,2]. "C" → [5]. "D" → [5,10]. "+" → [5,10,15]. Sum = 30.

Input: ops = ["5","-2","4","C","D","9","+","+"]

Output: 27

"5"→[5], "-2"→[5,-2], "4"→[5,-2,4], "C"→[5,-2], "D"→[5,-2,-4], "9"→[5,-2,-4,9], "+"→[5,-2,-4,9,5], "+"→[5,-2,-4,9,5,14]. Sum=27.

The brute-force approach

There is no brute force here, a stack (or list) is already the natural O(n) solution. Processing each operation once is the only reasonable approach.

scores = []

for op in ops:
    if op == 'C':
        scores.pop()
    elif op == 'D':
        scores.append(scores[-1] * 2)
    elif op == '+':
        scores.append(scores[-1] + scores[-2])
    else:
        scores.append(int(op))

return sum(scores)

This is already O(n). The stack approach is optimal.

Time: O(n)Space: O(n)

Spotting the pattern

This is a Stack problem. The key question to ask yourself:

When an operation needs to reference or undo the most recent score, what does that tell me about how to store them?

Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.

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