EasyArrayStackSimulation

Baseball Game

1 approach, code in all languages

You are keeping a running record of scores for a game. You are given a list of string operations, each of which modifies the record in one of four ways.

An integer string x records a new score equal to x. The string "+" records a new score equal to the sum of the previous two scores. The string "D" records a new score equal to double the previous score. The string "C" cancels the previous score, removing it from the record.

Each operation is guaranteed to be valid: when "+", "D", or "C" is applied there are enough prior scores present. Return the sum of all scores remaining in the record after processing every operation.

Example 1

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

Output: 30

Record 5, then 2 -> [5,2]. "C" removes 2 -> [5]. "D" doubles to 10 -> [5,10]. "+" adds 5+10=15 -> [5,10,15]. Sum = 30.

Example 2

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

Output: 27

Record 5, -2, 4 -> [5,-2,4]. "C" removes 4 -> [5,-2]. "D" doubles -2 to -4 -> [5,-2,-4]. Record 9 -> [5,-2,-4,9]. "+" -> -4+9=5 -> [...,5]. "+" -> 9+5=14. Sum = 5-2-4+9+5+14 = 27.

Constraints

1 <= operations.length <= 1000
operations[i] is "C", "D", "+", or a string representing an integer in the range [-3 * 10^4, 3 * 10^4]
For "+", the record has at least two prior scores; for "C" and "D" it has at least one

Score Stack Simulation

Keep recorded scores on a stack, apply each operation directly on top, then sum what remains.

Loading animation…

Intuition

Every operation only ever references the most recent one or two recorded scores, which is exactly the behavior a stack captures. A plain number is pushed as a new score. "+" needs the two scores below it, so you read the top two and push their sum. "D" needs only the top, which you double and push. "C" undoes the last record, so you pop. Because each operation acts on the end of the record, the stack stays in sync with the true state, and the answer is simply the total of everything still on it.

Algorithm walkthrough

1Create an empty stack to hold the current recorded scores.
2For each operation: if it is "+", push the sum of the top two scores.
3If it is "D", push double the top score.
4If it is "C", pop the top score to cancel it.
5Otherwise parse the integer and push it as a new score.
6After all operations, return the sum of the values left on the stack.

Implementation

class Solution:
    def calPoints(self, operations: List[str]) -> int:
        stack = []
        for op in operations:
            if op == "+":
                stack.append(stack[-1] + stack[-2])
            elif op == "D":
                stack.append(2 * stack[-1])
            elif op == "C":
                stack.pop()
            else:
                stack.append(int(op))
        return sum(stack)

Time

O(n)

Space

O(n)

Each operation is handled in constant time, and the stack holds at most n scores.

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