Is Baseball Game easy or hard?

Baseball Game is categorized as an Easy problem. The challenge mainly involves correctly simulating operations and choosing a stack to manage previous scores. Developers comfortable with basic array traversal and stack operations can usually solve it quickly.

Baseball Game Python/Java solution

In Python, a list is typically used as a stack with append() and pop() operations. In Java, developers commonly use a Stack or ArrayDeque to store scores. Both implementations follow the same O(n) simulation logic and maintain previous round results for future operations.

How to solve Baseball Game in O(n)?

Iterate through the operations array once and maintain a stack of valid round scores. Push numbers directly, pop the last score for "C", push double the last score for "D", and push the sum of the last two scores for "+". After processing all operations, sum the stack values to get the final score.

What is the best approach for Baseball Game?

The best approach is a stack-based simulation. Each valid round score is pushed onto a stack, while operations like "C", "D", and "+" manipulate the most recent scores. This matches the problem's rule that operations depend on the latest rounds. The solution runs in O(n) time with O(n) space.

Is Baseball Game asked at Google/Amazon/Meta?

Baseball Game is commonly used in coding interviews at companies that test fundamental data structure knowledge. It appears frequently in interview prep lists and practice platforms because it evaluates stack usage, array iteration, and simulation skills.

What data structure is used in Baseball Game?

A stack is the primary data structure used to track previous round scores. The stack allows efficient access to the most recent values, which is required for operations like removing the last score, doubling it, or summing the last two scores.

What is the time complexity of Baseball Game?

The optimal solution runs in O(n) time where n is the number of operations. Each operation is processed exactly once, and stack operations such as push, pop, and top access take constant time. The space complexity is O(n) because previous scores are stored for later operations.

Description Solution Video 7:02

682. Baseball Game

Easy80.8% acceptance6 companies

You are keeping the scores for a baseball game with strange rules. At the beginning of the game, you start with an empty record.

You are given a list of strings operations, where operations[i] is the i^th operation you must apply to the record and is one of the following:

An integer x.
- Record a new score of x.
'+'.
- Record a new score that is the sum of the previous two scores.
'D'.
- Record a new score that is the double of the previous score.
'C'.
- Invalidate the previous score, removing it from the record.

Return the sum of all the scores on the record after applying all the operations.

The test cases are generated such that the answer and all intermediate calculations fit in a 32-bit integer and that all operations are valid.

Example 1:

Input: ops = ["5","2","C","D","+"]
Output: 30
Explanation:
"5" - Add 5 to the record, record is now [5].
"2" - Add 2 to the record, record is now [5, 2].
"C" - Invalidate and remove the previous score, record is now [5].
"D" - Add 2 * 5 = 10 to the record, record is now [5, 10].
"+" - Add 5 + 10 = 15 to the record, record is now [5, 10, 15].
The total sum is 5 + 10 + 15 = 30.

Example 2:

Input: ops = ["5","-2","4","C","D","9","+","+"]
Output: 27
Explanation:
"5" - Add 5 to the record, record is now [5].
"-2" - Add -2 to the record, record is now [5, -2].
"4" - Add 4 to the record, record is now [5, -2, 4].
"C" - Invalidate and remove the previous score, record is now [5, -2].
"D" - Add 2 * -2 = -4 to the record, record is now [5, -2, -4].
"9" - Add 9 to the record, record is now [5, -2, -4, 9].
"+" - Add -4 + 9 = 5 to the record, record is now [5, -2, -4, 9, 5].
"+" - Add 9 + 5 = 14 to the record, record is now [5, -2, -4, 9, 5, 14].
The total sum is 5 + -2 + -4 + 9 + 5 + 14 = 27.

Example 3:

Input: ops = ["1","C"]
Output: 0
Explanation:
"1" - Add 1 to the record, record is now [1].
"C" - Invalidate and remove the previous score, record is now [].
Since the record is empty, the total sum is 0.

Constraints:

1 <= operations.length <= 1000
operations[i] is "C", "D", "+", or a string representing an integer in the range [-3 * 10⁴, 3 * 10⁴].
For operation "+", there will always be at least two previous scores on the record.
For operations "C" and "D", there will always be at least one previous score on the record.