How do you solve Baseball Game on LeetCode?

Baseball Game (LeetCode #682) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a stack allows efficient management of scores.

What is the brute force solution for Baseball Game?

The brute force approach for Baseball Game is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves simulating the operations one by one and maintaining a list of scores. We will process each operation sequentially, updating our scores list accordingly. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Baseball Game?

Baseball Game uses the following concepts: Array, Stack, Simulation. The recommended patterns to study are: Stack, Array.

What are the interview tips for Baseball Game?

Explain your thought process while coding. Discuss edge cases, such as consecutive 'C' operations. Practice similar problems to get comfortable with stack operations.

What are common mistakes when solving Baseball Game?

Not handling the 'C' operation correctly. Forgetting to check the stack size before accessing elements.

#682

Baseball Game

Easy

ArrayStackSimulationStackArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a stack to efficiently manage the scores. This allows us to handle operations in constant time, making the entire process faster and more efficient.

⚙️

Algorithm

4 steps

1Step 1: Initialize an empty stack to store scores.
2Step 2: Iterate through each operation in the input list.
3Step 3: For each operation, update the stack based on the operation type (integer, '+', 'D', or 'C').
4Step 4: At the end, sum all the scores in the stack and return the total.

solution.py14 lines

1# Full working Python code
2
3def calPoints(ops):
4    stack = []
5    for op in ops:
6        if op == 'C':
7            stack.pop()
8        elif op == 'D':
9            stack.append(2 * stack[-1])
10        elif op == '+':
11            stack.append(stack[-1] + stack[-2])
12        else:
13            stack.append(int(op))
14    return sum(stack)

ℹ

Complexity note: The time complexity is O(n) because we process each operation exactly once. The space complexity is O(n) due to the stack storing scores.

1Using a stack allows efficient management of scores.
2Operations can be handled in constant time.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.