Introduction

In this blog post, we will explore how to solve the LeetCode problem "682" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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 ith 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.

Explanation

• Use a stack to efficiently maintain the record of scores. This allows for O(1) operations for adding, removing (invalidating), and accessing the last two scores.
  • Iterate through the operations, performing the appropriate action based on the operation type. Integer operations are pushed to the stack. Other operations manipulate the stack accordingly.
  • After processing all operations, sum the elements remaining in the stack to get the final score.

• Runtime Complexity: O(n), where n is the number of operations. Storage Complexity: O(n) in the worst case where no 'C' operations are encountered.

Code

    def calPoints(ops):
    """
    Calculates the sum of scores after applying a series of operations.

    Args:
        ops: A list of strings representing the operations.

    Returns:
        The sum of the scores on the record after applying all operations.
    """
    record = []
    for op in ops:
        if op == "+":
            record.append(record[-1] + record[-2])
        elif op == "D":
            record.append(2 * record[-1])
        elif op == "C":
            record.pop()
        else:
            record.append(int(op))

    return sum(record)