Baseball Game

Companies:
  • Amazon Interview Questions

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Companies: Amazon

You're now a baseball game point recorder.

Given a list of strings, each string can be one of the 4 following types:

  • Integer (one round's score): Directly represents the number of points you get in this round.
  • "+" (one round's score): Represents that the points you get in this round are the sum of the last two valid round's points.
  • "D" (one round's score): Represents that the points you get in * this round are the doubled data of the last valid round's points.
  • "C" (an operation, which isn't a round's score): Represents the last valid round's points you get were invalid and should be removed.
  • Each round's operation is permanent and could have an impact on the round before and the round after.

You need to return the sum of the points you could get in all the rounds.

Example 1:

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

Output: 30

Explanation:

Round 1: You could get 5 points. The sum is: 5.

Round 2: You could get 2 points. The sum is: 7.

Operation 1: The round 2's data was invalid. The sum is: 5.

Round 3: You could get 10 points (the round 2's data has been removed). The sum is: 15.

Round 4: You could get 5 + 10 = 15 points. The sum is: 30.

Note:

  • The size of the input list will be between 1 and 1000.
  • Every integer represented in the list will be between -30000 and 30000.

Solution:

It is obvious that we need to remember the old rounds and their scores as well, so stack is the most efficient solution for this type of problem. Through stack we can also easily remove the previous round if operation C is performed.

Implementation:

Java:

class Solution1 {
    public int calPoints(String[] ops) {
        Stack<Integer> stack = new Stack<>();
        int sum = 0;
        int firstLast = Integer.MIN_VALUE;
        int secondLast = Integer.MIN_VALUE;
        for (String op : ops) {
            if (op.equals("+")) {
                if (!stack.isEmpty()) {
                    firstLast = stack.pop();
                }
                if (!stack.isEmpty()) {
                    secondLast = stack.pop();
                }
                int thisRoundPoints = firstLast + secondLast;

                if (secondLast != Integer.MIN_VALUE) {
                    stack.push(secondLast);
                }
                if (firstLast != Integer.MIN_VALUE) {
                    stack.push(firstLast);
                }
                stack.push(thisRoundPoints);
                sum += thisRoundPoints;

                firstLast = Integer.MIN_VALUE;
                secondLast = Integer.MIN_VALUE;
            } else if (op.equals("D")) {
                if (!stack.isEmpty()) {
                    int thisRoundPoints = stack.peek() * 2;
                    stack.push(thisRoundPoints);
                    sum += thisRoundPoints;
                }
            } else if (op.equals("C")) {
                if (!stack.isEmpty()) {
                    int removedData = stack.pop();
                    sum -= removedData;
                }
            } else {
                Integer val = Integer.parseInt(op);
                sum += val;
                stack.push(val);
            }
        }
        return sum;
    }
}

Complexity Analysis:

  • Time Complexity: O(N).

  • Space Complexity: O(N).

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