Stack - Baseball Game

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.

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.

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.

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.

Solution 1 - Using stack

In this approach, we are going to use a Stack datastructure to store elements, and perform input operations.

Implementation steps:

First, we will create a variable stack as Stack<Integer> that we will use to baseball game scores.
Next, loop through input string array operations and for each operation do the following:
- If the operation is '+' : look at top two elements, compute the sum of these two elements and add it back to the stack.
- If the operation is 'D' : look at top element, double it's value and add it back to the stack.
- If the operation is 'C' : remove the top element from stack.
- If the operation is a number: add it to the stack.
Now, loop through elements of the stack and return its sum.

package io.cscode.algorithms.stack; import java.util.Stack; public class BaseballGame { static int calPoints(String[] operations) { Stack<Integer> stack = new Stack<>(); for (String op: operations) { if("+".equals(op)) { // record a new score, sum of two previous scores if(stack.size()>=2) { int x1 = stack.pop(); int x2 = stack.peek(); int newRecord = x1+x2; stack.push(x1); stack.push(newRecord); } } else if ("D".equals(op)) { // record a new score, that is double of previous element if(stack.size()>=1) { stack.push(stack.peek() *2); } } else if ("C".equals(op)) { //invalidate previous if(stack.size()>=1) { stack.pop(); } } else { stack.push(Integer.parseInt(op)); } } int total = 0; for(Integer num: stack) { total += num; } return total; } public static void main(String[] args) { System.out.println(calPoints(new String[]{"5","2","C","D","+"})); System.out.println(calPoints(new String[]{"5","-2","4","C","D","9","+","+"})); System.out.println(calPoints(new String[]{"1","C"})); } }

Complexity Analysis:

Time complexity: Above code runs in O(n) time where n is the length of input string arrayoperations.
Space complexity: O(n), since we are storing operation elements into a Stack.

Above implementations source code can be found at GitHub link for Java code

Constraints:

Contents

Implementation steps:

Complexity Analysis: