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682. Baseball Game

Description

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.

Solutions

Solution 1: Stack + Simulation

We can use a stack to simulate this process.

Traverse $\textit{operations}$, for each operation:

  • If it is +, add the top two elements of the stack and push the result onto the stack;
  • If it is D, multiply the top element of the stack by 2 and push the result onto the stack;
  • If it is C, pop the top element of the stack;
  • If it is a number, push the number onto the stack.

Finally, sum all the elements in the stack to get the answer.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of $\textit{operations}$.

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class Solution:
    def calPoints(self, operations: List[str]) -> int:
        stk = []
        for op in operations:
            if op == "+":
                stk.append(stk[-1] + stk[-2])
            elif op == "D":
                stk.append(stk[-1] << 1)
            elif op == "C":
                stk.pop()
            else:
                stk.append(int(op))
        return sum(stk)

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class Solution {
    public int calPoints(String[] operations) {
        Deque<Integer> stk = new ArrayDeque<>();
        for (String op : operations) {
            if ("+".equals(op)) {
                int a = stk.pop();
                int b = stk.peek();
                stk.push(a);
                stk.push(a + b);
            } else if ("D".equals(op)) {
                stk.push(stk.peek() << 1);
            } else if ("C".equals(op)) {
                stk.pop();
            } else {
                stk.push(Integer.valueOf(op));
            }
        }
        return stk.stream().mapToInt(Integer::intValue).sum();
    }
}

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class Solution {
public:
    int calPoints(vector<string>& operations) {
        vector<int> stk;
        for (auto& op : operations) {
            int n = stk.size();
            if (op == "+") {
                stk.push_back(stk[n - 1] + stk[n - 2]);
            } else if (op == "D") {
                stk.push_back(stk[n - 1] << 1);
            } else if (op == "C") {
                stk.pop_back();
            } else {
                stk.push_back(stoi(op));
            }
        }
        return accumulate(stk.begin(), stk.end(), 0);
    }
};

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func calPoints(operations []string) (ans int) {
    var stk []int
    for _, op := range operations {
        n := len(stk)
        switch op {
        case "+":
            stk = append(stk, stk[n-1]+stk[n-2])
        case "D":
            stk = append(stk, stk[n-1]*2)
        case "C":
            stk = stk[:n-1]
        default:
            num, _ := strconv.Atoi(op)
            stk = append(stk, num)
        }
    }
    for _, x := range stk {
        ans += x
    }
    return
}

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function calPoints(operations: string[]): number {
    const stk: number[] = [];
    for (const op of operations) {
        if (op === '+') {
            stk.push(stk.at(-1)! + stk.at(-2)!);
        } else if (op === 'D') {
            stk.push(stk.at(-1)! << 1);
        } else if (op === 'C') {
            stk.pop();
        } else {
            stk.push(+op);
        }
    }
    return stk.reduce((a, b) => a + b, 0);
}

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impl Solution {
    pub fn cal_points(operations: Vec<String>) -> i32 {
        let mut stk = vec![];
        for op in operations {
            match op.as_str() {
                "+" => {
                    let n = stk.len();
                    stk.push(stk[n - 1] + stk[n - 2]);
                }
                "D" => {
                    stk.push(stk.last().unwrap() * 2);
                }
                "C" => {
                    stk.pop();
                }
                n => {
                    stk.push(n.parse::<i32>().unwrap());
                }
            }
        }
        stk.into_iter().sum()
    }
}

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