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.
    <ul>
	<li>Record a new score of <code>x</code>.</li>
</ul>
</li>
<li><code>&#39;+&#39;</code>.
<ul>
	<li>Record a new score that is the sum of the previous two scores.</li>
</ul>
</li>
<li><code>&#39;D&#39;</code>.
<ul>
	<li>Record a new score that is the double of the previous score.</li>
</ul>
</li>
<li><code>&#39;C&#39;</code>.
<ul>
	<li>Invalidate the previous score, removing it from the record.</li>
</ul>
</li>

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

Python Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13

class Solution:
    def calPoints(self, ops: List[str]) -> int:
        stk = []
        for op in ops:
            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)

Java Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20

class Solution {
    public int calPoints(String[] ops) {
        Deque<Integer> stk = new ArrayDeque<>();
        for (String op : ops) {
            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();
    }
}

C++ Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20

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

Go Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22

func calPoints(ops []string) int {
	var stk []int
	for _, op := range ops {
		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)
		}
	}
	ans := 0
	for _, score := range stk {
		ans += score
	}
	return ans
}

TypeScript Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16

function calPoints(ops: string[]): number {
    const stack = [];
    for (const op of ops) {
        const n = stack.length;
        if (op === '+') {
            stack.push(stack[n - 1] + stack[n - 2]);
        } else if (op === 'D') {
            stack.push(stack[n - 1] * 2);
        } else if (op === 'C') {
            stack.pop();
        } else {
            stack.push(Number(op));
        }
    }
    return stack.reduce((p, v) => p + v);
}

Rust Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23

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