LeetCode 0682. Baseball Game Solution in Java, C++, Python & More | Explanation + Code

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: Stack + Simulation

We can use a stack to simulate this process.

Traverse 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 operations.

PythonJavaC++GoTypeScriptRust

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)(code-box)

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(); } }(code-box)

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); } };(code-box)

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 }(code-box)

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); }(code-box)

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() } }(code-box)