LeetCode #682 — EASY

Baseball Game

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

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.

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Stack

Example 1

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

Example 2

["5","-2","4","C","D","9","+","+"]

Example 3

["1","C"]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #682: Baseball Game
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();
    }
}

// Accepted solution for LeetCode #682: Baseball Game
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
}

# Accepted solution for LeetCode #682: Baseball Game
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)

// Accepted solution for LeetCode #682: Baseball Game
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()
    }
}

// Accepted solution for LeetCode #682: Baseball Game
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);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.