Baseball Game

Identifying Problem Isomorphism

“Baseball Game” can be mapped to “Evaluate Reverse Polish Notation”.

Both problems involve processing a list of strings and applying certain operations depending on the string’s value.

In “Baseball Game”, you calculate the player’s total points. Each string in the input list can represent an integer (which means the player gets that many points in the round), “D” (which means the player gets double the score of the previous round), “C” (which means the last score is invalid and should be removed), or “+” (which means the player gets the sum of the last two valid round’s points).

Similarly, in “Evaluate Reverse Polish Notation”, each string in the list can represent an integer, or it can represent one of the four operators “+”, “-”, “*”, or “/”. If it’s an operator, it means that you need to apply this operation to the last two numbers in the stack and push the result back into the stack.

“Evaluate Reverse Polish Notation” is simpler, as it deals with less operation types (4 instead of the 5 in “Baseball Game”, considering the individual integers as unique operations). It also does not involve removing elements from the stack (which corresponds to the “C” operation in the “Baseball Game”).

“Evaluate Reverse Polish Notation” might require more careful handling of corner cases, as it involves division, which could lead to zero division errors if not handled correctly.

This problem involves using a stack data structure and understanding the rules of the game to simulate the scoring system. Below are 10 problems for preparing to solve this problem:

  1. 20. Valid Parentheses: A classic problem that involves using a stack to check for balanced parentheses.

  2. 155. Min Stack: This problem requires the implementation of a stack that supports push, pop, top, and retrieving the minimum element in constant time.

  3. 225. Implement Stack using Queues: This problem is a good exercise in understanding the difference between stack and queue data structures.

  4. 232. Implement Queue using Stacks: This problem is the opposite of problem 225 and can be a good follow-up.

  5. 682. Baseball Game: This problem requires you to understand and implement the rules of a simple game using a stack.

  6. 496. Next Greater Element I: This problem can help you understand how to use stacks in array manipulation problems.

  7. 394. Decode String: This problem involves the use of a stack to handle nested string encodings.

  8. 716. Max Stack: Similar to the Min Stack problem, but requires tracking the maximum element in the stack.

  9. 844. Backspace String Compare: This problem involves simulating the backspace operation, which can be solved using stacks.

  10. 1047. Remove All Adjacent Duplicates In String: This problem requires using a stack to remove all adjacent duplicates in a string.

These cover how to use stacks to solve various problems, and how to interpret and implement rules in a problem statement, which are essential for solving the “Baseball Game” problem.

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    def calPoints(self, ops: List[str]) -> int:
        stack = []
        s = 0
        for op in ops:
            if op == "C":
                s -= stack.pop()
            elif op == 'D':
                stack.append(2*stack[-1])
                s += stack[-1]
            elif op == "+":
                stack.append(stack[-1] + stack[-2])
                s += stack[-1]
            else:
                stack.append(int(op))
                s += stack[-1]
        return s

Problem Classification

‘What’ Components:

  1. Input: You are given a list of strings operations which represents the operations you must apply to the record. Each operation could either be an integer ‘x’, ‘+’, ‘D’, or ‘C’.

  2. Output: You need to return the sum of all the scores on the record after applying all the operations.

  3. Data Structures/Representation: The problem involves maintaining a record of scores which can be represented as a list or stack data structure in programming. Scores are added, summed, doubled, or removed based on the operations.

  4. Constraints: The problem constraints mention that all operations are valid, and all intermediate calculations and the final answer fit in a 32-bit integer.

Classification of the Problem:

  1. Algorithmic Problem: The problem involves applying a series of operations sequentially, which can be viewed as an algorithm.

  2. Data Structure Problem: The solution requires a dynamic record of scores that needs to be updated based on the operations, which can be viewed as a data structure problem.

This problem involves the use of a data structure (e.g., list, stack) to store the score history and an algorithm to process each operation in the input array. For each operation, we modify the data structure accordingly: if the operation is an integer, we append it; if it’s ‘+’, we add the last two scores and append the sum; if it’s ‘D’, we double the last score and append it; if it’s ‘C’, we remove the last score. Finally, we return the sum of all scores in the data structure.

Clarification Questions

What are the clarification questions we can ask about this problem?

Identifying Problem Isomorphism

Can you help me with finding the isomorphism for this problem?

Which problem does this problem map to the corresponding isomorphic problem on Leetcode ?

Language Agnostic Coding Drills

  1. Dissection of Code into Distinct Concepts

a) Variable and List Initialization: The code begins by initializing a list (stack) and an integer (s) to keep track of the scores and their total.

b) Looping: The code then enters a for loop to iterate through the input operations.

c) Conditionals: Inside the loop, there are several conditional checks to determine the appropriate operation to perform based on the current string from the operations list.

d) Stack Operations: The code utilizes stack operations such as push (append) and pop.

e) Arithmetic Operations: The code performs arithmetic operations like addition, subtraction, and multiplication.

f) Type Conversion: The code converts the string to an integer when the operation is a score.

g) Returning Values: Finally, the code returns the sum of the scores.

  1. Coding Concepts/Drills Ordered by Difficulty

a) Variable and List Initialization: A basic concept of declaring and initializing variables.

b) Looping: Iterating over elements in a list is a fundamental concept that involves control flow.

c) Arithmetic Operations: Performing addition, subtraction, and multiplication is a basic concept.

d) Conditionals: Using if-elif-else statements to control flow based on different conditions. It requires understanding of comparison and logical operators.

e) Type Conversion: Converting data types is a commonly used feature. Here, a string is converted to an integer, which is relatively straightforward.

f) Stack Operations: Pushing and popping elements from a list as if it were a stack. This involves understanding how a stack data structure works.

g) Returning Values: The concept of a function returning a result. It’s a fundamental concept in function definition but could be difficult for beginners to grasp.

  1. Problem-Solving Approach
  • Start by initializing a list (acting as a stack) and a variable to hold the total score.
  • Iterate through each operation in the input list.
  • For each operation, perform a check to determine what kind of operation it is.
  • If it’s ‘C’, pop the last score from the stack and subtract it from the total.
  • If it’s ‘D’, push onto the stack a score that’s double the last score and add it to the total.
  • If it’s ‘+’, push onto the stack a score that’s the sum of the last two scores and add it to the total.
  • If it’s an integer, push it onto the stack and add it to the total.
  • After all operations have been processed, return the total score.
  • Each of these steps corresponds to a coding drill and they’re combined in the given order to solve the problem.

Targeted Drills in Python

  1. Python-based Coding Drills for Each Identified Concept:

a) Variable and List Initialization:

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s = 0
stack = []

In this drill, we initialize a variable s to 0 and a list stack to be an empty list.

b) Looping:

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for i in range(10):
    print(i)

This drill introduces looping. Here, we print the numbers from 0 to 9 using a for loop.

c) Arithmetic Operations:

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a = 5
b = 3
print(a+b)  # Addition
print(a-b)  # Subtraction
print(a*b)  # Multiplication

This drill introduces basic arithmetic operations in Python.

d) Conditionals:

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a = 5
if a > 3:
    print("Greater than 3")
elif a < 3:
    print("Less than 3")
else:
    print("Equal to 3")

This drill introduces conditional checks. It prints a message based on the value of a.

e) Type Conversion:

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s = "123"
print(int(s))

This drill converts a string to an integer.

f) Stack Operations:

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stack = []
stack.append(1)  # Push
print(stack[-1])  # Top
stack.pop()  # Pop

This drill introduces basic operations with a stack implemented as a Python list.

g) Returning Values:

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def add(a, b):
    return a + b
print(add(3, 4))

This drill introduces the concept of a function returning a result.

  1. Problem-specific Concepts:

This problem doesn’t involve any concepts that are completely unique to it. However, the application of stack operations, looping, conditionals, and arithmetic operations is specific to this problem context. The ability to interpret and implement the given operation strings is critical to solving the problem.

  1. Integration of the Coding Drills

The above drills can be integrated together to solve the problem in the following manner:

  • Start by initializing an empty list as a stack and an integer to keep track of the total score (Drill a).
  • Loop over the input operation list (Drill b).
  • Inside the loop, use a series of conditional checks to determine what kind of operation the current string is (Drill d).
  • If the operation is ‘C’, pop the last score from the stack and subtract it from the total (Drills f and c).
  • If the operation is ‘D’, calculate the double of the last score, push it onto the stack and add it to the total (Drills f and c).
  • If the operation is ‘+’, calculate the sum of the last two scores, push it onto the stack and add it to the total (Drills f and c).
  • If the operation is an integer, convert it to an integer, push it onto the stack and add it to the total (Drills e, f, and c).
  • After the loop, return the total score (Drill g).

Here’s a simple Python solution for this problem that uses a list as a stack to keep track of the scores. The time complexity for this solution is O(n), where n is the length of the operations array.

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

In this code:

  1. We initialize an empty list stack to keep track of the scores.
  2. We iterate over each operation in operations:
    • If the operation is ‘C’, we remove the last score by using pop().
    • If the operation is ‘D’, we double the last score and append it to the stack.
    • If the operation is ‘+’, we append the sum of the last two scores to the stack.
    • Otherwise, the operation is a string representing an integer. We convert it to an integer and append it to the stack.
  3. After we finish processing all operations, we return the sum of all scores in stack.

Q&A

I am getting NumberFormatException for int temp = Integer.parseInt(operations[i]) Can any one help me out how to resolve it.

A NumberFormatException is thrown by Java methods that attempt to convert a String into a number, like Integer.parseInt(), when the String cannot be properly parsed into a number. In this case, the NumberFormatException is thrown because the code attempts to parse non-numeric strings (“C”, “D”, “+”) into integers.

To avoid the NumberFormatException, you can add checks for non-numeric strings in your code. Here’s a Java solution for your problem that handles the different types of operations correctly:

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class Solution {
    public int calPoints(String[] operations) {
        Stack<Integer> stack = new Stack<>();
        for (String op : operations) {
            if (op.equals("C")) {
                stack.pop();
            } else if (op.equals("D")) {
                stack.push(2 * stack.peek());
            } else if (op.equals("+")) {
                int top = stack.pop();
                int newtop = top + stack.peek();
                stack.push(top);
                stack.push(newtop);
            } else {
                stack.push(Integer.parseInt(op));
            }
        }
        int sum = 0;
        for(int score : stack) {
            sum += score;
        }
        return sum;
    }
}

In this code:

  1. We use a Stack to keep track of the scores.
  2. We iterate over each operation:
    • If the operation is ‘C’, we remove the last score by calling pop().
    • If the operation is ‘D’, we double the last score and push it onto the Stack.
    • If the operation is ‘+’, we add the last two scores and push the sum onto the Stack.
    • Otherwise, the operation is a String representing an integer. We parse it to an integer and push it onto the Stack.
  3. Finally, we return the sum of all scores in the Stack.

please help

for case 20 : ["-60",“D”,"-36",“30”,“13”,“C”,“C”,"-33",“53”,“79”] my output is -87 expected is:-117 but when i do the ansewer on paper my answer comes to be

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import java.util.*;
public class Solution {
	public static int calPoints(String[] operations) {
		int count=0;
		for(int i=0;i < operations.length;i++){
			if(operations[i].equals("C")){
				operations[i]=null;
				operations[i-1]=null;
				count+=1;
			}
		}
		String[] op=new String[operations.length-count];
		int r=0;
		for(int k=0;k < operations.length;k++){
			if(operations[k]==null){
				continue;
			}else{
				op[r]=operations[k];
				r++;
			}
		}
		int[] arr=new int[operations.length-count];
		int pos=0;
		for(int j=0;j < op.length;j++){
			if(op[j]==null){
				continue;
			}

			if(op[j].equals("D")){
				int vald=(arr[j-1]);
				arr[pos]=vald*2;
				pos++;

			}else if(op[j].equals("+")){
				int val1a=(arr[j-2]);
				int val2a=(arr[j-1]);
				arr[pos]=val1a+val2a;
				pos++;
			}else{
				int okval=Integer.parseInt(op[j]);
				arr[pos]=okval;
				pos++;
			}
		}

		int sumof=0;
		for(int o=0;o<arr.length;o++){
			sumof+=arr[o];
		}
     
		return sumof;
	}
}

The problem in your code is that when you encounter “C”, you just nullify the current operation and the previous operation, which does not correctly represent the described behavior. Instead, you should think about “C” as removing the last valid score from the record.

Also, when “D” and “+” occur, you’re accessing the array at index “j-1” and “j-2”, but you should access the array at index “pos-1” and “pos-2” because “pos” tracks the number of valid scores.

Here is the corrected version of your code:

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import java.util.*;
public class Solution {
    public static int calPoints(String[] operations) {
        int[] arr = new int[operations.length];
        int pos = 0;
        
        for(int j=0;j<operations.length;j++){
            if(operations[j].equals("D")){
                arr[pos] = 2 * arr[pos-1];
                pos++;
            } else if(operations[j].equals("+")){
                arr[pos] = arr[pos-1] + arr[pos-2];
                pos++;
            } else if(operations[j].equals("C")){
                arr[pos-1] = 0; // Invalidate the last score
                pos = pos > 0 ? pos-1 : 0; // Make sure not to go negative
            } else {
                arr[pos] = Integer.parseInt(operations[j]);
                pos++;
            }
        }
      
        int sumof = 0;
        for(int o = 0; o < arr.length; o++){
            sumof += arr[o];
        }
     
        return sumof;
    }
}

In this corrected version, we keep track of the valid scores in the arr array and use the pos variable to keep track of the position of the last valid score. We then handle each operation according to the problem’s description.

Instead of taking Integer in the stack, why can’t we take String? Anyone?

The stack is used to store the scores, which are integers. The operations “D” and “+” require numerical calculations, specifically doubling the last score or adding the last two scores. This is why it’s more convenient to store the scores as integers in the stack.

If you stored strings in the stack, you’d have to convert these strings to integers each time you wanted to perform a calculation. This would add unnecessary overhead and complicate the code. So, for simplicity and efficiency, it’s better to store integers in the stack.