227. basic calculator ii

Understanding and Implementing Basic Calculator II (LeetCode 227)

The "Basic Calculator II" problem, often encountered on platforms like LeetCode, is a classic challenge that tests your understanding of parsing arithmetic expressions and handling operator precedence. While seemingly straightforward, implementing a robust solution requires careful thought about how to process numbers and operators in the correct order without the use of built-in evaluation functions.

The Problem Statement

Given a string s which represents an arithmetic expression, evaluate it and return its integer value. The expression contains only non-negative integers, the four basic arithmetic operators +, -, *, /, and spaces. Integer division should truncate toward zero.

For example:

• "3+2*2" should evaluate to 7
• " 3/2 " should evaluate to 1
• " 3+5 / 2 " should evaluate to 5

Constraints typically include that the given expression is always valid, and intermediate results will not exceed the range of a 32-bit integer.

The Challenge: Operator Precedence

The primary difficulty in this problem lies in correctly handling operator precedence. Multiplication and division operations must be performed before addition and subtraction. A naive left-to-right evaluation would lead to incorrect results (e.g., 3 + 2 * 2 would become (3 + 2) * 2 = 10 instead of 3 + (2 * 2) = 7).

A Common Approach: Using a Stack

A highly effective and widely used method to solve this problem is by employing a stack. The stack helps us defer addition and subtraction until all higher-precedence operations (multiplication and division) have been resolved. Here's the general idea:

1. Iterate through the expression string, character by character.
2. Parse integers as they appear.
3. Maintain a variable for the last encountered operator.
4. When an operator or the end of the string is reached, perform an action based on the last operator and the current number.
5. Finally, sum all numbers remaining in the stack.

Detailed Stack Algorithm

Let's break down the stack-based algorithm step-by-step:

1. Initialize an empty stack, stack.
2. Initialize currentNumber = 0 to accumulate digits for the current number.
3. Initialize operation = '+' to keep track of the operator just before the currentNumber.
4. Iterate through the string s from left to right, including an implicit final operator (or by adding a dummy operator to the string end):
   • If the current character char is a digit:
     • Update currentNumber = currentNumber * 10 + parseInt(char).
   • If char is not a digit and not a space, or if it's the last character of the string: (This signifies an operator or end of a number sequence)
     • Based on the operation variable (which stores the operator *before* the currentNumber):
       • If operation == '+': Push currentNumber onto the stack.
       • If operation == '-': Push -currentNumber onto the stack.
       • If operation == '*': Pop the top element from the stack, multiply it by currentNumber, and push the result back onto the stack.
       • If operation == '/': Pop the top element from the stack, divide it by currentNumber (using integer division that truncates towards zero), and push the result back onto the stack.
     • Update operation = char (the current operator becomes the "last operator" for the next number).
     • Reset currentNumber = 0.
5. After the loop finishes, the stack will contain all intermediate results, with multiplication and division already resolved. Sum all elements in the stack to get the final result.

Example Walkthrough: "3 + 2 * 2"

Let's trace the algorithm with the expression "3 + 2 * 2":

Input: s = "3 + 2 * 2"
stack = []
currentNumber = 0
operation = '+'

1. char = '3': currentNumber = 3
2. char = ' ': Skip
3. char = '+': (Operator encountered)
   - operation was '+': stack.push(3)  => stack: [3]
   - operation = '+'
   - currentNumber = 0
4. char = ' ': Skip
5. char = '2': currentNumber = 2
6. char = ' ': Skip
7. char = '*': (Operator encountered)
   - operation was '+': stack.push(2)  => stack: [3, 2]
   - operation = '*'
   - currentNumber = 0
8. char = ' ': Skip
9. char = '2': currentNumber = 2
10. End of string (i === s.length - 1): (Implicit operator)
    - operation was '*':
      - val = stack.pop()  => val = 2, stack: [3]
      - stack.push(val * currentNumber) => stack.push(2 * 2) => stack.push(4) => stack: [3, 4]
    - operation = '?' (not relevant after last char)
    - currentNumber = 0

Loop ends.
Sum stack elements: 3 + 4 = 7.
Final Result: 7
                    

Implementation Notes

• Handling spaces: Simply skip them during iteration.
• Integer division: Be mindful of how your chosen language handles integer division, especially with negative numbers. JavaScript's Math.trunc() is suitable here as it truncates towards zero.
• Edge cases: The problem statement usually guarantees valid input, simplifying error handling. Empty strings or malformed expressions are typically not tested.

Conclusion

The "Basic Calculator II" problem is an excellent exercise in understanding parsing logic and applying data structures like stacks to manage operator precedence. By systematically processing the input string and using a stack to prioritize multiplication and division, you can efficiently arrive at the correct result, demonstrating a fundamental skill in algorithm design.