#736

Hard

auto_awesomeStack-based state management

LeetCode Problem Workspace

Parse Lisp Expression

Parse Lisp expressions using stack-based state management to evaluate variables and operations.

Problem Statement

You are given a string representing a Lisp-like expression that you need to evaluate and return its integer result. The syntax involves nested expressions where operations and variable assignments are defined using constructs like 'let', 'add', and 'mult'. The key is to correctly manage variable scopes and operations within nested expressions.

The expression will always be well-formed, with each part separated by a space. You'll need to manage the state of variables using a stack, evaluating innermost expressions first and using hash tables to store variable values. When an expression begins with a digit or '-', it should be interpreted as an integer. If it starts with a letter, it's a variable.

Examples

Example 1

Input: expression = "(let x 2 (mult x (let x 3 y 4 (add x y))))"

Output: 14

In the expression (add x y), when checking for the value of the variable x, we check from the innermost scope to the outermost in the context of the variable we are trying to evaluate. Since x = 3 is found first, the value of x is 3.

Example 2

Input: expression = "(let x 3 x 2 x)"

Output: 2

Assignment in let statements is processed sequentially.

Example 3

Input: expression = "(let x 1 y 2 x (add x y) (add x y))"

Output: 5

The first (add x y) evaluates as 3, and is assigned to x. The second (add x y) evaluates as 3+2 = 5.

Constraints

1 <= expression.length <= 2000
There are no leading or trailing spaces in expression.
All tokens are separated by a single space in expression.
The answer and all intermediate calculations of that answer are guaranteed to fit in a 32-bit integer.
The expression is guaranteed to be legal and evaluate to an integer.

Solution Approach

Recursive Evaluation with Stack Management

A recursive approach combined with stack-based state management works well for evaluating Lisp expressions. Each recursive call handles an expression or nested scope, and the stack helps manage different variable contexts. When encountering a 'let' statement, push a new scope on the stack to evaluate it within that scope.

Using Hash Table for Variable Storage

A hash table can be used to store variables and their values within the current scope. When a variable is encountered, check the stack for its value in the most recent scope. If not found, check previous scopes until the variable's value is located.

Handling Nested Expressions

Handling nested expressions requires careful state management. For each nested expression, push a new stack frame, evaluate it, and then pop the frame once it's complete. This ensures that each scope's variables are isolated and do not interfere with outer scopes.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the depth of the recursion and the number of expressions. For each token in the expression, there is a constant time operation for evaluation, but recursion and stack manipulation introduce an additional cost for nested expressions. The space complexity is primarily affected by the recursion depth and stack size, which can grow proportionally to the number of nested scopes.

What Interviewers Usually Probe

Can the candidate explain the trade-offs between recursion and iterative solutions in evaluating expressions?
Does the candidate handle nested scopes correctly with the stack and hash table?
How well does the candidate optimize for time and space complexity in managing the stack?

Common Pitfalls or Variants

Common pitfalls

Failing to correctly manage nested variable scopes can lead to incorrect results.
Not utilizing a stack efficiently to track the current variable context can cause scoping issues.
Misunderstanding the recursion depth and stack size, leading to performance issues on large expressions.

Follow-up variants

Modify the problem to evaluate a broader range of operations beyond just 'add' and 'mult'.
Change the expression format to handle different types of scopes or operations like 'if' or 'define'.
Optimize the solution to handle larger expressions by implementing iterative evaluation instead of recursion.

FAQ

What is the main challenge in solving the Parse Lisp Expression problem?

The main challenge is correctly handling variable scoping and nested expressions using stack-based state management and hash tables.

How do I manage variables in the Parse Lisp Expression problem?

Variables are managed using a stack where each scope has its own hash table for storing variable values, ensuring correct scoping for nested expressions.

What happens when an expression starts with a digit or '-'?

If an expression starts with a digit or '-', it is treated as an integer and should be returned directly.

How do I handle nested 'let' expressions in Parse Lisp Expression?

For nested 'let' expressions, push a new scope onto the stack, evaluate the expression, and then pop the scope when done.

Can recursion be avoided in solving Parse Lisp Expression?

While recursion is a natural fit for this problem, iterative solutions are possible, though they might require additional stack management or auxiliary data structures.

terminal

Solution

Solution 1

#### Python3

class Solution:
    def evaluate(self, expression: str) -> int:
        def parseVar():
            nonlocal i
            j = i
            while i < n and expression[i] not in " )":
                i += 1
            return expression[j:i]

        def parseInt():
            nonlocal i
            sign, v = 1, 0
            if expression[i] == "-":
                sign = -1
                i += 1
            while i < n and expression[i].isdigit():
                v = v * 10 + int(expression[i])
                i += 1
            return sign * v

        def eval():
            nonlocal i
            if expression[i] != "(":
                return scope[parseVar()][-1] if expression[i].islower() else parseInt()
            i += 1
            if expression[i] == "l":
                i += 4
                vars = []
                while 1:
                    var = parseVar()
                    if expression[i] == ")":
                        ans = scope[var][-1]
                        break
                    vars.append(var)
                    i += 1
                    scope[var].append(eval())
                    i += 1
                    if not expression[i].islower():
                        ans = eval()
                        break
                for v in vars:
                    scope[v].pop()
            else:
                add = expression[i] == "a"
                i += 4 if add else 5
                a = eval()
                i += 1
                b = eval()
                ans = a + b if add else a * b
            i += 1
            return ans

        i, n = 0, len(expression)
        scope = defaultdict(list)
        return eval()

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