Parse Lisp Expression

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RecursionStrings

You are given a string expression representing a Lisp-like expression to return the integer value of.

The syntax for these expressions is given as follows.

An expression is either an integer, let expression, add expression, mult expression, or an assigned variable. Expressions always evaluate to a single integer.
(An integer could be positive or negative.)
A let expression takes the form "(let v₁ e₁ v₂ e₂ ... v_n e_n expr)", where let is always the string "let", then there are one or more pairs of alternating variables and expressions, meaning that the first variable v₁ is assigned the value of the expression e₁, the second variable v₂ is assigned the value of the expression e₂, and so on sequentially; and then the value of this let expression is the value of the expression expr.
An add expression takes the form "(add e₁ e₂)" where add is always the string "add", there are always two expressions e₁, e₂ and the result is the addition of the evaluation of e₁ and the evaluation of e₂.
A mult expression takes the form "(mult e₁ e₂)" where mult is always the string "mult", there are always two expressions e₁, e₂ and the result is the multiplication of the evaluation of e1 and the evaluation of e2.
For this question, we will use a smaller subset of variable names. A variable starts with a lowercase letter, then zero or more lowercase letters or digits. Additionally, for your convenience, the names "add", "let", and "mult" are protected and will never be used as variable names.
Finally, there is the concept of scope. When an expression of a variable name is evaluated, within the context of that evaluation, the innermost scope (in terms of parentheses) is checked first for the value of that variable, and then outer scopes are checked sequentially. It is guaranteed that every expression is legal. Please see the examples for more details on the scope.

Example 1:

Input: expression = "(let x 2 (mult x (let x 3 y 4 (add x y))))"
Output: 14
Explanation: 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
Explanation: 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
Explanation: 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.

Try coding it in LeetCode

Parse Lisp Expression

Hard

Affirm

9 views

Topics:

RecursionStrings

You are given a string expression representing a Lisp-like expression to return the integer value of.

The syntax for these expressions is given as follows.

An expression is either an integer, let expression, add expression, mult expression, or an assigned variable. Expressions always evaluate to a single integer.
(An integer could be positive or negative.)
A let expression takes the form "(let v₁ e₁ v₂ e₂ ... v_n e_n expr)", where let is always the string "let", then there are one or more pairs of alternating variables and expressions, meaning that the first variable v₁ is assigned the value of the expression e₁, the second variable v₂ is assigned the value of the expression e₂, and so on sequentially; and then the value of this let expression is the value of the expression expr.
An add expression takes the form "(add e₁ e₂)" where add is always the string "add", there are always two expressions e₁, e₂ and the result is the addition of the evaluation of e₁ and the evaluation of e₂.
A mult expression takes the form "(mult e₁ e₂)" where mult is always the string "mult", there are always two expressions e₁, e₂ and the result is the multiplication of the evaluation of e1 and the evaluation of e2.
For this question, we will use a smaller subset of variable names. A variable starts with a lowercase letter, then zero or more lowercase letters or digits. Additionally, for your convenience, the names "add", "let", and "mult" are protected and will never be used as variable names.
Finally, there is the concept of scope. When an expression of a variable name is evaluated, within the context of that evaluation, the innermost scope (in terms of parentheses) is checked first for the value of that variable, and then outer scopes are checked sequentially. It is guaranteed that every expression is legal. Please see the examples for more details on the scope.

Example 1:

Input: expression = "(let x 2 (mult x (let x 3 y 4 (add x y))))"
Output: 14
Explanation: 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
Explanation: 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
Explanation: 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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What characters are allowed in the Lisp expression besides parentheses and numbers? Can I assume only integers, or should I handle other data types like floats, symbols, or strings?
What is the expected output format? Should I return the evaluated result as an integer, a string representation, or a more complex data structure representing the expression tree?
Are there any limitations on the depth or complexity of the Lisp expressions I should consider?
How should I handle invalid Lisp expressions, such as unbalanced parentheses or invalid operators? Should I throw an error, return a specific value like null, or print an error message?
Are whitespace characters significant in the expression? Can I assume consistent spacing between elements, or should I handle cases with varying or missing whitespace?

Brute Force Solution

Approach

The brute force approach to parsing a Lisp expression involves exploring every possible interpretation of the expression. We systematically try out all possible combinations of operations and values until we find a valid result or exhaust all possibilities. It's like trying every single way to understand the expression, one at a time.

Here's how the algorithm would work step-by-step:

Start by examining the expression from left to right.
Whenever you encounter an opening parenthesis, assume you're starting a new operation.
Look at what comes after the parenthesis. Is it a known operation like 'add' or 'multiply'? If so, remember it.
Next, look for the numbers or other expressions that will be used by that operation.
For each number, consider it a possible value to use in the operation.
If you find another opening parenthesis instead of a number, treat that whole section as another smaller expression that needs to be solved first.
Keep track of all the operations and numbers you find and how they relate to each other.
Once you reach the closing parenthesis, try to calculate the result of that operation using the numbers you found.
If you can calculate a result, replace the entire expression within the parentheses with that result.
Repeat this process until the entire expression is reduced to a single value.
If at any point you encounter something that doesn't make sense (like a missing parenthesis or an unknown operation), that attempt fails and you would ideally backtrack to try another possible interpretation if the original expression was ambiguous (however, standard Lisp syntax isn't typically ambiguous so we usually stop there).
This exhaustive process guarantees that if the expression is valid, you'll eventually find its result. If you have tried all possible combination and reached a dead end every single time, then the expression is invalid.

Code Implementation

def parse_lisp_expression_brute_force(expression):
    expression = expression.replace('(', ' ( ').replace(')', ' ) ').split()
    
    def evaluate_expression(tokens):
        if not tokens:
            return None, []
        
        token = tokens[0]
        remaining_tokens = tokens[1:]

        if token == '(': 
            operator = remaining_tokens[0]
            remaining_tokens = remaining_tokens[1:]
            operands = []
            
            while remaining_tokens[0] != ')':
                # Recursively evaluate operands
                operand, remaining_tokens = evaluate_expression(remaining_tokens)
                operands.append(operand)
                
            remaining_tokens = remaining_tokens[1:]
            
            if operator == 'add':
                result = sum(operands)
            elif operator == 'multiply':
                result = 1
                for operand in operands:
                    result *= operand
            else:
                return None, [] # Invalid operator
            
            return result, remaining_tokens
        elif token == ')':
            return None, remaining_tokens # Should not happen here
        else:
            # Convert token to integer
            return int(token), remaining_tokens

    result, remaining_tokens = evaluate_expression(expression)

    if remaining_tokens:
        return None # Invalid expression

    return result

Big(O) Analysis

Time Complexity

O(n!) – The provided brute force approach explores every possible interpretation of the Lisp expression. In the worst-case scenario, where the expression is highly nested and contains multiple operations, the algorithm might need to try all possible combinations of operations and values. This exhaustive exploration leads to a factorial time complexity because for an expression of length n, the number of possible interpretations can grow factorially as each element could potentially be involved in different operations or be part of different sub-expressions that need to be evaluated in various orders. Consequently, the total number of operations approximates n!, which represents the product of all integers from 1 to n, thus resulting in O(n!) time complexity.

Space Complexity

O(N) – The space complexity is primarily determined by the depth of the nested Lisp expressions and the storage required for intermediate results. The recursion stack will grow proportionally to the maximum level of nesting in the expression. In the worst-case scenario, the expression might be a deeply nested structure, requiring N stack frames where N is the length of the input expression. Additionally, the algorithm might use temporary storage to hold intermediate numerical results or operation types during evaluation which will also be bounded by the length of the input expression. Thus, the auxiliary space required is O(N).

Optimal Solution

Approach

The problem requires us to evaluate Lisp expressions which can be nested. We can solve this effectively by using recursion, evaluating the innermost expressions first and working our way outwards. A stack-like structure implicitly created by the call stack allows for managing nested expressions gracefully.

Here's how the algorithm would work step-by-step:

Start by identifying the operator at the beginning of the expression.
Then, find the operands which can be either numbers or other Lisp expressions enclosed in parentheses.
If you encounter an operand that is a nested expression, recursively evaluate that sub-expression first.
Once all operands are evaluated, perform the operation specified by the operator.
Return the result of the operation. This will be used by the calling expression or as the final result.
Handle edge cases like invalid expressions or division by zero to ensure robustness.

Code Implementation

def parse_lisp_expression(expression):
    tokens = expression.replace('(', ' ( ').replace(')', ' ) ').split()
    return evaluate_expression(tokens, 0)[0]

def evaluate_expression(tokens, position):
    if tokens[position] == '(': 
        position += 1
        operator = tokens[position]
        position += 1
        operands = []

        # Recursively evaluate operands
        while tokens[position] != ')':
            operand, position = evaluate_expression(tokens, position)
            operands.append(operand)

        position += 1

        # Perform the operation
        if operator == '+':
            result = sum(operands)
        elif operator == '-':
            result = operands[0] - sum(operands[1:])
        elif operator == '*':
            result = 1
            for operand in operands:
                result *= operand
        elif operator == '/':
            # Handle division by zero
            if any(operand == 0 for operand in operands[1:]):
                raise ZeroDivisionError("Division by zero")
            result = operands[0]
            for operand in operands[1:]:
                result /= operand
        else:
            raise ValueError("Invalid operator: " + operator)
        
        return result, position
    else:
        # Convert the token to a number
        return int(tokens[position]), position + 1

Big(O) Analysis

Time Complexity

O(n) – The time complexity is primarily driven by the length of the input Lisp expression string, denoted as n. While recursion is used to handle nested expressions, each character in the input string is visited and processed at most a constant number of times to identify operators, operands, and parenthesis. The depth of the recursion is bounded by the level of nesting in the Lisp expression, but the total work done across all recursive calls is still proportional to the length of the input string. Therefore, the overall time complexity is linear, or O(n).

Space Complexity

O(N) – The primary driver of space complexity is the recursion depth. Each recursive call evaluates a nested Lisp expression, creating a new stack frame. In the worst-case scenario, the expression can be deeply nested, with the depth proportional to the length of the input expression N, where N is the number of tokens in the expression. This results in a call stack that can grow up to N frames deep. Therefore, the auxiliary space required for the call stack can be O(N).

Edge Cases

Case	How to Handle
Empty input string	Return an empty list or raise an exception indicating invalid input.
Input string with only whitespace	Treat as an empty string and return an empty list or an appropriate error.
Unbalanced parentheses (more opening than closing)	Detect and report an error indicating an incomplete expression.
Unbalanced parentheses (more closing than opening)	Detect and report an error indicating an invalid expression.
Nested expressions with deeply nested structures leading to stack overflow (recursion depth)	Consider iterative approaches or techniques like tail-call optimization if the language supports it, to prevent excessive recursion.
Invalid characters within the expression (e.g., special characters not part of the Lisp syntax)	Validate each character and throw an error upon encountering invalid characters.
Arithmetic overflow during evaluation (e.g., very large numbers in the expression)	Use appropriate data types (e.g., BigInteger) or check for overflow conditions before performing calculations and throw an exception if needed.
Division by zero	Check if the divisor is zero before performing the division operation, and return an error or handle the exception accordingly.