Leetcode #736: Parse Lisp Expression
In this guide, we solve Leetcode #736 Parse Lisp Expression in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Stack, Recursion, Hash Table, String
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
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()
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.