Solving Leetcode Interviews in Seconds with AI: Mini Parser
Introduction
In this blog post, we will explore how to solve the LeetCode problem "385" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.
Problem Statement
Given a string s represents the serialization of a nested list, implement a parser to deserialize it and return the deserialized NestedInteger. Each element is either an integer or a list whose elements may also be integers or other lists. Example 1: Input: s = "324" Output: 324 Explanation: You should return a NestedInteger object which contains a single integer 324. Example 2: Input: s = "[123,[456,[789]]]" Output: [123,[456,[789]]] Explanation: Return a NestedInteger object containing a nested list with 2 elements: 1. An integer containing value 123. 2. A nested list containing two elements: i. An integer containing value 456. ii. A nested list with one element: a. An integer containing value 789 Constraints: 1 <= s.length <= 5 * 104 s consists of digits, square brackets "[]", negative sign '-', and commas ','. s is the serialization of valid NestedInteger. All the values in the input are in the range [-106, 106].
Explanation
Here's the breakdown of the solution:
- Recursive Descent Parsing: The problem naturally lends itself to recursive descent parsing. We process the string character by character. If we encounter a '[', we recursively parse the nested list. If we encounter a number, we parse it as an integer.
- Handling nested structures: Utilize the call stack to keep track of the current nested level and manage element additions to the appropriate NestedInteger object.
String processing optimization: Iterate through the string efficiently using an index to avoid unnecessary string slicing, which improves performance.
Time & Space Complexity: Time: O(N), where N is the length of the string. Space: O(D), where D is the maximum depth of the nested list due to the recursive call stack.
Code
# """
# This is the interface that allows for creating nested lists.
# You should not implement it, or speculate about its implementation
# """
class NestedInteger:
def __init__(self, value=None):
"""
If value is not specified, initializes an empty list.
Otherwise initializes a single integer equal to value.
"""
if value is None:
self.integer = None
self.list = []
else:
self.integer = value
self.list = None
def isInteger(self):
"""
@return True if this NestedInteger holds a single integer, rather than a nested list.
:rtype bool
"""
return self.list is None
def add(self, elem):
"""
Set this NestedInteger to hold a nested list and adds a nested integer elem to it.
:rtype void
"""
if self.integer is not None:
self.list = []
self.list.append(NestedInteger(self.integer)) #copy current int to list
self.integer = None
self.list.append(elem)
def setInteger(self, value):
"""
Set this NestedInteger to hold a single integer equal to value.
:rtype void
"""
self.integer = value
self.list = None
def getInteger(self):
"""
@return the single integer that this NestedInteger holds, if it holds a single integer
Return None if this NestedInteger holds a nested list
:rtype int
"""
return self.integer
def getList(self):
"""
@return the nested list that this NestedInteger holds, if it holds a nested list
Return None if this NestedInteger holds a single integer
:rtype List[NestedInteger]
"""
return self.list
class Solution:
def deserialize(self, s: str) -> NestedInteger:
def parse(s, index):
if s[index].isdigit() or s[index] == '-':
negative = False
if s[index] == '-':
negative = True
index += 1
num = 0
while index < len(s) and s[index].isdigit():
num = num * 10 + int(s[index])
index += 1
if negative:
num = -num
return NestedInteger(num), index
else:
ni = NestedInteger()
index += 1 # Skip '['
while index < len(s) and s[index] != ']':
nested_integer, index = parse(s, index)
ni.add(nested_integer)
if index < len(s) and s[index] == ',':
index += 1
index += 1 # Skip ']'
return ni, index
return parse(s, 0)[0]