How do you solve Mini Parser on LeetCode?

Mini Parser (LeetCode #385) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how to manage nested structures is crucial.

What is the brute force solution for Mini Parser?

The brute force approach for Mini Parser is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves parsing the string character by character to build the NestedInteger structure. We will manually track the current number and the nested lists as we encounter brackets and commas. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Mini Parser?

Mini Parser uses the following concepts: String, Stack, Depth-First Search. The recommended patterns to study are: Stack, Recursion.

What are the interview tips for Mini Parser?

Think about edge cases, such as empty lists or negative numbers. Practice writing out the structure of the NestedInteger class before coding. Explain your thought process clearly while coding.

What are common mistakes when solving Mini Parser?

Not handling negative numbers properly. Forgetting to reset the number string after adding it to a NestedInteger.

#385

Mini Parser

Medium

StringStackDepth-First SearchStackRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a stack to manage the current context of nested lists and integers. This approach efficiently builds the NestedInteger structure in a single pass through the string.

⚙️

Algorithm

5 steps

1Step 1: Initialize a stack to keep track of NestedInteger objects and a variable to hold the current number.
2Step 2: Iterate through the string character by character. If a digit or negative sign is found, build the current number.
3Step 3: On encountering '[', push a new NestedInteger onto the stack.
4Step 4: On encountering ']', pop the top NestedInteger from the stack and add it to the last NestedInteger on the stack.
5Step 5: On encountering a ',', if there's a number, add it to the current NestedInteger.

solution.py42 lines

1class NestedInteger:
2    def __init__(self, value=None):
3        self.value = value
4        self.is_integer = value is not None
5        self.list = [] if value is None else None
6    
7    def add(self, ni):
8        self.list.append(ni)
9
10    def setInteger(self, value):
11        self.value = value
12        self.is_integer = True
13
14    def getInteger(self):
15        return self.value
16
17    def getList(self):
18        return self.list
19
20
21def deserialize(s):
22    stack = []
23    num = ''
24    for char in s:
25        if char.isdigit() or char == '-':
26            num += char
27        elif char == '[':
28            stack.append(NestedInteger())
29        elif char == ']':
30            if num:
31                stack[-1].add(NestedInteger(int(num)))
32                num = ''
33            ni = stack.pop()
34            if stack:
35                stack[-1].add(ni)
36            else:
37                return ni
38        elif char == ',':
39            if num:
40                stack[-1].add(NestedInteger(int(num)))
41                num = ''
42    return NestedInteger(int(num)) if num else stack[0]

ℹ

Complexity note: The time complexity is O(n) because we make a single pass through the string, processing each character once. The space complexity is O(n) due to the stack used for storing NestedInteger objects.

1Understanding how to manage nested structures is crucial.
2Recognizing the need for a stack to keep track of context in nested lists.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.