#385Medium~35 min
Mini Parser
Time O(n) · Space O(h) · Official statement on LeetCode
Solutions
// Time: O(n)
// Space: O(h)
/**
* // This is the interface that allows for creating nested lists.
* // You should not implement it, or speculate about its implementation
* class NestedInteger {
* public:
* // Constructor initializes an empty nested list.
* NestedInteger();
*
* // Constructor initializes a single integer.
* NestedInteger(int value);
*
* // Return true if this NestedInteger holds a single integer, rather than a nested list.
* bool isInteger() const;
*
* // Return the single integer that this NestedInteger holds, if it holds a single integer
* // The result is undefined if this NestedInteger holds a nested list
* int getInteger() const;
*
* // Set this NestedInteger to hold a single integer.
* void setInteger(int value);
*
* // Set this NestedInteger to hold a nested list and adds a nested integer to it.
* void add(const NestedInteger &ni);
*
* // Return the nested list that this NestedInteger holds, if it holds a nested list
* // The result is undefined if this NestedInteger holds a single integer
* const vector<NestedInteger> &getList() const;
* };
*/
// Iterative solution.
class Solution {
public:
NestedInteger deserialize(string s) {
if (s.empty()) {
return NestedInteger();
}
if (s[0] != '[') {
return NestedInteger(stoi(s));
}
stack<NestedInteger> stk;
for (int i = 0, j = 0; j < s.length(); ++j) {
if (s[j] == '[') {
stk.emplace(NestedInteger());
i = j + 1;
} else if (s[j] == ',' || s[j] == ']') {
if (isdigit(s[j - 1])) {
stk.top().add(NestedInteger(stoi(s.substr(i, j - i))));
}
if (s[j] == ']' && stk.size() > 1) {
NestedInteger cur = stk.top();
stk.pop();
stk.top().add(cur);
}
i = j + 1;
}
}
return stk.top();
}
};
// Time: O(n)
// Space: O(h)
// Recursive solution.
class Solution2 {
public:
NestedInteger deserialize(string s) {
if (s.empty()) {
return NestedInteger();
}
int i = 0;
return deserializeHelper(s, &i);
}
private:
NestedInteger deserializeHelper(const string& s, int *i) {
NestedInteger result;
if (s[*i] != '[') {
int j = *i;
while (j < s.length() && (s[j] == '-' || isdigit(s[j]))) {
++j;
}
result.setInteger(stoi(s.substr(*i, j - *i + 1)));
*i = j;
} else {
++(*i);
while (*i < s.length() && s[*i] != ']') {
result.add(deserializeHelper(s, i));
if (*i < s.length() && s[*i] == ',') {
++(*i);
}
}
++(*i);
}
return result;
}
};
// Time: O(n)
// Space: O(n)
// Recursive solution.
class Solution3 {
public:
NestedInteger deserialize(string s) {
if (s.empty()) {
return NestedInteger();
}
istringstream in(s); // copy string: extra O(n) space
return deserializeHelper(in);
}
private:
NestedInteger deserializeHelper(istringstream &in) {
NestedInteger result;
int num = 0;
if (in >> num) {
result.setInteger(num);
} else {
in.clear();
in.get();
while (in.peek() != ']') {
result.add(deserializeHelper(in));
if (in.peek() == ',') {
in.get();
}
}
in.get();
}
return result;
}
};
Beginner Explanation
What is Mini Parser?
Mini Parser (LeetCode #385) is a Medium problem that primarily trains stack.
How to think about it
- Restate the goal in your own words before coding.
- Work a tiny example by hand so the invariant becomes obvious.
- Identify the pattern — this problem aligns with stack.
- Only then translate the idea into code.
Why this problem matters
It sits in the sweet spot of interview difficulty: multiple valid approaches, clear trade-offs.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Mini Parser
Opening (30–60 seconds)
- Clarify inputs/outputs and edge cases (empty input, single element, duplicates, overflow).
- State a brute force so the interviewer knows you can solve it naively.
- Propose the optimal direction tied to stack.
Core solution narrative
- Define the state you track (pointers, DP cell, set membership, stack top, etc.).
- Explain the transition when you process the next element.
- Call out time (O(n)) and space (O(h)) before coding.
- Code cleanly; narrate variable names.
What interviewers listen for
- Correctness on edge cases
- Complexity honesty
- Ability to discuss trade-offs (e.g., hash map space vs. sort + two pointers)
Follow-up questions they may ask
- Can you solve it with less memory?
- What if the input stream is infinite / doesn't fit in RAM?
- How would tests look for adversarial inputs?
Optimized Approach
Optimized solution notes
The reference solutions on AlgoForge target O(n) time and O(h) space.
Pattern focus: stack
Use the pattern as a checklist:
- stack — confirm the invariant holds after each step
Multiple methods appear in the source solutions — compare them and explain when each is preferable.
Implementation tips
- Prefer readable names over micro-optimizations in interviews.
- Extract helpers only when they clarify (e.g., expand-around-center, DFS visit).
- After AC-level logic, re-scan for off-by-one and null checks.
Complexity Analysis
Complexity
| Measure | Bound |
|---|---|
| Time | O(n) |
| Space | O(h) |
How to justify this in an interview
- Time: count loops, map/set operations, and recursive branching; state average vs worst case if relevant.
- Space: include hash maps, recursion stack, and output allocation when the problem asks for it.
If your implementation differs from the reference, re-derive big-O from your code — never memorize a complexity you cannot defend.
Common Mistakes
Common mistakes on Mini Parser
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for stack — updating state too early or too late.
- Mutating input unexpectedly when the problem forbids it.
- Off-by-one in windows, ranges, or binary search bounds.
- Ignoring overflow / precision for integer arithmetic problems.
- Overengineering — jumping to an advanced structure when a simpler approach works.
Alternative Approaches
Alternatives
The source file includes more than one method. Compare:
- Primary optimized path — best complexity for typical interviews.
- Secondary approach — often brute force, sorting-based, or space-optimized variant.
Practice articulating when you would pick each (constraints, readability, follow-ups).
Edge Cases
Edge cases checklist
- Minimum input size
- Maximum input size / time limits
- Duplicates and already-sorted input
- Negative numbers / zeros (if applicable)
- Disconnected structures (graphs/trees)
- Single path vs branching recursion depth
Pattern Recognition
Spotting this pattern
Signal phrases that point to stack:
- Sorted input or ability to sort without changing the answer class
- Need for contiguous subarray / substring → consider sliding window
- Need for O(1) membership → hash set/map
- Optimal substructure + overlapping subproblems → DP
- Connectivity / components → graph DFS/BFS or Union-Find
Primary topics: stack.
Follow-up Interview Questions
Follow-ups
- How does the solution change if the input is a stream?
- Can you solve it in-place?
- What if duplicates must be handled differently?
- How would you parallelize the approach?
- Design tests that would break a buggy implementation.
Practice Recommendations
What to practice next
- Re-solve Mini Parser in a second language (cpp, python).
- Drill 3–5 more problems tagged stack.
- Teach the solution out loud in under 5 minutes.
- Add this problem to your revision calendar in 3 days and 14 days.
Visualization
Conceptual diagram for Mini Parser: show input structure (stack), highlight the moving parts of the stack approach, and annotate each step with the maintained invariant and complexity.
Study checklist
- Read the official problem statement on LeetCode
- Solve on paper / whiteboard first
- Implement the stack approach
- Verify edge cases from the checklist
- State time and space complexity aloud
- Compare with the AlgoForge reference solution
- Schedule a revision session
Revision notes
Mini Parser (#385) — Medium. Pattern: stack. Complexity: O(n) time / O(h) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Mini Parser?+
The reference solutions aim for O(n) time and O(h) space. Always re-derive complexity from the code you write in the interview.
What pattern does Mini Parser use?+
It primarily maps to stack, within the broader topic of stack.
Is Mini Parser good for interviews?+
Yes — as a Medium problem it is a solid practice target. Pair it with related problems in the same pattern family for spaced repetition.
Where can I read the official statement?+
Open the official LeetCode page for constraints and examples: https://leetcode.com/problems/mini-parser/