Check if There Is a Valid Parentheses String Path
Time O(m * n * (m + n) / 32) · Space O(n * (m + n) / 32) · Official statement on LeetCode
Solutions
// Time: O((m * n) * (m + n) / 32)
// Space: O(n * (m + n) / 32)
// dp with bitsets
class Solution {
public:
bool hasValidPath(vector<vector<char>>& grid) {
if ((size(grid) + size(grid[0]) - 1) % 2 == 1) {
return false;
}
vector<bitset<100>> dp(size(grid[0]) + 1);
for (int i = 0; i < size(grid); ++i) {
dp[0][0] = !i;
for (int j = 0; j < size(grid[0]); ++j) {
dp[j + 1] = (grid[i][j] == '(') ? (dp[j] | dp[j + 1]) << 1: (dp[j] | dp[j + 1]) >> 1;
}
}
return dp.back()[0];
}
};
// Time: O(m * n)
// Space: O(n)
// dp, optimized from solution1 (wrong answer)
class Solution_WA {
public:
bool hasValidPath(vector<vector<char>>& grid) {
const int MAX_M = 100;
const int MAX_N = 100;;
if ((size(grid) + size(grid[0]) - 1) % 2 == 1) {
return false;
}
vector<pair<int, int>> dp(size(grid[0]) + 1, {MAX_M + MAX_N, -(MAX_M + MAX_N)});
for (int i = 0; i < size(grid); ++i) {
dp[0] = !i ? pair(0, 0) : pair(MAX_M + MAX_N, -(MAX_M + MAX_N));
for (int j = 0; j < size(grid[0]); ++j) {
const int d = (grid[i][j] == '(') ? 1 : -1;
dp[j + 1] = {min(dp[j + 1].first, dp[j].first) + d, max(dp[j + 1].second, dp[j].second) + d};
// bitset pattern is like xxx1010101xxxx (in fact, it is not always true in this problem where some paths are invalid)
if (dp[j + 1].second < 0) {
dp[j + 1] = {MAX_M + MAX_N, -(MAX_M + MAX_N)};
} else {
dp[j + 1].first = max(dp[j + 1].first, dp[j + 1].second % 2);
}
}
}
return dp.back().first == 0;
}
};
Beginner Explanation
What is Check if There Is a Valid Parentheses String Path?
Check if There Is a Valid Parentheses String Path (LeetCode #2267) is a Hard problem that primarily trains dynamic programming.
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 dynamic programming.
- Only then translate the idea into code.
Why this problem matters
Hard problems force you to combine patterns and prove complexity carefully — interview gold. Official solution notes mention: DP, Bitsets.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Check if There Is a Valid Parentheses String Path
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 dynamic programming.
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(m * n * (m + n) / 32)) and space (O(n * (m + n) / 32)) 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(m * n * (m + n) / 32) time and O(n * (m + n) / 32) space.
Pattern focus: dynamic programming
Use the pattern as a checklist:
- dynamic programming — 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(m * n * (m + n) / 32) |
| Space | O(n * (m + n) / 32) |
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 Check if There Is a Valid Parentheses String Path
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for dynamic programming — 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 dynamic programming:
- 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: dynamic programming.
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 Check if There Is a Valid Parentheses String Path in a second language (cpp, python).
- Drill 3–5 more problems tagged dynamic programming.
- Teach the solution out loud in under 5 minutes.
- Add this problem to your revision calendar in 3 days and 14 days.
Visualization
Study checklist
- Read the official problem statement on LeetCode
- Solve on paper / whiteboard first
- Implement the dynamic programming 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
Check if There Is a Valid Parentheses String Path (#2267) — Hard. Pattern: dynamic programming. Complexity: O(m * n * (m + n) / 32) time / O(n * (m + n) / 32) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Check if There Is a Valid Parentheses String Path?+
The reference solutions aim for O(m * n * (m + n) / 32) time and O(n * (m + n) / 32) space. Always re-derive complexity from the code you write in the interview.
What pattern does Check if There Is a Valid Parentheses String Path use?+
It primarily maps to dynamic programming, within the broader topic of dynamic programming.
Is Check if There Is a Valid Parentheses String Path good for interviews?+
Yes — as a Hard 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/check-if-there-is-a-valid-parentheses-string-path/