Count Non Adjacent Subsets in a Rooted Tree
Time O(n * k^2) · Space O(n * k) · Official statement on LeetCode
Solutions
// Time: O(n * k^2)
// Space: O(n * k)
// if k is large, the per-merge cyclic convolution can be done in O(klogk) via NTT + CRT, Time: O(n * klogk), Space: O(n * k) in total
// topological sort, tree dp
class Solution {
public:
int countValidSubsets(vector<int>& parent, vector<int>& nums, int k) {
static const int MOD = 1e9 + 7;
const auto& merge = [&](const auto& a, const auto& b) {
vector<int> result(size(a));
for (int i = 0; i < k; ++i) {
if (!a[i]) {
continue;
}
for (int j = 0; j < k; ++j) {
if (!b[j]) {
continue;
}
result[(i + j) % k] = ((result[(i + j) % k] + static_cast<int64_t>(a[i]) * b[j]) % MOD);
}
}
return result;
};
vector<vector<int>> adj(size(nums));
for (int v = 1; v < size(parent); ++v) {
adj[parent[v]].emplace_back(v);
}
vector<vector<vector<int>>> dp(size(nums), vector<vector<int>>(2, vector<int>(k)));
for (int u = size(nums) - 1; u >= 0; --u) {
vector<vector<int>> new_dp(2, vector<int>(k));
new_dp[0][0] = 1;
new_dp[1][nums[u] % k] = 1;
for (const auto& v : adj[u]) {
vector<int> dp2(k);
for (int i = 0; i < k; ++i) {
dp2[i] = (dp[v][0][i] + dp[v][1][i]) % MOD;
}
new_dp[0] = merge(new_dp[0], dp2);
new_dp[1] = merge(new_dp[1], dp[v][0]);
}
dp[u] = move(new_dp);
}
return ((dp[0][0][0] + dp[0][1][0] - 1LL) + MOD) % MOD;
}
};
Beginner Explanation
What is Count Non Adjacent Subsets in a Rooted Tree?
Count Non Adjacent Subsets in a Rooted Tree (LeetCode #3939) 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 topological sort and tree dp.
- 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: Topological Sort, Tree DP.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Count Non Adjacent Subsets in a Rooted Tree
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 topological sort and tree dp.
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 * k^2)) and space (O(n * k)) 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 * k^2) time and O(n * k) space.
Pattern focus: topological sort and tree dp
Use the pattern as a checklist:
- topological sort — confirm the invariant holds after each step
- tree dp — 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 * k^2) |
| Space | O(n * k) |
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 Count Non Adjacent Subsets in a Rooted Tree
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for topological sort and tree dp — 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 topological sort and tree dp:
- 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 Count Non Adjacent Subsets in a Rooted Tree 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 topological sort and tree dp 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
Count Non Adjacent Subsets in a Rooted Tree (#3939) — Hard. Pattern: topological sort and tree dp. Complexity: O(n * k^2) time / O(n * k) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Count Non Adjacent Subsets in a Rooted Tree?+
The reference solutions aim for O(n * k^2) time and O(n * k) space. Always re-derive complexity from the code you write in the interview.
What pattern does Count Non Adjacent Subsets in a Rooted Tree use?+
It primarily maps to topological sort and tree dp, within the broader topic of dynamic programming.
Is Count Non Adjacent Subsets in a Rooted Tree 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/count-non-adjacent-subsets-in-a-rooted-tree/