Number of Integers With Popcount-Depth Equal to K I
Time precompute: O((logr)^2) runtime: O((logn)^2) · Space O((logr)^2) · Official statement on LeetCode
Solutions
// Time: precompute: O((logr)^2), r = max(n)
// runtime: O((logn)^2)
// Space: O((logr)^2)
// combinatorics
int bit_length(int64_t x) {
return (x ? std::__lg(x) : -1) + 1;
}
pair<vector<vector<int64_t>>, vector<int>> init() {
static const int64_t MAX_N = 1e15;
static const int MAX_BIT_LEN = bit_length(MAX_N);
vector<vector<int64_t>> NCR(MAX_BIT_LEN + 1, vector<int64_t>(MAX_BIT_LEN + 1));
for (int i = 0; i < size(NCR); ++i) {
for (int j = 0; j <= i; ++j) {
NCR[i][j] = 0 < j && j < i ? NCR[i - 1][j] + NCR[i - 1][j - 1] : 1;
}
}
vector<int> D(MAX_BIT_LEN + 1, 0);
for (int i = 2; i < size(D); ++i) {
D[i] = D[__builtin_popcount(i)] + 1;
}
return {NCR, D};
}
const auto& [NCR, D] = init();
class Solution {
public:
long long popcountDepth(long long n, int k) {
if (k == 0) {
return 1;
}
const int l = bit_length(n);
if (k == 1) {
return l - 1;
}
const auto& count = [&](int c) {
int64_t result = 0;
int cnt = 0;
for (int i = l - 1; i >= 0; --i) {
if (!(n & (1ll << i))) {
continue;
}
if (0 <= c - cnt && c - cnt <= i) {
result += NCR[i][c - cnt];
}
++cnt;
}
if (cnt == c) {
++result;
}
return result;
};
int64_t result = 0;
for (int c = 2; c <= l; ++c) {
if (D[c] == k - 1) {
result += count(c);
}
}
return result;
}
};
Beginner Explanation
What is Number of Integers With Popcount-Depth Equal to K I?
Number of Integers With Popcount-Depth Equal to K I (LeetCode #3621) is a Hard problem that primarily trains math.
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 general problem-solving.
- 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: Combinatorics.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Number of Integers With Popcount-Depth Equal to K I
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 general problem-solving.
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 (precompute: O((logr)^2) runtime: O((logn)^2)) and space (O((logr)^2)) 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 precompute: O((logr)^2) runtime: O((logn)^2) time and O((logr)^2) space.
Pattern focus: general problem-solving
Use the pattern as a checklist:
- Identify the dominant pattern and stick to one clear invariant
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 | precompute: O((logr)^2) runtime: O((logn)^2) |
| Space | O((logr)^2) |
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 Number of Integers With Popcount-Depth Equal to K I
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for general problem-solving — 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 general problem-solving:
- 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: math.
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 Number of Integers With Popcount-Depth Equal to K I in a second language (cpp, python).
- Drill 3–5 more problems tagged math.
- 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 general problem-solving 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
Number of Integers With Popcount-Depth Equal to K I (#3621) — Hard. Pattern: general problem-solving. Complexity: precompute: O((logr)^2) runtime: O((logn)^2) time / O((logr)^2) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Number of Integers With Popcount-Depth Equal to K I?+
The reference solutions aim for precompute: O((logr)^2) runtime: O((logn)^2) time and O((logr)^2) space. Always re-derive complexity from the code you write in the interview.
What pattern does Number of Integers With Popcount-Depth Equal to K I use?+
It primarily maps to general problem-solving, within the broader topic of math.
Is Number of Integers With Popcount-Depth Equal to K I 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/number-of-integers-with-popcount-depth-equal-to-k-i/