Falling Squares
Time O(nlogn) · Space O(n) · Official statement on LeetCode
Solutions
// Time: O(nlogn)
// Space: O(n)
class Solution {
public:
vector<int> fallingSquares(vector<vector<int>>& positions) {
vector<int> result;
map<int, int> heights;
int maxH = heights[-1] = 0;
for (const auto& p : positions) {
auto it0 = heights.upper_bound(p[0]);
auto it1 = heights.lower_bound(p[0] + p[1]);
int h0 = prev(it0)->second;
int h1 = prev(it1)->second;
for (auto it = it0; it != it1; ++it) {
h0 = max(h0, it->second);
}
heights.erase(it0, it1);
heights[p[0]] = h0 + p[1];
heights[p[0] + p[1]] = h1;
maxH = max(maxH, h0 + p[1]);
result.emplace_back(maxH);
}
return result;
}
};
// Time: O(nlogn)
// Space: O(n)
// Segment Tree solution.
class Solution2 {
public:
vector<int> fallingSquares(vector<vector<int>>& positions) {
set<int> index;
for (const auto& position : positions) {
index.emplace(position[0]);
index.emplace(position[0] + position[1] - 1);
}
SegmentTree tree(index.size());
auto max_height = 0;
vector<int> result;
for (const auto& position : positions) {
const auto L = distance(index.begin(), index.find(position[0]));
const auto R = distance(index.begin(), index.find(position[0] + position[1] - 1));
const auto h = tree.query(L, R) + position[1];
tree.update(L, R, h);
max_height = max(max_height, h);
result.emplace_back(max_height);
}
return result;
}
private:
class SegmentTree {
public:
SegmentTree(int N)
: N_(N),
tree_(2 * N),
lazy_(N)
{
H_ = 1;
while ((1 << H_) < N) {
++H_;
}
}
void update(int L, int R, int h) {
L += N_; R += N_;
int L0 = L, R0 = R;
while (L <= R) {
if ((L & 1) == 1) {
apply(L++, h);
}
if ((R & 1) == 0) {
apply(R--, h);
}
L >>= 1; R >>= 1;
}
pull(L0); pull(R0);
}
int query(int L, int R) {
L += N_; R += N_;
auto result = 0;
push(L); push(R);
while (L <= R) {
if ((L & 1) == 1) {
result = max(result, tree_[L++]);
}
if ((R & 1) == 0) {
result = max(result, tree_[R--]);
}
L >>= 1; R >>= 1;
}
return result;
}
private:
int N_, H_;
vector<int> tree_, lazy_;
void apply(int x, int val) {
tree_[x] = max(tree_[x], val);
if (x < N_) {
lazy_[x] = max(lazy_[x], val);
}
}
void pull(int x) {
while (x > 1) {
x >>= 1;
tree_[x] = max(tree_[x * 2], tree_[x * 2 + 1]);
tree_[x] = max(tree_[x], lazy_[x]);
}
}
void push(int x) {
for (int h = H_; h > 0; --h) {
int y = x >> h;
if (lazy_[y] > 0) {
apply(y * 2, lazy_[y]);
apply(y * 2 + 1, lazy_[y]);
lazy_[y] = 0;
}
}
}
};
};
// Time: O(n * sqrt(n))
// Space: O(n)
class Solution3 {
public:
vector<int> fallingSquares(vector<vector<int>>& positions) {
set<int> index;
for (const auto& position : positions) {
index.emplace(position[0]);
index.emplace(position[0] + position[1] - 1);
}
const auto W = index.size();
const auto B = static_cast<int>(sqrt(W));
vector<int> heights(W);
vector<int> blocks(B + 2), blocks_read(B + 2);
auto max_height = 0;
vector<int> result;
for (const auto& position : positions) {
const auto L = distance(index.begin(), index.find(position[0]));
const auto R = distance(index.begin(), index.find(position[0] + position[1] - 1));
const auto h = query(B, L, R, heights, blocks, blocks_read) + position[1];
update(B, h, L, R, &heights, &blocks, &blocks_read);
max_height = max(max_height, h);
result.emplace_back(max_height);
}
return result;
}
private:
int query(const int B,
int left, int right,
const vector<int>& heights,
const vector<int>& blocks, const vector<int>& blocks_read) {
int result = 0;
while (left % B > 0 && left <= right) {
result = max(result, max(heights[left], blocks[left / B]));
result = max(result, blocks[left / B]);
++left;
}
while (right % B != B - 1 && left <= right) {
result = max(result, max(heights[right], blocks[right / B]));
--right;
}
while (left <= right) {
result = max(result, max(blocks[left / B], blocks_read[left / B]));
left += B;
}
return result;
}
void update(const int B, const int h,
int left, int right,
vector<int> *heights,
vector<int> *blocks, vector<int> *blocks_read) {
while (left % B > 0 && left <= right) {
(*heights)[left] = max((*heights)[left], h);
(*blocks_read)[left / B] = max((*blocks_read)[left / B], h);
++left;
}
while (right % B != B - 1 && left <= right) {
(*heights)[right] = max((*heights)[right], h);
(*blocks_read)[right / B] = max((*blocks_read)[right / B], h);
--right;
}
while (left <= right) {
(*blocks)[left / B] = max((*blocks)[left / B], h);
left += B;
}
}
};
// Time: O(n^2)
// Space: O(n)
class Solution4 {
public:
vector<int> fallingSquares(vector<vector<int>>& positions) {
vector<int> heights(positions.size());
for (int i = 0; i < positions.size(); ++i) {
int left_i = positions[i][0], size_i = positions[i][1];
int right_i = left_i + size_i;
heights[i] += size_i;
for (int j = i + 1; j < positions.size(); ++j) {
int left_j = positions[j][0], size_j = positions[j][1];
int right_j = left_j + size_j;
if (left_j < right_i and left_i < right_j) { // intersect
heights[j] = max(heights[j], heights[i]);
}
}
}
vector<int> result;
for (const auto& height : heights) {
result.emplace_back(result.empty() ? height : max(result.back(), height));
}
return result;
}
};
Beginner Explanation
What is Falling Squares?
Falling Squares (LeetCode #699) is a Hard problem that primarily trains tree.
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 segment tree.
- 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: Segment Tree.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Falling Squares
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 segment tree.
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(nlogn)) and space (O(n)) 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(nlogn) time and O(n) space.
Pattern focus: segment tree
Use the pattern as a checklist:
- segment tree — 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(nlogn) |
| Space | O(n) |
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 Falling Squares
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for segment tree — 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 segment tree:
- 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: tree.
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 Falling Squares in a second language (cpp, python).
- Drill 3–5 more problems tagged tree.
- 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 segment tree 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
Falling Squares (#699) — Hard. Pattern: segment tree. Complexity: O(nlogn) time / O(n) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Falling Squares?+
The reference solutions aim for O(nlogn) time and O(n) space. Always re-derive complexity from the code you write in the interview.
What pattern does Falling Squares use?+
It primarily maps to segment tree, within the broader topic of tree.
Is Falling Squares 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/falling-squares/