What is the time complexity of Falling Squares?

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Falling Squares?

This problem uses Array, Segment Tree, Ordered Set. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Falling Squares?

Insight 1: The crucial realization is that only the 'landing' of the square matters, not the intermediate falling process. We only care about the height of the existing squares in the interval covered by the new square. Insight 2: The X-coordinates can be large, but the number of squares is limited. We can use coordinate compression to map these large X-coordinates to smaller indices to efficiently use data structures like segment trees.

Which companies ask LeetCode 699. Falling Squares in interviews?

This problem is frequently asked at Block.

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Falling Squares - LeetCode 699 Solution

Falling Squares - Complete Solution Guide

Falling Squares is LeetCode problem 699, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There are several squares being dropped onto the X-axis of a 2D plane. You are given a 2D integer array positions where positions[i] = [left i , sideLength i ] represents the i th square with a side length of sideLength i that is dropped with its left edge aligned with X-coordinate left i . Each square is dropped one at a time from a height above any landed squares. It then falls downward (negative Y direction) until it either lands on the top side of another square or on the X-axis . A square b

Detailed Explanation

The Falling Squares problem involves dropping squares onto the X-axis. Each square is defined by its leftmost X-coordinate and its side length. Squares fall until they hit either the X-axis or another square. The goal is to determine the maximum height of the stack of squares after each square is dropped and return a list containing these heights.

Solution Approach

The solution employs a Segment Tree with Lazy Propagation to efficiently track the maximum height within any given range on the X-axis. Coordinate compression is used to handle large X-coordinate values. After dropping each square, the algorithm queries the segment tree to find the maximum height within the range covered by the new square, adds the side length of the new square to that height, updates the segment tree with the new height for that range, and then records the maximum height encountered so far.

Step-by-Step Algorithm

Step 1: Collect all X-coordinates (left and right edges) of all squares.
Step 2: Sort the collected X-coordinates and remove duplicates to perform coordinate compression. This maps the X-coordinates to indices in a smaller range.
Step 3: Build a segment tree with lazy propagation. The segment tree stores the maximum height in each range of compressed coordinates.
Step 4: Iterate through each square:
Step 5: For each square, query the segment tree for the maximum height in the range [left_index, right_index], where the indices are based on compressed coordinates.
Step 6: Calculate the new height by adding the side length of the current square to the maximum height found in the previous step.
Step 7: Update the segment tree with the new height for the range [left_index, right_index].
Step 8: Update the global maximum height if the new height exceeds the current global maximum.
Step 9: Store the global maximum height in the answer list.
Step 10: Return the answer list.

Key Insights

Insight 1: The crucial realization is that only the 'landing' of the square matters, not the intermediate falling process. We only care about the height of the existing squares in the interval covered by the new square.
Insight 2: The X-coordinates can be large, but the number of squares is limited. We can use coordinate compression to map these large X-coordinates to smaller indices to efficiently use data structures like segment trees.
Insight 3: A Segment Tree with Lazy Propagation is the right data structure. Because we update ranges of coordinates with a new height, the lazy propagation optimizes the update operation.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Segment Tree, Ordered Set.

Companies

Asked at: Block.

Falling Squares | LeetCode 699

Hard

Array Segment Tree Ordered Set

Time: O(n log n)Space: O(n)

Solution Code

Available in:

import java.util.*;

class Solution {
    private int[] tree;
    private int[] lazy;
    private int size;

    public List<Integer> fallingSquares(int[][] positions) {
        Set<Integer> coords = new HashSet<>();
        for (int[] position : positions) {
            int left = position[0];
            int side = position[1];
            coords.add(left);
            coords.add(left + side);
        }

        List<Integer> sortedCoords = new ArrayList<>(coords);
        Collections.sort(sortedCoords);

        Map<Integer, Integer> coordMap = new HashMap<>();
        for (int i = 0; i < sortedCoords.size(); i++) {
            coordMap.put(sortedCoords.get(i), i);
        }

        int numIndices = sortedCoords.size();
        this.tree = new int[4 * numIndices];
        this.lazy = new int[4 * numIndices];
        this.size = numIndices;

        List<Integer> ans = new ArrayList<>();

        for (int[] position : positions) {
            int left = position[0];
            int side = position[1];
            int right = left + side;
            int lIdx = coordMap.get(left);
            int rIdx = coordMap.get(right) - 1;

            int baseHeight = _query(1, 0, this.size - 1, lIdx, rIdx);
            int newHeight = baseHeight + side;

            _update(1, 0, this.size - 1, lIdx, rIdx, newHeight);

            ans.add(this.tree[1]);
        }

        return ans;
    }

    private void _push(int v) {
        if (this.lazy[v] != 0 && v * 2 + 1 < this.tree.length) {
            this.tree[v * 2] = this.lazy[v];
            this.lazy[v * 2] = this.lazy[v];
            this.tree[v * 2 + 1] = this.lazy[v];
            this.lazy[v * 2 + 1] = this.lazy[v];
            this.lazy[v] = 0;
        }
    }

    private void _update(int v, int tl, int tr, int l, int r, int h) {
        if (l > r) {
            return;
        }
        if (l == tl && r == tr) {
            this.tree[v] = h;
            this.lazy[v] = h;
            return;
        }

        _push(v);
        int tm = (tl + tr) / 2;
        _update(v * 2, tl, tm, l, Math.min(r, tm), h);
        _update(v * 2 + 1, tm + 1, tr, Math.max(l, tm + 1), r, h);
        this.tree[v] = Math.max(this.tree[v * 2], this.tree[v * 2 + 1]);
    }

    private int _query(int v, int tl, int tr, int l, int r) {
        if (l > r) {
            return 0;
        }
        if (l == tl && r == tr) {
            return this.tree[v];
        }

        _push(v);
        int tm = (tl + tr) / 2;
        int leftH = _query(v * 2, tl, tm, l, Math.min(r, tm));
        int rightH = _query(v * 2 + 1, tm + 1, tr, Math.max(l, tm + 1), r);
        return Math.max(leftH, rightH);
    }
}

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Problem Info

Description

Problem 699: Falling Squares - Detailed explanation and solution approaches.

Topics

Array Segment Tree Ordered Set

Companies

Block

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:699

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Collect all X-coordinates (left and right edges) of all squares.
2
Step 2: Sort the collected X-coordinates and remove duplicates to perform coordinate compression. This maps the X-coordinates to indices in a smaller range.
3
Step 3: Build a segment tree with lazy propagation. The segment tree stores the maximum height in each range of compressed coordinates.
4
Step 4: Iterate through each square:
5
Step 5: For each square, query the segment tree for the maximum height in the range [left_index, right_index], where the indices are based on compressed coordinates.
6
Step 6: Calculate the new height by adding the side length of the current square to the maximum height found in the previous step.
7
Step 7: Update the segment tree with the new height for the range [left_index, right_index].
8
Step 8: Update the global maximum height if the new height exceeds the current global maximum.
9
Step 9: Store the global maximum height in the answer list.
10
Step 10: Return the answer list.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The coordinate compression involves sorting a set of at most 2*n coordinates. The segment tree is of size 4*n where n is the number of unique coordinate values. The loop iterates through the positions array once, and within each iteration performs segment tree queries and updates (log n operations).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Iterate through all previously dropped squares for each new square and check for overlaps. Keep track of the maximum height.

Time: O(n^2), where n is the number of squares. For each square, we iterate through all existing squares.Space: O(n), to store the heights of the squares.

Trade-offs: Much simpler to implement, but significantly slower for larger inputs. This is not a feasible solution for the given constraints.

Ordered Set (Interval Tree)

Use an ordered set (e.g., TreeSet in Java, set in C++) to represent the intervals and their heights. When a new square is added, find all overlapping intervals, calculate the new height, and update the ordered set.

Time: O(n log n), due to the operations on the ordered set.Space: O(n), to store the intervals in the ordered set.

Trade-offs: Similar time complexity to the segment tree approach, but might be slightly more complex to implement correctly.

✨

Key Insights

1Insight 1: The crucial realization is that only the 'landing' of the square matters, not the intermediate falling process. We only care about the height of the existing squares in the interval covered by the new square.
2Insight 2: The X-coordinates can be large, but the number of squares is limited. We can use coordinate compression to map these large X-coordinates to smaller indices to efficiently use data structures like segment trees.
3Insight 3: A Segment Tree with Lazy Propagation is the right data structure. Because we update ranges of coordinates with a new height, the lazy propagation optimizes the update operation.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: Overlapping squares. The algorithm correctly handles overlapping squares by querying the current maximum height within the relevant range and updating it accordingly.
Edge case 2: Squares with same left coordinate but different side lengths. The solution finds the maximum height correctly since it updates the covered interval with the maximum height.
Edge case 3: Empty input (positions is empty). The code implicitly handles this, and will return an empty list without issues.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using coordinate compression. Without coordinate compression, the segment tree could be very large, leading to memory issues and slower performance.

✗Common mistake 2: Forgetting to push lazy updates down the tree. This can lead to incorrect query results, especially if updates are not immediately reflected in the tree nodes.

✗Common mistake 3: Incorrectly calculating indices during query/update operations after coordinate compression. It's important to use the mapped indices and not the original coordinates.

✗Common mistake 4: Not handling the case where l > r in query/update calls. The base case should return 0.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation. Imagine allocating blocks of memory to processes. This algorithm could find the largest contiguous block already in use within a range.

🏦Real-world scenario 2: Skyline calculation. Determining the skyline of a city after adding buildings of varying heights and widths.

🎮Real-world scenario 3: Range queries in databases. Optimizing range queries in large databases by using segment trees for efficient data retrieval.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 7, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Falling Squares"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether segment tree techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?