What is the time complexity of Block Placement Queries?

The optimal solution has a time complexity of O(q * log(m)).

What data structures are used to solve Block Placement Queries?

This problem uses Array, Binary Search, Binary Indexed Tree, Segment Tree. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Block Placement Queries?

Insight 1: The core challenge lies in efficiently determining the largest contiguous free space within a given range after multiple obstacles have been placed. Insight 2: Segment Tree data structure is suitable for solving the range queries efficiently. The segment tree will store information about maximum contiguous free space within each segment.

Which companies ask LeetCode 3161. Block Placement Queries in interviews?

This problem is frequently asked at Autodesk, Capital One, PayPay, Roblox, SIG and 1 more companies.

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Block Placement Queries - LeetCode 3161 Solution

Block Placement Queries - Complete Solution Guide

Block Placement Queries is LeetCode problem 3161, 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 exists an infinite number line, with its origin at 0 and extending towards the positive x-axis. You are given a 2D array queries , which contains two types of queries: For a query of type 1, queries[i] = [1, x] . Build an obstacle at distance x from the origin. It is guaranteed that there is no obstacle at distance x when the query is asked. For a query of type 2, queries[i] = [2, x, sz] . Check if it is possible to place a block of size sz anywhere in the range [0, x] on the line, such th

Detailed Explanation

The problem asks us to simulate placing obstacles on an infinite number line and then answering queries about whether a block of a given size can be placed within a specified range, avoiding the obstacles. There are two types of queries: type 1 places an obstacle at a given coordinate, and type 2 asks if a block of a given size can be placed entirely within a specified range without overlapping any obstacles. The output should be a boolean array, where each element indicates whether the corresponding type 2 query can be satisfied.

Solution Approach

The solution uses a Segment Tree to efficiently handle the obstacle placements and block placement queries. Each node in the segment tree stores the length of the longest contiguous free segment, the length of the free segment starting from the left end (prefix), the length of the free segment ending at the right end (suffix) and total length of the interval represented by that node. The `_build` function constructs the segment tree initially with all segments considered free. The `_update` function marks a point (obstacle) in the tree, which involves updating the tree recursively from the root to the leaf node corresponding to that point and then updating the affected parent nodes. The `_query` function finds the maximum contiguous free segment length in the given range. The main function `getResults` iterates through the queries, updating the segment tree for type 1 queries and querying the tree for type 2 queries. The results of type 2 queries are stored in the `results` list and returned.

Step-by-Step Algorithm

Step 1: Initialize the Segment Tree: Build a segment tree representing the number line from 0 to the maximum coordinate specified in the input queries. Each node in the tree will store the maximum contiguous free space, prefix free space, suffix free space and total length of the interval.
Step 2: Process Queries: Iterate through the queries one by one.
Step 3: Handle Type 1 Queries (Obstacle Placement): For a query of type 1, update the segment tree to mark the specified coordinate as occupied by an obstacle. This involves traversing the tree to the leaf node representing the coordinate and updating the node's information to indicate no free space. Then, update the parent nodes recursively.
Step 4: Handle Type 2 Queries (Block Placement Check): For a query of type 2, query the segment tree to find the maximum contiguous free space within the specified range. This involves traversing the tree and merging results from relevant nodes to calculate the maximum free space in the given range.
Step 5: Check Placement Possibility: Compare the maximum contiguous free space found in Step 4 with the block size specified in the query. If the maximum free space is greater than or equal to the block size, the block can be placed; otherwise, it cannot.
Step 6: Store Results: Add the result (true or false) of the block placement check to the `results` list.
Step 7: Return Results: After processing all queries, return the `results` list.

Key Insights

Insight 1: The core challenge lies in efficiently determining the largest contiguous free space within a given range after multiple obstacles have been placed.
Insight 2: Segment Tree data structure is suitable for solving the range queries efficiently. The segment tree will store information about maximum contiguous free space within each segment.
Insight 3: The merge operation on the segment tree is crucial. It needs to combine information from the left and right children to correctly calculate the maximum free space, prefix free space and suffix free space for the parent node.

Complexity Analysis

Time Complexity: O(q * log(m))

Space Complexity: O(m)

Topics

This problem involves: Array, Binary Search, Binary Indexed Tree, Segment Tree.

Companies

Asked at: Autodesk, Capital One, PayPay, Roblox, SIG, Visa.

Block Placement Queries | LeetCode 3161

Hard

Array Binary Search Binary Indexed Tree Segment Tree

Time: O(q * log(m))Space: O(m)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {

    private int MAX_COORD;
    private int[][] tree;

    public List<Boolean> getResults(int[][] queries) {
        int max_coord = 0;
        for (int[] q : queries) {
            max_coord = Math.max(max_coord, q[1]);
        }

        this.MAX_COORD = max_coord + 1;
        this.tree = new int[4 * this.MAX_COORD][4];

        _build(1, 0, this.MAX_COORD - 1);

        List<Boolean> results = new ArrayList<>();
        for (int[] q : queries) {
            if (q[0] == 1) {
                int x = q[1];
                _update(1, 0, this.MAX_COORD - 1, x);
            } else {
                int x = q[1], sz = q[2];

                if (x == 0) {
                    results.add(sz == 0);
                    continue;
                }

                int[] res = _query(1, 0, this.MAX_COORD - 1, 0, x);
                int p = res[0], s = res[1], m = res[2];

                int max_block_size = Math.max(Math.max(p, s), m);
                results.add(max_block_size >= sz);
            }
        }

        return results;
    }

    private int[] _merge(int[] left_res, int[] right_res) {
        int lp = left_res[0], ls = left_res[1], lm = left_res[2], llen = left_res[3];
        int rp = right_res[0], rs = right_res[1], rm = right_res[2], rlen = right_res[3];

        int pref = lp;
        if (lp == llen) {
            pref += rp;
        }

        int suff = rs;
        if (rs == rlen) {
            suff += ls;
        }

        int max_free = Math.max(Math.max(lm, rm), ls + rp);

        int total_len = llen + rlen;
        return new int[] {pref, suff, max_free, total_len};
    }

    private void _build(int v, int tl, int tr) {
        int length = tr - tl + 1;
        if (tl == tr) {
            this.tree[v] = new int[] {1, 1, 1, 1};
            return;
        }

        int tm = (tl + tr) / 2;
        _build(2 * v, tl, tm);
        _build(2 * v + 1, tm + 1, tr);

        this.tree[v] = new int[] {length, length, length, length};
    }

    private void _update(int v, int tl, int tr, int pos) {
        if (tl == tr) {
            this.tree[v] = new int[] {0, 0, 0, 1};
            return;
        }

        int tm = (tl + tr) / 2;
        if (pos <= tm) {
            _update(2 * v, tl, tm, pos);
        } else {
            _update(2 * v + 1, tm + 1, tr, pos);
        }

        this.tree[v] = _merge(this.tree[2 * v], this.tree[2 * v + 1]);
    }

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

        int tm = (tl + tr) / 2;

        if (r <= tm) {
            return _query(2 * v, tl, tm, l, r);
        }
        if (l > tm) {
            return _query(2 * v + 1, tm + 1, tr, l, r);
        }

        int[] res_l = _query(2 * v, tl, tm, l, tm);
        int[] res_r = _query(2 * v + 1, tm + 1, tr, tm + 1, r);

        return _merge(res_l, res_r);
    }
}

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Problem Info

Description

Problem 3161: Block Placement Queries - Detailed explanation and solution approaches.

Topics

Array Binary Search Binary Indexed Tree Segment Tree

Companies

AutodeskCapital OnePayPayRobloxSIGVisa

Stats

Difficulty:Hard

Languages:4

Companies:6

Problem #:3161

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

7 steps

1
Step 1: Initialize the Segment Tree: Build a segment tree representing the number line from 0 to the maximum coordinate specified in the input queries. Each node in the tree will store the maximum contiguous free space, prefix free space, suffix free space and total length of the interval.
2
Step 2: Process Queries: Iterate through the queries one by one.
3
Step 3: Handle Type 1 Queries (Obstacle Placement): For a query of type 1, update the segment tree to mark the specified coordinate as occupied by an obstacle. This involves traversing the tree to the leaf node representing the coordinate and updating the node's information to indicate no free space. Then, update the parent nodes recursively.
4
Step 4: Handle Type 2 Queries (Block Placement Check): For a query of type 2, query the segment tree to find the maximum contiguous free space within the specified range. This involves traversing the tree and merging results from relevant nodes to calculate the maximum free space in the given range.
5
Step 5: Check Placement Possibility: Compare the maximum contiguous free space found in Step 4 with the block size specified in the query. If the maximum free space is greater than or equal to the block size, the block can be placed; otherwise, it cannot.
6
Step 6: Store Results: Add the result (true or false) of the block placement check to the `results` list.
7
Step 7: Return Results: After processing all queries, return the `results` 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(q * log(m))

💾

Space Complexity

O(m)

📝

Why?

The segment tree has a height of log(m), so traversing from the root to a leaf during update or query takes O(log(m)) time. The space complexity is determined by the size of the segment tree array, which is proportional to the range of coordinates [0, m], leading to O(m) space complexity.

📖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 possible positions within the given range and check if the block can be placed without overlapping any obstacles. This would involve iterating from 0 to x-sz, and for each starting position, iterating from start to start+sz-1 to check if any obstacle falls in that range.

Time: O(q * x * sz) in the worst case, where q is the number of queries, x is the range end, and sz is the block size. This can become O(q * m^2) in the worst case.Space: O(k) where k is the number of obstacles, to store their locations. This can be improved to O(1) if obstacles are passed in as input and not stored.

Trade-offs: This approach is simple to implement but inefficient for a large number of queries or large ranges. The segment tree approach provides significantly better time complexity. This approach is viable only for small ranges and a small number of obstacles.

Binary Indexed Tree (BIT)

A Binary Indexed Tree can be used to maintain the number of obstacles up to a given index. This allows determining if there are any obstacles in a range using prefix sums in O(log m) time.

Time: O(q * x * log(m)) in the worst case. The x factor is because we still need to iterate through potential starting positions for each query. However, the presence check in each iteration becomes O(log m) instead of O(sz).Space: O(m) for the BIT

Trade-offs: BIT might be simpler to implement than a segment tree. However, it is still not optimal because of the x factor in the time complexity making it slower than the Segment tree solution. In the brute force, the inner loop iterates across 'sz', so if sz is smaller than log(m), then BIT provides no advantage.

✨

Key Insights

1Insight 1: The core challenge lies in efficiently determining the largest contiguous free space within a given range after multiple obstacles have been placed.
2Insight 2: Segment Tree data structure is suitable for solving the range queries efficiently. The segment tree will store information about maximum contiguous free space within each segment.
3Insight 3: The merge operation on the segment tree is crucial. It needs to combine information from the left and right children to correctly calculate the maximum free space, prefix free space and suffix free space for the parent node.

📚

Key Terms Explained

For Students

Array

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

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: Query range starts or ends at the edge of the number line (0 or x). The segment tree handles this correctly as it considers the prefix and suffix free spaces.
Edge case 2: Block size is zero. If the block size is zero and x is non-zero, we can place a block of size zero anywhere in [0, x]. If both size and x are zero, it's placed at 0, hence returning true. The code handles the case with `if (x == 0)` which directly evaluates if `sz == 0`.
Edge case 3: No obstacles are present within the query range. In this case, the query function will return the length of the range as the maximum free space.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly merging the results from the left and right subtrees in the segment tree. This is a common mistake because it requires careful consideration of prefix, suffix, and maximum free space.

✗Common mistake 2: Not considering the case when the query range is empty (l > r) or the coordinates are zero. Handling zero coordinates correctly is important to avoid errors.

✗Common mistake 3: Forgetting to initialize segment tree correctly, leading to incorrect results on initial queries.

✅

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 contiguous blocks of memory or disk space. Obstacles would be already allocated regions. The queries would be whether a new block of a certain size can be allocated in a specified range.

🏦Real-world scenario 2: Scheduling: Consider scheduling tasks on a timeline where certain slots are already occupied (obstacles). The queries would be if a new task of a given duration can be scheduled within a specific time window.

🎮Real-world scenario 3: Placing advertisements on a webpage: Webpages have sections where advertisements can be inserted. Some sections might be occupied, some not. The query will check if an advertisement of given dimension can be placed within some section.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ 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: September 10, 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

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

@AlgoMasterJul 12

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 5

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 28

❓

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.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

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 "Block Placement Queries"?

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 binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?