What is the time complexity of Range Module?

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

What data structures are used to solve Range Module?

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

What are the key insights for solving Range Module?

Insight 1: The core idea is to maintain a sorted list of non-overlapping intervals. This allows efficient searching and manipulation of the ranges. Insight 2: Binary search is crucial for quickly locating the position to insert, query, or remove intervals within the sorted list.

Which companies ask LeetCode 715. Range Module in interviews?

This problem is frequently asked at Coupang, Machine Zone.

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Range Module - LeetCode 715 Solution

Q: How to solve Range Module?

The solution utilizes a sorted list (or array in C) to store the tracked ranges. The `addRange`, `queryRange`, and `removeRange` operations all rely on binary search to find the relevant intervals. During `addRange`, overlapping intervals are merged. During `removeRange`, intervals are split if necessary, and any completely removed intervals are deleted. `queryRange` checks if a given range is completely covered by any existing interval.

Range Module - Complete Solution Guide

Range Module is LeetCode problem 715, 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

A Range Module is a module that tracks ranges of numbers. Design a data structure to track the ranges represented as half-open intervals and query about them. A half-open interval [left, right) denotes all the real numbers x where left <= x < right . Implement the RangeModule class: RangeModule() Initializes the object of the data structure. void addRange(int left, int right) Adds the half-open interval [left, right) , tracking every real number in that interval. Adding an interval that partiall

Detailed Explanation

The Range Module problem requires designing a data structure to efficiently track and manage ranges of numbers, represented as half-open intervals [left, right). The data structure should support three operations: adding a range, querying if a range is fully tracked, and removing a range. The constraints specify that the left and right bounds are integers between 1 and 10^9, and there are at most 10^4 calls to the addRange, queryRange, and removeRange methods.

Solution Approach

The solution utilizes a sorted list (or array in C) to store the tracked ranges. The `addRange`, `queryRange`, and `removeRange` operations all rely on binary search to find the relevant intervals. During `addRange`, overlapping intervals are merged. During `removeRange`, intervals are split if necessary, and any completely removed intervals are deleted. `queryRange` checks if a given range is completely covered by any existing interval.

Step-by-Step Algorithm

Step 1: **Initialization:** The `RangeModule` is initialized with an empty list to store the ranges.
Step 2: **addRange(left, right):**
a. Find the index `i` where the new range should be inserted (or where a merge might start) using binary search.
b. Find the index `j` where the merge should stop using binary search.
c. Adjust `left` and `right` to encompass any overlapping intervals that are to be merged.
d. Remove the merged intervals.
e. Insert the resulting interval [left, right).
Step 3: **queryRange(left, right):**
a. Find the interval that might contain the start of the query range using binary search.
b. Check if this interval covers the entire query range (i.e., its end is greater than or equal to `right`).
Step 4: **removeRange(left, right):**
a. Find the index `i` where the removal should start (or where a split might occur) using binary search.
b. Find the index `j` where the removal should stop (or where a split might occur) using binary search.
c. Split any overlapping intervals into smaller intervals that remain after the removal.
d. Remove the intervals that are completely within the removal range.
e. Insert the new, smaller intervals created by the split.

Key Insights

Insight 1: The core idea is to maintain a sorted list of non-overlapping intervals. This allows efficient searching and manipulation of the ranges.
Insight 2: Binary search is crucial for quickly locating the position to insert, query, or remove intervals within the sorted list.
Insight 3: Overlapping intervals need to be merged during addition and split during removal to maintain the non-overlapping property and correctly track the ranges.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

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

Companies

Asked at: Coupang, Machine Zone.

Range Module | LeetCode 715

Hard

Design Segment Tree Ordered Set

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

Solution Code

Available in:

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

class RangeModule {

    private List<int[]> ranges;

    public RangeModule() {
        this.ranges = new ArrayList<>();
    }

    public void addRange(int left, int right) {
        int i = binarySearchLeft(this.ranges, left);
        int j = binarySearchRight(this.ranges, right);

        int startReplace = i;
        if (i > 0 && this.ranges.get(i - 1)[1] >= left) {
            startReplace--;
        }

        int endReplace = j;

        if (startReplace < endReplace) {
            left = Math.min(left, this.ranges.get(startReplace)[0]);
            right = Math.max(right, this.ranges.get(endReplace - 1)[1]);
        }

        List<int[]> replacement = new ArrayList<>();
        replacement.add(new int[]{left, right});
        this.ranges.subList(startReplace, endReplace).clear();
        this.ranges.addAll(startReplace, replacement);
    }

    public boolean queryRange(int left, int right) {
        int i = binarySearchRight(this.ranges, left);
        if (i == 0) {
            return false;
        }

        int[] prev = this.ranges.get(i - 1);
        int prevStart = prev[0];
        int prevEnd = prev[1];
        return prevEnd >= right;
    }

    public void removeRange(int left, int right) {
        int i = binarySearchLeft(this.ranges, left);
        int j = binarySearchRight(this.ranges, right);

        int startReplace = i;
        if (i > 0 && this.ranges.get(i - 1)[1] > left) {
            startReplace--;
        }

        int endReplace = j;

        List<int[]> newSegments = new ArrayList<>();
        if (startReplace < endReplace) {
            if (this.ranges.get(startReplace)[0] < left) {
                newSegments.add(new int[]{this.ranges.get(startReplace)[0], left});
            }
            if (this.ranges.get(endReplace - 1)[1] > right) {
                newSegments.add(new int[]{right, this.ranges.get(endReplace - 1)[1]});
            }
        }

        this.ranges.subList(startReplace, endReplace).clear();
        this.ranges.addAll(startReplace, newSegments);
    }

    private int binarySearchLeft(List<int[]> ranges, int value) {
        int low = 0;
        int high = ranges.size();

        while (low < high) {
            int mid = low + (high - low) / 2;
            if (ranges.get(mid)[0] < value) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        return low;
    }

    private int binarySearchRight(List<int[]> ranges, int value) {
        int low = 0;
        int high = ranges.size();

        while (low < high) {
            int mid = low + (high - low) / 2;
            if (ranges.get(mid)[0] <= value) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        return low;
    }
}

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Problem Info

Description

Problem 715: Range Module - Detailed explanation and solution approaches.

Topics

Design Segment Tree Ordered Set

Companies

CoupangMachine Zone

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:715

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

16 steps

1
Step 1: **Initialization:** The `RangeModule` is initialized with an empty list to store the ranges.
2
Step 2: **addRange(left, right):**
3
a. Find the index `i` where the new range should be inserted (or where a merge might start) using binary search.
4
b. Find the index `j` where the merge should stop using binary search.
5
c. Adjust `left` and `right` to encompass any overlapping intervals that are to be merged.
6
d. Remove the merged intervals.
7
e. Insert the resulting interval [left, right).
8
Step 3: **queryRange(left, right):**
9
a. Find the interval that might contain the start of the query range using binary search.
10
b. Check if this interval covers the entire query range (i.e., its end is greater than or equal to `right`).
11
Step 4: **removeRange(left, right):**
12
a. Find the index `i` where the removal should start (or where a split might occur) using binary search.
13
b. Find the index `j` where the removal should stop (or where a split might occur) using binary search.
14
c. Split any overlapping intervals into smaller intervals that remain after the removal.
15
d. Remove the intervals that are completely within the removal range.
16
e. Insert the new, smaller intervals created by the split.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

Each operation, `addRange`, `queryRange`, and `removeRange`, performs binary searches, which take O(log n) time. However, `addRange` and `removeRange` might require merging/splitting intervals, which can involve iterating through the entire list of intervals in the worst case. Therefore, the overall time complexity is dominated by the worst-case merging/splitting operations, leading to O(n).

📖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

Segment Tree

A segment tree can be used to represent the ranges. Each node in the tree represents an interval, and the tree is constructed such that overlapping intervals are efficiently queried and updated.

Time: O(log n) for each operation, where n is the range of possible values (10^9 in this case).Space: O(n), where n is the range of possible values (10^9 in this case).

Trade-offs: Segment trees provide faster query and update operations but consume significantly more memory, especially when the range of possible values is large. This approach is efficient if the number of operations is very high, but the memory usage makes it less practical for the given constraints because the entire range 1 to 10^9 is stored in the tree even if just 1 to 2 intervals are created and it leads to memory limit exceeded error. The provided sorted list approach is better due to its lower memory footprint and decent performance.

Ordered Set (e.g., using TreeSet in Java or set in C++)

Using an ordered set of endpoints. Store the left and right endpoints of the ranges in the set. Operations involve finding the relevant endpoints and updating the set.

Time: O(log n) for each operation, where n is the number of endpoints.Space: O(n), where n is the number of endpoints.

Trade-offs: Ordered sets offer good performance, especially with balanced tree implementations, but can be more complex to implement compared to the sorted list approach. Requires more work to make sure the intervals are non-overlapping

✨

Key Insights

1Insight 1: The core idea is to maintain a sorted list of non-overlapping intervals. This allows efficient searching and manipulation of the ranges.
2Insight 2: Binary search is crucial for quickly locating the position to insert, query, or remove intervals within the sorted list.
3Insight 3: Overlapping intervals need to be merged during addition and split during removal to maintain the non-overlapping property and correctly track the ranges.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: Adding/Removing a range that completely covers all existing ranges. This requires clearing the entire existing range list and inserting the new range appropriately.
Edge case 2: Adding/Removing a range that falls entirely before or after all existing ranges. This requires inserting the new range at the beginning or end of the list, respectively.
Edge case 3: Empty RangeModule at the start. The solution needs to initialize properly and handle cases when the ranges list is empty.
Edge case 4: Adjacent intervals. The code should merge or split these appropriately, handling inclusive/exclusive boundary conditions correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect binary search implementation leading to off-by-one errors or incorrect interval selection.

✗Common mistake 2: Failing to handle overlapping intervals correctly during `addRange` and `removeRange` leading to incorrect tracking of the ranges.

✗Common mistake 3: Memory leaks in C implementation by not properly freeing dynamically allocated memory.

✗Common mistake 4: Ignoring edge cases like empty RangeModule or ranges covering all existing ranges.

✗Common mistake 5: Integer overflow when calculating midpoints in binary search, especially important in languages like C and C++.

✅

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 - Tracking available and allocated time slots or memory blocks.

🏦Real-world scenario 2: Calendar scheduling - Managing meeting times and detecting conflicts.

🎮Real-world scenario 3: Network address management - Allocating and tracking IP address ranges.

📱Real-world scenario 4: Video Editing software - Managing timelines of different cuts or edits.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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 23, 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

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

@DevExpert92Jul 14

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

@AlgoMasterJul 7

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

@InterviewProJun 30

❓

Frequently Asked Questions

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 "Range Module"?

For this hard-level Design 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?