Range Module

We can implement the RangeModule class using an interval tree or a sorted list to efficiently manage the intervals. Below, I’ll present a solution that uses a sorted list.

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class RangeModule:

    def __init__(self):
        self.ranges = []

    def addRange(self, left: int, right: int) -> None:
        new_ranges = []
        i = 0
        # Skip the intervals that end before the new interval starts
        while i < len(self.ranges) and self.ranges[i][1] < left:
            new_ranges.append(self.ranges[i])
            i += 1
        # Merge overlapping intervals with the new interval
        while i < len(self.ranges) and self.ranges[i][0] <= right:
            left = min(left, self.ranges[i][0])
            right = max(right, self.ranges[i][1])
            i += 1
        new_ranges.append((left, right))
        # Append the remaining intervals
        new_ranges += self.ranges[i:]
        self.ranges = new_ranges

    def queryRange(self, left: int, right: int) -> bool:
        for start, end in self.ranges:
            if end < left:
                continue
            if start <= left and end >= right:
                return True
            return False
        return False

    def removeRange(self, left: int, right: int) -> None:
        new_ranges = []
        for start, end in self.ranges:
            if end <= left or start >= right:
                new_ranges.append((start, end))
            else:
                if start < left:
                    new_ranges.append((start, left))
                if end > right:
                    new_ranges.append((right, end))
        self.ranges = new_ranges

This code represents the intervals as a sorted list of tuples (start, end). The addRange method merges overlapping intervals with the new one, the queryRange method checks whether the given range is completely covered, and the removeRange method removes or splits the intervals that overlap with the given range.

The time complexity of the methods depends on the number of intervals and can be up to O(n) where n is the number of intervals in the worst case. There are more optimized solutions using more complex data structures like interval trees or segment trees, but this implementation should be enough for the given constraints.

Identifying Problem Isomorphism

“Range Module” has an approximate isomorphic problem “My Calendar III”.

In both problems, the main task involves handling ranges (or intervals) and adding new ranges to existing ones. The “Range Module” problem asks you to design a data structure to track ranges, supporting addRange, queryRange, and removeRange methods. The “My Calendar III” problem, on the other hand, requires implementing a booking system that supports booking an event and returns the K-booking (maximum overlap between events) at any given time.

In “Range Module”, the addRange method can be seen as the book method in the “My Calendar III” problem, where you are adding a new range to the existing ranges. The queryRange method in the “Range Module” problem can be somewhat seen as determining the K-booking in the “My Calendar III” problem, as both involve querying the state of the intervals.

This mapping is approximate, because the “Range Module” problem involves the removeRange operation, which doesn’t have an equivalent in the “My Calendar III” problem. Furthermore, the specific task of finding the K-booking in the “My Calendar III” problem is also unique to it and doesn’t have an exact equivalent in the “Range Module” problem. Nevertheless, the central theme of handling intervals and adding new intervals is common in both problems, making this an approximate isomorphism.

10 Prerequisite LeetCode Problems

“Range Module” (LeetCode Problem #715) involves designing a data structure that efficiently supports operations to add, query, and remove ranges. This requires understanding of data structures and interval manipulation. Here are problems as preparation:

1. “Merge Intervals” (LeetCode Problem #56): This is a basic interval problem that introduces the concept of merging overlapping intervals.

2. “Insert Interval” (LeetCode Problem #57): This problem is a bit more challenging than “Merge Intervals” as you need to insert a new interval into an existing set of intervals.

3. “Non-overlapping Intervals” (LeetCode Problem #435): This problem takes the concept of intervals a step further and asks you to minimize the number of intervals so that none of them overlap.

4. “Interval List Intersections” (LeetCode Problem #986): This problem requires finding the intersection of two lists of intervals.

5. “Data Stream as Disjoint Intervals” (LeetCode Problem #352): This problem is similar to the “Range Module” but with ranges coming in as a data stream.

6. “Find Right Interval” (LeetCode Problem #436): This problem involves searching for a specific interval in an array of intervals.

7. “Meeting Rooms II” (LeetCode Problem #253): This problem requires you to find the minimum number of meeting rooms given an array of meeting time intervals.

8. “Partition Labels” (LeetCode Problem #763): This problem deals with partitioning a string into as many parts as possible so that each letter appears in at most one part.

9. “My Calendar I” (LeetCode Problem #729): This problem involves scheduling without overlaps, similar to some of the operations in the “Range Module”.

10. “Find And Replace in String” (LeetCode Problem #833): Although not dealing with numerical ranges, this problem exercises similar skills in manipulating intervals within a string.

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class RangeModule:
    def __init__(self):
        self.track = []

    def addRange(self, left, right):
        start = bisect.bisect_left(self.track, left)
        end = bisect.bisect_right(self.track, right)

        subtrack = []
        if start % 2 == 0:
            subtrack.append(left)
        if end % 2 == 0:
            subtrack.append(right)

        self.track[start:end] = subtrack

    def removeRange(self, left, right):
        start = bisect.bisect_left(self.track, left)
        end = bisect.bisect_right(self.track, right)

        subtrack = []
        if start % 2 == 1:
            subtrack.append(left)
        if end % 2 == 1:
            subtrack.append(right)

        self.track[start:end] = subtrack

    def queryRange(self, left, right):
        start = bisect.bisect_right(self.track, left)
        end = bisect.bisect_left(self.track, right)

        return start == end and start % 2 == 1

Problem Classification

The problem falls into the domain of Intervals. It is about implementing a data structure to manage and query ranges of numbers efficiently. This problem involves working with half-open intervals, which is a common concept in mathematical range operations and data structures like interval trees.

Here are the ‘What’ components of the problem:

1. Data Structure Initialization: This task involves designing a data structure that can handle range-based operations.

2. Adding a Range: We need to create a method to add a half-open interval to the data structure. If there are overlaps with existing ranges, we should merge them into a single range.

3. Querying a Range: We must implement a method to query whether all the numbers in a given half-open interval are being tracked by our data structure.

4. Removing a Range: Finally, we need to devise a method to stop tracking all the numbers in a given half-open interval.

This can be further classified as a Data Structure Design and Interval Management problem. It is about designing a data structure with a particular set of operations (adding, querying, removing ranges) and optimizing it for efficient range tracking and management.

This problem is common in scenarios that involve tracking ranges or intervals, such as time scheduling, resource allocation in computing systems, or even in genome interval queries in bioinformatics. The categorizations provide a way to focus on the fundamental operations needed and also to compare this problem with other similar problems in data structures and algorithms.

Thought Process

The thought process behind coming up with the explanation and the code is largely iterative and involves both top-down and bottom-up thinking.

1. Understanding the Problem: The first step is always to thoroughly understand the problem and its requirements. This involves not only understanding what the problem is asking for but also the constraints under which the problem needs to be solved.

2. Initial Design: Based on the problem statement, the necessity for an efficient way to track intervals and to handle overlapping intervals is identified. The nature of the problem suggests that intervals need to be kept sorted. With these insights, an initial design involving a list to store the intervals in sorted order is made.

3. Deeper Insights: Now we think about how to efficiently perform the operations addRange, removeRange, and queryRange. This is where the binary search (bisect in Python) comes in. Binary search can find the positions to add or remove ranges in logarithmic time, which significantly enhances the performance.

4. Handling Edge Cases: At this point, the use of odd and even indices is considered to determine whether a range is inside or outside of the current tracked ranges. This is a clever way to handle the possible overlap and interval extension cases.

5. Coding: With a clear design in mind, the actual coding becomes more straightforward. Python’s built-in bisect module is used to perform binary search operations, and the remainder operation (%) is used to determine if an index is even or odd.

6. Explanation: The explanation is then based on the final code and the thought process behind it. It breaks down what each part of the code is doing and how it contributes to the overall solution, highlighting the use of binary search and the handling of intervals with odd and even indices.

Remember, coming up with such solutions and explanations often involves a lot of trial and error, practice, and familiarity with different data structures and algorithms. Over time, you will develop an intuition for identifying effective strategies for different types of problems.

The even or odd nature of the index represents whether the value at that index represents the start of a range (even index) or the end of a range (odd index) within the tracking list.

To provide some context, let’s visualize this with an example: Suppose we have a tracking list like this [a, b, c, d, e, f]. In this list, a and b represent a range, c and d represent another range, and e and f represent a third range. Notice how the start of every range is at an even index (0, 2, 4) and the end of every range is at an odd index (1, 3, 5). This is how we distinguish between the start and end of ranges in our tracking list.

So, when we get the index using the bisect function and check if the index is even or odd, we are determining if the value we’re dealing with falls within a range (an odd index, indicating it’s not beyond the end of an existing range) or outside of a range (an even index, indicating it’s not within the start and end of an existing range).

In addRange, if start or end index is even, it means left or right falls outside an existing range, so they should be included in the tracking list to extend the existing range or start a new range.

In removeRange, if start or end index is odd, it means left or right falls inside an existing range, so they should be included in the tracking list to reduce the existing range.

Understanding the role of even and odd indices in this context is crucial for understanding how the addRange and removeRange methods work.

It’s indeed a clever way to tackle the problem, and it nicely demonstrates the power of applying algorithmic thinking and knowledge of data structures.

The even length of the tracking array is indeed a critical aspect to grasp. It’s based on the idea that we always store intervals (or ranges) which always have a start and an end, hence two elements. So, no matter how many ranges we store, we always end up with an even number of elements in the array.

Remember, solutions like these often come from a deep understanding of the problem at hand and a strong grasp of fundamental computer science concepts. This involves not just coding, but also thinking about different ways to represent and manipulate data to meet our requirements. It’s this mix of creative problem solving and technical skill that makes programming such a rewarding endeavor!

The key thing to understand is how we’re representing the ranges in the self.track array. We’re representing ranges as pairs of start and end values. If we had a range from 5 to 10 and another range from 15 to 20, self.track would look like this: [5, 10, 15, 20].

The array stores the start and end of each range as pairs, so for any given pair, the start is at an even index (0-based) and the end is at an odd index. If we were to visualize it, it would look something like this:

Index:  0   1   2   3
Value:  5  10  15  20

So if start or end is an even number, it corresponds to the beginning of a range or a spot in between ranges, and if it’s odd, it corresponds to inside a range. That’s why if start or end is even in addRange(), we know that the left or right value is not currently in any range, and we add it to the subtrack array.

Similarly, in removeRange(), if start or end is odd, we know that left or right is currently in a range, and we add it to the subtrack array.

In other words, the parity of the index is used to infer whether we are inside a range (odd index) or outside a range (even index) in the self.track array. This understanding helps to correctly update the self.track array when we are adding or removing ranges.

The time complexity of the final solution is as follows:

1. addRange(): The time complexity is O(n) due to the slicing operation in Python lists, which requires shifting elements and can take up to O(n) time in the worst case. Here, n is the number of elements in the tracking list.

2. removeRange(): Similar to addRange(), the slicing operation results in a time complexity of O(n).

3. queryRange(): The time complexity is O(log n) as it involves binary search operations, which is what bisect_right and bisect_left perform. Here, n is the number of elements in the tracking list.

It’s important to note that the main source of complexity in addRange() and removeRange() comes from manipulating the list, not from finding the insertion points. Although the bisect module allows us to find the insertion points in O(log n) time, the list slicing operation (e.g., self.track[start:end] = subtrack) is what results in the O(n) complexity.

For the space complexity, it’s O(m), where m is the number of unique ranges being tracked. In the worst-case scenario, every addition could be a new unique range not overlapping with any existing range.

Language Agnostic Coding Drills

1. Dissecting the Code into Distinct Concepts

   a. Object-Oriented Programming (OOP): This concept is shown in the creation of the class RangeModule and the methods it contains. In OOP, data and functions that manipulate the data are bundled together as objects.

   b. Initializer method: In Python, the __init__ method is what’s known as a constructor. This method is called when an object is created from a class and it allows the class to initialize the attributes of the class.

   c. List Manipulation: In the code, lists are used to keep track of ranges. Methods like append are used to add elements to the list.

   d. Binary Search: The bisect module in Python implements binary search and insertion into a sorted list. The bisect_left function finds the insertion point for a given element which would be appropriate to maintain sorted order of the list. The bisect_right (or bisect) function works similarly, but returns an insertion point which comes after (to the right) of any existing entries of the element in the list.

   e. Interval Manipulation: The main problem this code is solving revolves around manipulation of intervals. This includes adding intervals, removing intervals, and querying intervals.

2. Difficulty Ordering and Description

   a. List Manipulation (Easy): Basic understanding of lists (arrays) is essential for almost all coding problems.

   b. Object-Oriented Programming (Intermediate): This requires an understanding of classes, objects, methods, and properties, which are fundamental to Python and many other modern languages.

   c. Initializer method (Intermediate): Understanding the use of constructors is crucial for object-oriented programming and is used extensively in codebases.

   d. Binary Search (Intermediate): This is a bit more complex, requiring understanding of divide-and-conquer strategies, but is still a fundamental algorithmic concept.

   e. Interval Manipulation (Advanced): This is a higher level concept that involves managing multiple ranges of values, including how they overlap, and is the most difficult concept here.

3. Problem-Solving Approach

This problem is essentially about managing a set of intervals, or ranges. The overall solution works by maintaining a sorted list of disjoint intervals, and then manipulating this list whenever a new interval is added or an existing interval is removed.

Each of the concepts/drills plays a key role in the overall solution:

a. List Manipulation: The `addRange` and `removeRange` methods manipulate the `track` list by adding or removing elements to represent the current set of ranges.

b. Object-Oriented Programming: The problem is naturally framed as an object that tracks a set of ranges, and the operations are methods that manipulate this object's state.

c. Initializer method: This sets up the initial state of the `RangeModule` object, which starts off tracking no ranges.

d. Binary Search: This is used to efficiently find the positions in the `track` list where the `addRange` and `removeRange` methods should manipulate the list.

e. Interval Manipulation: This is the heart of the problem - representing a set of ranges as a list of endpoints, and then adding or removing ranges by inserting or deleting endpoints in the list.

Targeted Drills in Python

1. Python-Based Coding Drills:

   a. List Manipulation

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   # Initialization my_list = [] # Adding elements my_list.append(10) my_list.append(20) # Removing elements my_list.remove(10) # Accessing elements print(my_list[0]) # Outputs: 20

   b. Object-Oriented Programming

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   class MyClass: def __init__(self, x): self.x = x def increment(self): self.x += 1 obj = MyClass(10) obj.increment() print(obj.x) # Outputs: 11

   c. Initializer Method

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   class MyClass: def __init__(self, x): self.x = x obj = MyClass(10) print(obj.x) # Outputs: 10

   d. Binary Search

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   import bisect # Initialize a sorted list sorted_list = [10, 20, 30, 40] # Find insertion point to maintain sorted list print(bisect.bisect_left(sorted_list, 25)) # Outputs: 2

   e. Interval Manipulation

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   # Initialize list of intervals intervals = [[10, 20], [30, 40]] # Adding interval new_interval = [25, 35] intervals.append(new_interval) # Querying interval query = [15, 35] # Check if this interval overlaps with any in the list for interval in intervals: if query[0] < interval[1] and query[1] > interval[0]: print("Overlaps with interval:", interval)

2. Problem-Specific Concepts:

   The primary problem-specific concept in this problem is the interval manipulation. As the problem deals with ranges and how they overlap or don’t overlap, understanding how to work with these ranges is critical to solving the problem. For instance, when adding a range, it is necessary to understand how to add that range in a way that keeps the list sorted and non-overlapping.

3. Integration of the Concepts:

   The concepts listed above would be integrated together in the following way:

   a. Object-Oriented Programming and Initializer Method: First, define a class RangeModule and initialize an empty list track in the constructor (__init__ method).

   b. List Manipulation: Use list manipulation in the methods of the class to add or remove elements from track list.

   c. Binary Search: Use binary search in addRange, removeRange, and queryRange methods to efficiently find positions in the track list where to add or remove elements.

   d. Interval Manipulation: The core part of the problem is managing intervals. When adding or removing a range, use the interval manipulation techniques to adjust the track list accordingly. When querying a range, use interval manipulation to check if every number in the queried range is being tracked.