Range Module

Hard

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Topics:

Arrays

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 partially overlaps with currently tracked numbers should add any numbers in the interval [left, right) that are not already tracked.
boolean queryRange(int left, int right) Returns true if every real number in the interval [left, right) is currently being tracked, and false otherwise.
void removeRange(int left, int right) Stops tracking every real number currently being tracked in the half-open interval [left, right).

Example:

Input
["RangeModule", "addRange", "removeRange", "queryRange", "queryRange", "queryRange"]
[[], [10, 20], [14, 16], [10, 14], [13, 15], [16, 17]]
Output
[null, null, null, true, false, true]

Explanation
RangeModule rangeModule = new RangeModule();
rangeModule.addRange(10, 20);
rangeModule.removeRange(14, 16);
rangeModule.queryRange(10, 14); // return True,(Every number in [10, 14) is being tracked)
rangeModule.queryRange(13, 15); // return False,(Numbers like 14, 14.03, 14.17 in [13, 15) are not being tracked)
rangeModule.queryRange(16, 17); // return True, (The number 16 in [16, 17) is still being tracked, despite the remove operation)

Constraints:

1 <= left < right <= 10^9
At most 10^4 calls will be made to addRange, queryRange, and removeRange.

Range Module

Hard

Google

5 views

Topics:

Arrays

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 partially overlaps with currently tracked numbers should add any numbers in the interval [left, right) that are not already tracked.
boolean queryRange(int left, int right) Returns true if every real number in the interval [left, right) is currently being tracked, and false otherwise.
void removeRange(int left, int right) Stops tracking every real number currently being tracked in the half-open interval [left, right).

Example:

Input
["RangeModule", "addRange", "removeRange", "queryRange", "queryRange", "queryRange"]
[[], [10, 20], [14, 16], [10, 14], [13, 15], [16, 17]]
Output
[null, null, null, true, false, true]

Explanation
RangeModule rangeModule = new RangeModule();
rangeModule.addRange(10, 20);
rangeModule.removeRange(14, 16);
rangeModule.queryRange(10, 14); // return True,(Every number in [10, 14) is being tracked)
rangeModule.queryRange(13, 15); // return False,(Numbers like 14, 14.03, 14.17 in [13, 15) are not being tracked)
rangeModule.queryRange(16, 17); // return True, (The number 16 in [16, 17) is still being tracked, despite the remove operation)

Constraints:

1 <= left < right <= 10^9
At most 10^4 calls will be made to addRange, queryRange, and removeRange.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Are the ranges provided to the addRange, queryRange, and removeRange methods always integers? What is the valid range for these integers?
What should I return from queryRange if the given range is only partially covered by tracked ranges? Should I return true only if the entire range is covered?
If multiple ranges overlap or are adjacent during an addRange or removeRange operation, should I merge or split them accordingly to maintain a disjoint set of ranges?
How frequently will each operation (addRange, queryRange, removeRange) be called relative to the others? Should I optimize for a specific operation?
Should I assume that the input ranges [left, right) for addRange, queryRange, and removeRange are always valid (i.e., left < right)?

Brute Force Solution

Approach

The Range Module problem involves keeping track of numerical intervals. A brute force approach directly manipulates the intervals by checking every number within the given ranges for insertions, deletions, or queries.

Here's how the algorithm would work step-by-step:

To add a range, go through each number within the specified range, one by one.
For each number, check if it's already present in any existing range.
If the number isn't already present, add it to the overall collection of numbers.
To remove a range, go through each number within the specified range, one by one.
For each number, check if it's present in our overall collection of numbers.
If the number is present, remove it.
To check if a range is tracked, go through each number within the specified range, one by one.
For each number, check if it's present in our overall collection of numbers.
If any number is missing, then the whole range isn't tracked, so we can say it's false.
If we get through all numbers in the specified range and they're all present, then the whole range is tracked, so we can say it's true.

Code Implementation

class RangeModule:

    def __init__(self):
        self.tracked_numbers = set()

    def addRange(self, left_bound: int, right_bound: int) -> None:
        for number in range(left_bound, right_bound):
            # We add the number if it's not tracked yet
            if number not in self.tracked_numbers:

                self.tracked_numbers.add(number)

    def removeRange(self, left_bound: int, right_bound: int) -> None:
        for number in range(left_bound, right_bound):
            # Remove tracked numbers within the range
            if number in self.tracked_numbers:

                self.tracked_numbers.remove(number)

    def queryRange(self, left_bound: int, right_bound: int) -> bool:
        for number in range(left_bound, right_bound):
            # If any number is missing, the range isn't tracked
            if number not in self.tracked_numbers:

                return False

        return True

Big(O) Analysis

Time Complexity

O(n*r) – The addRange and removeRange operations iterate through each number within the given range, where r represents the size of the range. For each number within the range, these operations check if the number is present in the existing data structure, which in the worst case could involve checking all n numbers currently being tracked. The queryRange operation also iterates through each number in the range (r), and for each number checks its presence amongst the n numbers. Thus, the time complexity of each operation (add, remove, query) becomes O(n*r), where n is the number of elements tracked and r is the range size.

Space Complexity

O(N) – The brute force approach outlined maintains an overall collection of numbers. In the worst case, we might add all numbers within the range of possible inputs to this collection. If the range of possible numbers is related to the number of add operations, and thus indirectly to 'N' (where 'N' represents the total number of operations), then this collection can grow linearly with 'N'. Thus, the auxiliary space required is proportional to the number of added numbers, leading to O(N) space complexity.

Optimal Solution

Approach

The problem deals with keeping track of intervals of numbers. The best way is to maintain a collection of sorted intervals and efficiently merge or split them when new ranges are added or removed. This avoids checking every single number within the range.

Here's how the algorithm would work step-by-step:

Store the existing intervals in a sorted manner. This allows us to quickly find where new intervals should go.
When adding a new range, find any existing intervals that overlap with it.
If overlapping intervals exist, merge them into a single larger interval that includes the new range.
If no overlaps, simply add the new range to the sorted list of intervals in the correct position.
When removing a range, find any existing intervals that overlap with the range to be removed.
If an interval overlaps, split it into two new intervals on either side of the removed range, or remove the interval entirely if it's completely covered by the removed range.
Always maintain the sorted order of the intervals after each addition or removal to ensure efficient operations.

Code Implementation

class RangeModule:

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

    def addRange(self, left, right):
        new_interval = [left, right]
        new_intervals = []
        insert_position = 0

        for interval_start, interval_end in self.intervals:
            # No overlap, new interval is completely before
            if new_interval[1] < interval_start:
                new_intervals.append([interval_start, interval_end])
            # No overlap, new interval is completely after
            elif new_interval[0] > interval_end:
                new_intervals.append([interval_start, interval_end])
                insert_position += 1
            # Overlap, merge the intervals
            else:
                new_interval[0] = min(new_interval[0], interval_start)
                new_interval[1] = max(new_interval[1], interval_end)

        # Insert the merged/new interval in the correct sorted position
        new_intervals.insert(insert_position, new_interval)
        self.intervals = new_intervals

    def queryRange(self, left, right):
        for interval_start, interval_end in self.intervals:
            # If any interval contains the range, return True
            if interval_start <= left and interval_end >= right:
                return True
        return False

    def removeRange(self, left, right):
        new_intervals = []

        for interval_start, interval_end in self.intervals:
            # No overlap, keep the existing interval
            if right < interval_start or left > interval_end:
                new_intervals.append([interval_start, interval_end])
            # Interval is completely contained in the range to remove
            elif left <= interval_start and right >= interval_end:
                continue
            # Partial overlap, split the interval into two
            else:
                if left > interval_start:
                    new_intervals.append([interval_start, left])

                if right < interval_end:
                    new_intervals.append([right, interval_end])

        # Updating the intervals is necessary after removal
        self.intervals = new_intervals

Big(O) Analysis

Time Complexity

O(n) – The core operations involve finding overlapping intervals and merging or splitting them. Since the intervals are maintained in sorted order, binary search can be used to locate the intervals that potentially overlap with a new range or a range to be removed. Binary search has a time complexity of O(log n), where n is the number of intervals. However, in the worst-case scenario, adding or removing a range might require merging or splitting a significant number of existing intervals, potentially iterating through them. This iteration could involve shifting or modifying a substantial portion of the interval list which will take linear time. Therefore, the addRange, removeRange, and queryRange methods have O(n) since a linear number of intervals may need to be shifted in the intervals list during merges or splits.

Space Complexity

O(N) – The primary space usage comes from storing the sorted intervals. In the worst-case scenario, where ranges are added and removed in a way that causes maximal fragmentation, we might end up with a number of non-overlapping intervals proportional to the number of add operations. Therefore, the space required to store these intervals can grow linearly with the number of range operations, N, where N represents the total number of add and remove operations. Thus, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Empty range module (no intervals added yet)	Ensure query and remove functions handle empty module gracefully, returning appropriate boolean values or doing nothing without errors.
Adding a range that completely overlaps an existing range	The new range should completely replace the existing range, effectively merging them.
Adding a range that partially overlaps an existing range (left or right)	The existing range should be expanded to include the new range, creating a single merged range.
Adding a range that is adjacent to an existing range	The adjacent ranges should be merged into a single range.
Removing a range that completely contains existing ranges	Split the containing range into at most two ranges, one before the removed range and one after, if necessary.
Removing a range that partially overlaps existing ranges (left or right)	The partially overlapped ranges should be truncated to exclude the removed range.
Querying for a range that falls exactly on the boundary of existing ranges	The query function should correctly determine if the range is contained within or intersects with existing ranges considering closed interval boundaries.
Integer overflow when calculating the sum or difference of range endpoints	Use appropriate data types (e.g., long) to prevent potential integer overflows when performing arithmetic operations on range endpoints.