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Count Integers in Intervals

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ArraysTrees

Given an empty set of intervals, implement a data structure that can:

  • Add an interval to the set of intervals.
  • Count the number of integers that are present in at least one interval.

Implement the CountIntervals class:

  • CountIntervals() Initializes the object with an empty set of intervals.
  • void add(int left, int right) Adds the interval [left, right] to the set of intervals.
  • int count() Returns the number of integers that are present in at least one interval.

Note that an interval [left, right] denotes all the integers x where left <= x <= right.

Example 1:

Input
["CountIntervals", "add", "add", "count", "add", "count"]
[[], [2, 3], [7, 10], [], [5, 8], []]
Output
[null, null, null, 6, null, 8]

Explanation
CountIntervals countIntervals = new CountIntervals(); // initialize the object with an empty set of intervals. 
countIntervals.add(2, 3);  // add [2, 3] to the set of intervals.
countIntervals.add(7, 10); // add [7, 10] to the set of intervals.
countIntervals.count();    // return 6
                           // the integers 2 and 3 are present in the interval [2, 3].
                           // the integers 7, 8, 9, and 10 are present in the interval [7, 10].
countIntervals.add(5, 8);  // add [5, 8] to the set of intervals.
countIntervals.count();    // return 8
                           // the integers 2 and 3 are present in the interval [2, 3].
                           // the integers 5 and 6 are present in the interval [5, 8].
                           // the integers 7 and 8 are present in the intervals [5, 8] and [7, 10].
                           // the integers 9 and 10 are present in the interval [7, 10].

Constraints:

  • 1 <= left <= right <= 109
  • At most 105 calls in total will be made to add and count.
  • At least one call will be made to count.

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:

  1. The `count()` method's return type is `int`. Given that interval coordinates can be up to 10^9, the total count can easily exceed the capacity of a 32-bit integer. Should I use a 64-bit integer type for my calculations and the return value?
  2. To clarify the core requirement, does the problem want me to maintain a set of disjoint intervals by merging any new interval that overlaps with existing ones?
  3. What is the expected distribution of `add` versus `count` calls? For example, will all `add` calls occur before any `count` calls, or should I expect them to be frequently interleaved? This might influence performance trade-offs in my data structure choice.
  4. If the `add` method is called with an interval that is completely contained within an already covered range, is it safe to assume that this operation should result in no change to the total count?
  5. Are there any specific memory constraints I should consider? My initial thought is to store the disjoint intervals, which in the worst case could be up to 10^5 intervals.

Brute Force Solution

Approach

The most straightforward way to solve this is to maintain a master collection of every single integer that has been covered by any of the ranges. Whenever a new range is added, we add all its integers to our collection, and to get the total count, we just check the size of this collection.

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

  1. Start with an empty collection to hold all the unique numbers we've seen so far.
  2. When a new range of numbers is given, we will look at every single whole number within it, one by one.
  3. For each of these individual numbers, we add it to our master collection.
  4. Our collection will automatically handle duplicates, so if a number is already there, it won't be added a second time.
  5. We repeat this process for every new range that is provided, continuously adding numbers to our master collection.
  6. To find the total count at any time, we simply count how many distinct numbers are currently in our master collection.

Code Implementation

class CountIntervals:
    def __init__(self):
        # A set is used as our special collection to automatically handle duplicate numbers.

        self.covered_numbers = set()

    def add(self, left_boundary: int, right_boundary: int) -> None:
        # We must iterate through every integer in the new range to ensure each one is individually added.

        for current_number in range(left_boundary, right_boundary + 1):
            self.covered_numbers.add(current_number)

    def count(self) -> int:
        # The size of the set directly gives the total count of unique integers covered by all intervals.

        return len(self.covered_numbers)

Big(O) Analysis

Time Complexity
O(n)The time complexity is determined by the `add` operation, as the `count` operation is O(1). For an input interval of length `n` (where `n = right - left + 1`), the described method iterates through each of the `n` integers. Inside this loop, adding an integer to the master collection (like a hash set) is an O(1) operation on average. Therefore, the total operations for a single `add` call are directly proportional to the length of the interval, which simplifies to O(n).
Space Complexity
O(N)The algorithm's primary space consumption comes from the 'master collection' used to store every unique integer. This collection, behaving like a hash set, holds one entry for each distinct number added from the input ranges. If N is the total count of unique integers covered by the union of all intervals added so far, the collection will grow to store these N items. Consequently, the auxiliary space required is directly proportional to the number of unique integers, resulting in a space complexity of O(N).

Optimal Solution

Approach

The core idea is to avoid counting every single number. Instead, we maintain a simplified collection of non-overlapping number ranges by merging any new range with existing ones it touches, which makes calculating the total covered length much simpler.

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

  1. The goal is to find the total number of individual spots covered by all the given ranges, but without counting the spots in any overlapping areas more than once.
  2. The clever approach is to not deal with overlaps directly. Instead, we maintain a simplified view of the number line, made up of distinct, non-touching segments.
  3. When a new range arrives, we look at our simplified view and identify all the existing segments that this new range touches or covers.
  4. We then effectively 'melt' the new range and all the segments it touches together into one continuous, larger segment. This new segment stretches from the earliest start point to the latest end point of all the combined pieces.
  5. This new, larger segment replaces the old, smaller ones it was made from, keeping our simplified view clean and easy to manage.
  6. Finding the total count becomes easy: we just add up the lengths of all the separate, non-touching segments currently in our view.
  7. This method is efficient because we only manage the start and end points of these consolidated segments, rather than potentially dealing with every individual number.

Code Implementation

class CountIntervals:
    def __init__(self):
        self.intervals = []
        self.total_covered_count = 0

    def add(self, left_boundary: int, right_boundary: int) -> None:
        indices_of_conflicting_intervals = []
        
        # Find all existing ranges that overlap or are directly next to the new range.
        for index, current_interval in enumerate(self.intervals):
            current_start, current_end = current_interval
            if current_start <= right_boundary + 1 and current_end >= left_boundary - 1:
                indices_of_conflicting_intervals.append(index)

        merged_start, merged_end = left_boundary, right_boundary
        
        # If conflicts exist, combine them all into a single new continuous range.
        if indices_of_conflicting_intervals:
            for index in indices_of_conflicting_intervals:
                start_to_merge, end_to_merge = self.intervals[index]
                merged_start = min(merged_start, start_to_merge)
                merged_end = max(merged_end, end_to_merge)

            # We must remove the old intervals and subtract their sizes from the total count.
            for index in reversed(indices_of_conflicting_intervals):
                old_start, old_end = self.intervals.pop(index)
                self.total_covered_count -= (old_end - old_start + 1)

        self.total_covered_count += (merged_end - merged_start + 1)

        # To keep the collection neat and simplified, insert the new range in sorted order.
        insertion_point = 0
        while insertion_point < len(self.intervals) and self.intervals[insertion_point][0] < merged_start:
            insertion_point += 1
        self.intervals.insert(insertion_point, [merged_start, merged_end])

    def count(self) -> int:
        return self.total_covered_count

Big(O) Analysis

Time Complexity
O(n²)The complexity is determined by the total work performed across `n` calls to add an interval. For each new interval being added, the algorithm must scan through the entire existing collection of disjoint intervals to find any that overlap. Since the number of disjoint intervals can grow with each addition, the work for `n` additions is proportional to 1 + 2 + ... + (n-1). This summation approximates to n * n / 2 operations, which simplifies to a time complexity of O(n²).
Space Complexity
O(N)The main auxiliary space is used for the 'simplified collection of non-overlapping number ranges' which stores the merged intervals. Let N represent the total number of `add` operations performed. In the worst-case scenario, each of the N added intervals is disjoint from all others, preventing any merging from occurring. Consequently, the collection must store all N intervals individually, causing the space required to grow linearly with the number of additions, resulting in O(N) space complexity.

Edge Cases

Adding a new interval that is completely contained within an existing interval.
How to Handle:
The merge logic correctly identifies that no new integers are covered, thus the set of intervals and the total count remain unchanged.
Adding a new interval that completely contains one or more existing intervals.
How to Handle:
The algorithm removes all fully contained existing intervals and merges them into the new, larger interval while correctly updating the total count.
A new interval bridges the gap and connects multiple previously disjoint intervals.
How to Handle:
An iterative merging process must combine the new interval with all overlapping and adjacent existing intervals into a single contiguous block.
Adding an interval that is perfectly adjacent to an existing one, for example [1, 5] and [6, 10].
How to Handle:
The overlap detection logic must treat adjacent intervals as mergeable to ensure they are combined correctly into a single interval.
The total count of covered integers exceeds the maximum value of a 32-bit signed integer.
How to Handle:
The variable storing the total count must be a 64-bit integer type to prevent overflow given the large coordinate values and number of intervals.
A worst-case sequence of additions where a new interval merges all previously added intervals.
How to Handle:
Using a balanced binary search tree for storage ensures efficient lookups and amortizes the cost of merging across all add operations.
Adding an interval that is an exact duplicate of an already existing disjoint interval.
How to Handle:
This case is handled correctly by the general logic for contained intervals, resulting in no change to the data structure or the total count.
Adding an interval that represents a single point where the left and right boundaries are identical.
How to Handle:
The standard length calculation `right - left + 1` correctly evaluates to one for single-point intervals, requiring no special case logic.