Count Integers in Intervals

Hard

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Database Problems

Design a data structure to manage a set of intervals and efficiently count the number of integers covered by these intervals. Implement the CountIntervals class with the following methods:

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

Note: An interval [left, right] includes all integers x where left <= x <= right.

Example:

CountIntervals countIntervals = new CountIntervals();
countIntervals.add(2, 3);  // Adds interval [2, 3]
countIntervals.add(7, 10); // Adds interval [7, 10]
countIntervals.count();    // Returns 6 (integers 2, 3, 7, 8, 9, 10)
countIntervals.add(5, 8);  // Adds interval [5, 8]
countIntervals.count();    // Returns 8 (integers 2, 3, 5, 6, 7, 8, 9, 10)

Explain the time and space complexity of your solution. Consider edge cases such as overlapping intervals and large input ranges (1 <= left <= right <= 10^9). At most 10^5 calls to add and count will be made.

Count Integers in Intervals

Hard

Google

4 views

Topics:

Database Problems

Design a data structure to manage a set of intervals and efficiently count the number of integers covered by these intervals. Implement the CountIntervals class with the following methods:

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

Note: An interval [left, right] includes all integers x where left <= x <= right.

Example:

CountIntervals countIntervals = new CountIntervals();
countIntervals.add(2, 3);  // Adds interval [2, 3]
countIntervals.add(7, 10); // Adds interval [7, 10]
countIntervals.count();    // Returns 6 (integers 2, 3, 7, 8, 9, 10)
countIntervals.add(5, 8);  // Adds interval [5, 8]
countIntervals.count();    // Returns 8 (integers 2, 3, 5, 6, 7, 8, 9, 10)

Explain the time and space complexity of your solution. Consider edge cases such as overlapping intervals and large input ranges (1 <= left <= right <= 10^9). At most 10^5 calls to add and count will be made.

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:

What are the possible ranges for the integer values within the intervals?
Can the intervals overlap, and if so, how should overlapping intervals be handled when counting?
If an interval is completely contained within another, should I only count the integers in the outer interval, or count those integers covered by both?
What is the expected behavior if I try to add an interval that is invalid (e.g., start > end)? Should I throw an exception or simply ignore it?
Are there any limitations on the total number of intervals that could be added over the lifetime of the object?

Brute Force Solution

Approach

The brute force method to count integers in intervals means we meticulously check every number within the given ranges. We look at each interval individually and see which whole numbers fall inside. Then, we simply add up the count from each interval.

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

Consider each interval given to us, one at a time.
For each interval, look at every whole number between its starting and ending points.
If a number hasn't already been counted from a previous interval, count it.
Keep a running total of all the unique numbers you've counted.
After checking all the intervals, the running total is your final answer: the total number of distinct integers covered by the intervals.

Code Implementation

def count_integers_in_intervals_brute_force(intervals):
    count = 0
    seen_integers = set()

    for interval in intervals:
        start_of_interval = interval[0]
        end_of_interval = interval[1]

        # Iterate through the integers in the interval.
        for integer in range(start_of_interval, end_of_interval + 1):

            # Ensures we only count each integer once.
            if integer not in seen_integers:
                count += 1
                seen_integers.add(integer)

    return count

Big(O) Analysis

Time Complexity

O(n*m) – Let n be the number of intervals and m be the maximum range size of an interval (difference between its end and start points). For each of the n intervals, we iterate through a range of numbers of size at most m. In the worst case, for each interval, we may iterate close to m times, and each such iteration does O(1) operations. Thus the overall time complexity is O(n*m), where n is the number of intervals and m is the maximum length of any interval.

Space Complexity

O(N) – The brute force method involves keeping track of the distinct integers that have been counted. This implicitly requires storing these integers, potentially in a set or a similar data structure. In the worst-case scenario where all integers within the intervals are distinct, we may need to store a number of integers proportional to the range spanned by the intervals. If we let N represent the total number of integers covered by the intervals, then the auxiliary space will be O(N).

Optimal Solution

Approach

The core idea is to maintain a collection of non-overlapping intervals, merging them when new intervals overlap existing ones. By keeping track of the merged intervals, we can easily count the number of integers covered and efficiently update the collection.

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

Maintain a collection that stores all the intervals that have been added so far.
When asked to add a new interval, check if it overlaps with any of the existing intervals.
If it overlaps, merge the new interval with the overlapping ones to create a larger, combined interval.
Repeat merging until the new interval no longer overlaps with any other interval in the collection. Insert the merged/new interval into the collection.
When asked to count the integers, add up the lengths of all the intervals in the collection.

Code Implementation

class CountIntervals:

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

    def add(self, left_bound: int, right_bound: int) -> None:
        new_interval = [left_bound, right_bound]
        index = 0
        while index < len(self.intervals):
            # Check for overlaps and merge intervals if necessary
            if new_interval[1] < self.intervals[index][0]:
                break
            elif new_interval[0] > self.intervals[index][1]:
                index += 1
                continue
            else:
                # Intervals overlap, merge them
                new_interval[0] = min(new_interval[0], self.intervals[index][0])
                new_interval[1] = max(new_interval[1], self.intervals[index][1])
                self.intervals.pop(index)
                continue

        self.intervals.insert(index, new_interval)

    def count(self) -> int:
        total_count = 0
        # Calculate the total length of all intervals
        for interval in self.intervals:
            total_count += (interval[1] - interval[0] + 1)

        return total_count

# Your CountIntervals object will be instantiated and called as such:
# count_intervals = CountIntervals()
# count_intervals.add(left, right)
# result = count_intervals.count()

Big(O) Analysis

Time Complexity

O(n) – Adding a new interval involves iterating through the existing intervals (let's say there are up to 'n' of them). The while loop continues merging as long as there are overlapping intervals, potentially merging with all existing intervals. The count operation iterates through all the intervals, which can be up to 'n' in the worst case. Therefore, the addNum operation has a worst-case time complexity of O(n), and count() is O(n). Over multiple calls to addNum, the average may degrade towards O(n) for each call depending on input patterns, but each addNum operation and the count operation will not exceed O(n).

Space Complexity

O(N) – The algorithm maintains a collection of non-overlapping intervals. In the worst-case scenario, where each added interval does not overlap with any existing intervals, the collection could store up to N intervals, where N is the number of added intervals. Each interval requires storing two integers (start and end). Therefore, the auxiliary space used is proportional to the number of intervals stored. This leads to a space complexity of O(N).

Edge Cases

Case	How to Handle
Null or invalid input intervals (e.g., null array, intervals with start > end)	Throw an IllegalArgumentException or return an appropriate error code to indicate invalid input.
Empty intervals array	Initialize count to 0 as there are no intervals present.
Intervals with identical start and end points (single-point intervals)	These intervals should be treated as valid and contribute 1 to the count.
Overlapping intervals	Merge overlapping intervals to avoid double-counting integers.
Intervals with negative numbers or zero	The data structure and algorithm should correctly handle negative or zero values within the intervals.
Large intervals spanning a significant range of integers, potentially causing memory issues if storing all integers	Use interval merging to efficiently represent the covered range without explicitly storing each integer, and consider segment trees.
Maximum number of intervals, exceeding memory limitations	Consider using an external sorting algorithm or a disk-based data structure to handle a large number of intervals if memory is a constraint.
Integer overflow when calculating the count of integers within intervals	Use long data type or appropriate overflow checks to prevent integer overflow during count calculations.