Is Count Integers in Intervals easy or hard?

Count Integers in Intervals is classified as a Hard problem. The challenge comes from designing a data structure that supports dynamic interval insertion while avoiding double counting and keeping operations efficient.

Count Integers in Intervals Python/Java solution

Python solutions often use a sorted list with binary search or external libraries like SortedContainers to maintain ordered intervals. Java implementations commonly rely on TreeMap to efficiently find and merge overlapping intervals while updating the total count.

How to solve Count Integers in Intervals efficiently?

Maintain a set of non-overlapping intervals sorted by start position. When adding a new interval, locate overlapping intervals, merge them into a single range, and update the stored total length. Ordered structures like TreeMap or balanced BSTs keep these operations around O(log n) time.

What is the best approach for Count Integers in Intervals?

The most practical approach uses a balanced tree (ordered set or TreeMap) to store non-overlapping intervals. Each add operation merges overlapping intervals and updates a running total of covered integers. This keeps insertion around O(log n) and count queries O(1). It scales well when many intervals are added dynamically.

Is Count Integers in Intervals asked at Google/Amazon/Meta?

Interval data structure design problems similar to this appear in interviews at companies like Google, Amazon, and Meta. They test understanding of interval merging, ordered maps, and maintaining aggregate values under dynamic updates.

What data structure is used in Count Integers in Intervals?

The typical solution uses an ordered set or balanced binary search tree to store disjoint intervals. Alternatives include segment trees with coordinate compression, but ordered maps are simpler and more memory efficient for this problem.

What is the time complexity of Count Integers in Intervals?

Using a balanced tree structure, the add(left, right) operation typically runs in O(log n) time for searching and merging neighboring intervals, while count() is O(1) because the total covered length is maintained incrementally. Space complexity is O(n) for storing disjoint intervals.

Description Hints 3 Solution Video 11:44

2276. Count Integers in Intervals

Hard36.1% acceptance4 companies

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: