Count Integers in Intervals solution leetcode – 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
.

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Example 1:
Count Integers in Intervals solution leetcode
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:
Count Integers in Intervals solution leetcode
1 <= left <= right <= 109
 At most
105
calls in total will be made toadd
andcount
.  At least one call will be made to
count
.