2276 - Count Integers in Intervals (Hard)
Problem Link
https://leetcode.com/problems/count-integers-in-intervals
Problem Statement
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 <= 10^9
- At most
105
calls in total will be made toadd
andcount
. - At least one call will be made to
count
.
Approach 1: Sweep Line & Merge
class CountIntervals {
public:
CountIntervals() {
modified = 0;
res = 0;
}
void add(int left, int right) {
// in
m[left] += 1;
// out
m[right + 1] -= 1;
// mark as modified to check if we need to recalculate the count
modified = 1;
}
int count() {
if (modified) {
res = 0;
map<int, int> m2;
int l = 0, n = 0;
for (auto& x : m) {
// start of interval
if (n == 0) l = x.first;
n += x.second;
// end of interval
if (n == 0) {
// update the new map
m2[l] += 1;
m2[x.first] -= 1;
// calculate the range
res += x.first - l;
}
}
// replace the map
m = m2;
}
modified = 0;
return res;
}
private:
map<int, int> m;
int modified, res;
};
/**
* Your CountIntervals object will be instantiated and called as such:
* CountIntervals* obj = new CountIntervals();
* obj->add(left,right);
* int param_2 = obj->count();
*/