LeetCode The Hard Way
2200 - 2299

2276 - Count Integers in Intervals (Hard)

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 1e5 calls in total will be made to add and count.
  • 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();
 */

On this page