710143. Maximum number of overlapping Intervals
Maximum number of overlapping Intervals
Slug: intersecting-intervals
Difficulty: Hard
Id: 710143
Topic Tags: None
Company Tags: None
Summary
Given a set of intervals, find the maximum number of overlapping intervals. This problem involves understanding and manipulating intervals in a way that maximizes overlap.
Detailed Explanation
To solve this problem, we will use a greedy approach to keep track of the current maximum count of overlapping intervals. We initialize our count to 1, as there is at least one interval. Then, for each subsequent interval, we check if it overlaps with the previous one (i.e., its start time is less than or equal to the end time of the previous interval). If it does overlap, we increment our count; otherwise, we reset the count to 1.
Here's a step-by-step breakdown:
- Initialize
maxCountto 1. - Iterate through each interval in the set.
- For each interval, check if it overlaps with the previous one (i.e., its start time is less than or equal to the end time of the previous interval).
- If it does overlap, increment
maxCount. - If not, reset
maxCountto 1. - Return
maxCount.
Time complexity: O(n), where n is the number of intervals. Space complexity: O(1), as we only use a constant amount of space to keep track of our count.
Optimized Solutions
Java
public int maxOverlap(int[][] intervals) {
if (intervals.length == 0) return 0;
int maxCount = 1;
Interval prevInterval = new Interval(intervals[0][0], intervals[0][1]);
for (int i = 1; i < intervals.length; i++) {
Interval currInterval = new Interval(intervals[i][0], intervals[i][1]);
if (currInterval.start <= prevInterval.end) {
maxCount++;
} else {
maxCount = 1;
}
prevInterval = currInterval;
}
return maxCount;
}
class Interval {
int start;
int end;
public Interval(int start, int end) {
this.start = start;
this.end = end;
}
}