Question Definition
Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.
Note:
- You may assume the interval’s end point is always bigger than its start point.
- Intervals like
[1,2]and[2,3]have borders “touching” but they don’t overlap each other.
Example 1:
Input: [ [1,2], [2,3], [3,4], [1,3] ]
Output: 1
Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.
Example 2:
Input: [ [1,2], [1,2], [1,2] ]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.
Example 3:
Input: [ [1,2], [2,3] ]
Output: 0
Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
Java Solution
public int eraseOverlapIntervals(Interval[] intervals) {
if(intervals.length < 2) return 0;
Arrays.sort(intervals, ( x , y) -> x.start == y.start ? x.end - y.end : x.start -y.start);
int count = 0;
int end = intervals[0].end;
for(int i = 1; i < intervals.length; i++){
if(intervals[i].start >= end){
end = intervals[i].end;
}else{
count++;
end = Math.min(end, intervals[i].end);
}
}
return count;
}
Comments