435. Non-overlapping Intervals

Given an array of intervals intervals where intervals[i] = [starti, endi], return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

Example 1:

Input: intervals = [[1,2],[2,3],[3,4],[1,3]]
Output: 1
Explanation: [1,3] can be removed and the rest of the intervals are non-overlapping.

Example 2:

Input: intervals = [[1,2],[1,2],[1,2]]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of the intervals non-overlapping.

Example 3:

Input: intervals = [[1,2],[2,3]]
Output: 0
Explanation: You don't need to remove any of the intervals since they're already non-overlapping.

Constraints:

• 1 <= intervals.length <= 105
• intervals[i].length == 2
• -5 * 104 <= starti < endi <= 5 * 104

Solution:

class Solution {
    public int eraseOverlapIntervals(int[][] intervals) {
        int n = intervals.length;
        if (n <= 1){
            return 0;
        }

        Arrays.sort(intervals, (a,b) -> Integer.compare(a[1], b[1]));

        int prev = 0;
        int count = 1; // not overlap
        for (int i = 1; i < n; i++){
            if (intervals[prev][1] <= intervals[i][0]){
                prev = i;
                count++;
            }
        }


        return n - count;
    }
}

// 1

// [1, 11] p
// [2, 12]
// [11, 22]
// [1,10]

// TC: O(nlogn)
// SC: O(1)

class Solution {
    public int eraseOverlapIntervals(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> (a[1] - b[1]));
        // [1, 2] [1,3] [2,3] [3, 4]
        // [[0,2],[1,3],[2,4],[3,5],[4,6]]


        int result = intervals.length;

        PriorityQueue<Integer> pq = new PriorityQueue<>((a,b) -> (b-a));
        // max Heap

        for (int[] interval : intervals){
            if (!pq.isEmpty() && pq.peek() > interval[0]){
                // 2 > 1 
                continue;
            }
            pq.offer(interval[1]);//2 
        }

        result = result - pq.size();
        return result;


    }
}
// TC: O(nlogn)
// SC: O(n)