The greedy approach involves sorting the intervals by their end times. Then, we iterate through the sorted list and count the number of overlapping intervals to remove. The idea is to always pick the interval with the earliest end time, which leaves more room for the remaining intervals.
Steps:
Time Complexity: O(n log n) due to sorting.
Space Complexity: O(1), excluding the input space for the intervals.
1function eraseOverlapIntervals(intervals) {
2 intervals.sort((a, b) => a[1] - b[1]);
3 let end = intervals[0][1];
4 let count = 0;
5 for (let i = 1; i < intervals.length; i++) {
6 if (intervals[i][0] < end) {
7 count++;
8 } else {
9 end = intervals[i][1];
10 }
11 }
12 return count;
13}The JavaScript solution uses the default sort function with a custom comparator for sorting based on the end times of intervals. As it iterates, it counts overlapping intervals to remove by adjusting the 'end' variable low.
This approach utilizes dynamic programming to solve the problem. It is typically less efficient than the greedy method but serves as an illustrative illustration of tackling overlap problems using dp.
Time Complexity: O(n^2) due to nested loops for dp updates.
Space Complexity: O(n) required for the dp array.
1import java.util.Arrays;
2
3public class Solution {
4 public int eraseOverlapIntervals(int[][] intervals) {
5 if (intervals.length == 0) return 0;
6 Arrays.sort(intervals, (a, b) -> Integer.compare(a[0], b[0]));
7 int n = intervals.length;
8 int[] dp = new int[n];
9 Arrays.fill(dp, 1);
10 int maxCount = 1;
11 for (int i = 1; i < n; i++) {
12 for (int j = 0; j < i; j++) {
13 if (intervals[j][1] <= intervals[i][0]) {
14 dp[i] = Math.max(dp[i], dp[j] + 1);
15 }
16 }
17 maxCount = Math.max(maxCount, dp[i]);
18 }
19 return n - maxCount;
20 }
21}Java's dynamic programming approach similarly initializes the dp table and sorts intervals by start time. The nested loop checks intervals for compatibility of non-overlapping arrangement and updates maximum possible count found.