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.
1#include <algorithm>
2#include <vector>
3
4using namespace std;
5
6int eraseOverlapIntervals(vector<vector<int>>& intervals) {
7 if (intervals.empty()) return 0;
8 sort(intervals.begin(), intervals.end(), [](const vector<int>& a, const vector<int>& b) {
9 return a[1] < b[1];
10 });
11 int count = 0;
12 int end = intervals[0][1];
13 for (int i = 1; i < intervals.size(); ++i) {
14 if (intervals[i][0] < end) {
15 ++count;
16 } else {
17 end = intervals[i][1];
18 }
19 }
20 return count;
21}This C++ implementation uses the sort function with a custom comparator to sort the intervals by their end times. It tracks the end of the last non-overlapping interval and increments the overlap count if a new interval overlaps.
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.
1def eraseOverlapIntervals(intervals):
2 intervals.sort(key=lambda x: x[0])
3 n = len(intervals)
4 dp = [1] * n
5 for i in range(n):
6 for j in range(i):
7 if intervals[j][1] <= intervals[i][0]:
8 dp[i] = max(dp[i], dp[j] + 1)
9 max_non_overlapping = max(dp)
10 return n - max_non_overlappingThe Python calculation sorts intervals and applies nested loops to fill a dp array with the maximum number of non-overlapping subsequences possible within the designated comparisons among start and end times.