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435. Non-overlapping Intervals
Description
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 <= 105intervals[i].length == 2-5 * 104 <= starti < endi <= 5 * 104
Solutions
Greedy.
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class Solution { public int eraseOverlapIntervals(int[][] intervals) { Arrays.sort(intervals, Comparator.comparingInt(a -> a[1])); int t = intervals[0][1], ans = 0; for (int i = 1; i < intervals.length; ++i) { if (intervals[i][0] >= t) { t = intervals[i][1]; } else { ++ans; } } return ans; } } -
class Solution { public: int eraseOverlapIntervals(vector<vector<int>>& intervals) { sort(intervals.begin(), intervals.end(), [](const auto& a, const auto& b) { return a[1] < b[1]; }); int ans = 0, t = intervals[0][1]; for (int i = 1; i < intervals.size(); ++i) { if (t <= intervals[i][0]) t = intervals[i][1]; else ++ans; } return ans; } }; -
class Solution: def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int: intervals.sort(key=lambda x: x[1]) ans, t = 0, intervals[0][1] for s, e in intervals[1:]: if s >= t: t = e else: ans += 1 return ans -
func eraseOverlapIntervals(intervals [][]int) int { sort.Slice(intervals, func(i, j int) bool { return intervals[i][1] < intervals[j][1] }) t, ans := intervals[0][1], 0 for i := 1; i < len(intervals); i++ { if intervals[i][0] >= t { t = intervals[i][1] } else { ans++ } } return ans } -
function eraseOverlapIntervals(intervals: number[][]): number { intervals.sort((a, b) => a[1] - b[1]); let end = intervals[0][1], ans = 0; for (let i = 1; i < intervals.length; ++i) { let cur = intervals[i]; if (end > cur[0]) { ans++; } else { end = cur[1]; } } return ans; }