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Medium
Greedy
Array
Dynamic Programming
Sorting

435. Non-overlapping Intervals

中文文档

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.

Note that intervals which only touch at a point are non-overlapping. For example, [1, 2] and [2, 3] are 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

Solutions

Solution 1: Sorting + Greedy

We first sort the intervals in ascending order by their right boundary. We use a variable $\textit{pre}$ to record the right boundary of the previous interval and a variable $\textit{ans}$ to record the number of intervals that need to be removed. Initially, $\textit{ans} = \textit{intervals.length}$.

Then we iterate through the intervals. For each interval:

  • If the left boundary of the current interval is greater than or equal to $\textit{pre}$, it means that this interval does not need to be removed. We directly update $\textit{pre}$ to the right boundary of the current interval and decrement $\textit{ans}$ by one;
  • Otherwise, it means that this interval needs to be removed, and we do not need to update $\textit{pre}$ and $\textit{ans}$.

Finally, we return $\textit{ans}$.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$, where $n$ is the number of intervals.

Python3

class Solution:
    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
        intervals.sort(key=lambda x: x[1])
        ans = len(intervals)
        pre = -inf
        for l, r in intervals:
            if pre <= l:
                ans -= 1
                pre = r
        return ans

Java

class Solution {
    public int eraseOverlapIntervals(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> a[1] - b[1]);
        int ans = intervals.length;
        int pre = Integer.MIN_VALUE;
        for (var e : intervals) {
            int l = e[0], r = e[1];
            if (pre <= l) {
                --ans;
                pre = r;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int eraseOverlapIntervals(vector<vector<int>>& intervals) {
        ranges::sort(intervals, [](const vector<int>& a, const vector<int>& b) {
            return a[1] < b[1];
        });
        int ans = intervals.size();
        int pre = INT_MIN;
        for (const auto& e : intervals) {
            int l = e[0], r = e[1];
            if (pre <= l) {
                --ans;
                pre = r;
            }
        }
        return ans;
    }
};

Go

func eraseOverlapIntervals(intervals [][]int) int {
	sort.Slice(intervals, func(i, j int) bool {
		return intervals[i][1] < intervals[j][1]
	})
	ans := len(intervals)
	pre := math.MinInt32
	for _, e := range intervals {
		l, r := e[0], e[1]
		if pre <= l {
			ans--
			pre = r
		}
	}
	return ans
}

TypeScript

function eraseOverlapIntervals(intervals: number[][]): number {
    intervals.sort((a, b) => a[1] - b[1]);
    let [ans, pre] = [intervals.length, -Infinity];
    for (const [l, r] of intervals) {
        if (pre <= l) {
            --ans;
            pre = r;
        }
    }
    return ans;
}