435. Non-overlapping Intervals

435. Non-overlapping Intervals (Medium)

Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

 

Example 1:

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

Example 2:

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

Example 3:

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

 

Note:

1. You may assume the interval's end point is always bigger than its start point.
2. Intervals like [1,2] and [2,3] have borders "touching" but they don't overlap each other.

Related Topics:
Greedy

Similar Questions:

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Solution 1. Interval Scheduling Maximization Problem

// OJ: https://leetcode.com/problems/non-overlapping-intervals/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int eraseOverlapIntervals(vector<vector<int>>& A) {
        sort(begin(A), end(A), [](auto &a, auto &b) { return a[1] < b[1]; });
        int ans = 0, e = INT_MIN;
        for (auto &v : A) {
            if (v[0] >= e) e = v[1];
            else ++ans;
        }
        return ans;
    }
};