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435.non-overlapping-intervals.cpp
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57 lines (52 loc) · 1.49 KB
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// Tag: Greedy, Array, Dynamic Programming, Sorting, Line Sweep
// Time: O(NlogN)
// Space: O(1)
// Ref: -
// Note: -
// 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 <= 105
// intervals[i].length == 2
// -5 * 104 <= starti < endi <= 5 * 104
//
//
class Solution {
public:
int eraseOverlapIntervals(vector<vector<int>>& intervals) {
int n = intervals.size();
sort(intervals.begin(), intervals.end(), [](vector<int> &a, vector<int> &b) {
return a[1] < b[1];
});
int pre = intervals[0][1];
int res = 0;
for (int i = 1; i < n; i++) {
if (intervals[i][0] >= pre){
pre = intervals[i][1];
} else {
res++;
}
}
return res;
}
};