Algorithm
Problem Name: 435. Non-overlapping Intervals
Problem Link: https://leetcode.com/problems/non-overlapping-intervals/
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
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
int eraseOverlapIntervals(vector<Interval>& intervals) {
sort(intervals.begin(), intervals.end(), [](Interval& a, Interval& b){ return a.start < b.start; });
int count = 0;
for(int i = 1, j = 0; i < intervals.size(); i++){
int pre = i;
if(intervals[i].start < intervals[j].end>{
count++;
if(intervals[i].end > intervals[j].end) pre = j;
}
j = pre;
}
return count;
}
};
Input
intervals = [[1,2],[2,3],[3,4],[1,3]]
Output
1
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int eraseOverlapIntervals(int[][] intervals) {
if (intervals.length == 0) {
return 0;
}
Arrays.sort(intervals, Comparator.comparingInt(o -> o[1]));
int notRemoved = 1;
int end = intervals[0][1];
for (int i = 1; i < intervals.length; i++) {
if (intervals[i][0] >= end) {
notRemoved++;
end = intervals[i][1];
}
}
return intervals.length - notRemoved;
}
}
Input
intervals = [[1,2],[2,3],[3,4],[1,3]]
Output
1
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const eraseOverlapIntervals = function(intervals) {
if(intervals == null || intervals.length === 0) return 0
intervals.sort((a, b) => a[1] - b[1])
let res = 1, end = intervals[0][1]
const len = intervals.length
for(let i = 1; i < len; i++> {
if(intervals[i][0] >= end) {
end = intervals[i][1]
res++
}
}
return len - res
};
Input
intervals = [[1,2],[1,2],[1,2]]
Output
2
#4 Code Example with Python Programming
Code -
Python Programming
class Solution:
def eraseOverlapIntervals(self, intervals):
intervals.sort(key = lambda x: x.end); res, curr = 0, -float("inf")
for i in intervals:
if curr > i.start: res += 1
else: curr = i.end
return res
Input
intervals = [[1,2],[1,2],[1,2]]
Output
2