LeetCode 435: Non-overlapping Intervals Solution
Master LeetCode problem 435 (Non-overlapping Intervals), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
435. Non-overlapping Intervals
Problem Explanation
Explanation
To solve this problem, we can use a greedy approach. We first sort the intervals based on their end points. Then, we iterate through the sorted intervals and maintain an end point variable representing the end of the last non-overlapping interval. If the current interval overlaps with the previous one, we need to remove one of them. We always choose to remove the interval with the larger end point since this would maximize the chances of fitting more intervals.
By following this approach, we can find the minimum number of intervals that need to be removed to make the rest non-overlapping.
- Time complexity: O(nlogn) where n is the number of intervals
- Space complexity: O(1)
Solution Code
import java.util.Arrays;
class Solution {
public int eraseOverlapIntervals(int[][] intervals) {
if(intervals.length == 0) return 0;
Arrays.sort(intervals, (a, b) -> Integer.compare(a[1], b[1]));
int count = 1;
int end = intervals[0][1];
for(int i = 1; i < intervals.length; i++) {
if(intervals[i][0] >= end) {
count++;
end = intervals[i][1];
}
}
return intervals.length - count;
}
}Try It Yourself
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