README.md

435. 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 <= 105
• intervals[i].length == 2
• -5 * 104 <= starti < endi <= 5 * 104

Solutions (Rust)

1. Solution

impl Solution {
    pub fn erase_overlap_intervals(intervals: Vec<Vec<i32>>) -> i32 {
        let mut intervals = intervals;
        let mut stack = vec![];

        intervals.sort_unstable();

        for i in 0..intervals.len() {
            let (start0, end0) = (intervals[i][0], intervals[i][1]);
            let &(start1, end1) = stack.last().unwrap_or(&(0, start0));

            if start0 >= end1 {
                stack.push((start0, end0));
            } else if end0 <= end1 {
                stack.pop();
                stack.push((start0, end0));
            }
        }

        (intervals.len() - stack.len()) as i32
    }
}