57. Insert Interval

Description of Problem

You are given an array of non-overlapping intervals intervals where intervals[i] = [starti, endi] represent the start and the end of the ith interval and intervals is sorted in ascending order by starti. You are also given an interval newInterval = [start, end] that represents the start and end of another interval.

Insert newInterval into intervals such that intervals is still sorted in ascending order by starti and intervals still does not have any overlapping intervals (merge overlapping intervals if necessary).

Return intervals after the insertion.

Example 1:

Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]

Example 2:

Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].

Constraints:

• 0 <= intervals.length <= 10^4
• intervals[i].length == 2
• 0 <= starti <= endi <= 10^5
• intervals is sorted by starti in ascending order.
• newInterval.length == 2
• 0 <= start <= end <= 10^5

Solution

Tags: Stack

Explanation

Since intervals is sorted, we can insert new element at proper position in O(n). And call solution of 56. Merge Intervals to solve the problem.

Code (Rust)

impl Solution {
    pub fn insert(intervals: Vec<Vec<i32>>, new_interval: Vec<i32>) -> Vec<Vec<i32>> {
        let mut intervals = intervals;
        let y = new_interval[0];
        let index = intervals.partition_point(|i| i[0] < y);
        intervals.insert(index, new_interval);
        return Self::merge(intervals);
    }

    // Copy solution of [56. Merge Intervals]
    #[inline(always)]
    fn merge(intervals: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
        //let mut intervals = intervals;
        //intervals.sort_by(|a,b| a[0].cmp(&b[0]));

        let mut stack : Vec<Vec<i32>> = Vec::new();
        for int in intervals.into_iter() {
            let (c, d) = (int[0], int[1]);
            if let Some(v) = stack.last(){
                let (a, b) = (v[0], v[1]);
                if b >= c {
                    stack.pop();
                    stack.push(vec![a.min(c), b.max(d)]);
                }else{
                    stack.push(int);
                }
            }else{
                stack.push(int);
            }
        }

        return stack;
    }
}

Complexity

• n is length of intervals

Time Complexity

• \( T(n) = \Theta(n) \)

Auxiliary Space

• \( S(n) = O(1) \)