Insert Interval

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 <= 104
• intervals[i].length == 2
• 0 <= starti <= endi <= 105
• intervals is sorted by starti in ascending order.
• newInterval.length == 2
• 0 <= start <= end <= 105

Solution

/**
 * @param {number[][]} intervals
 * @param {number[]} newInterval
 * @return {number[][]}
 */
var insert = function(intervals, newInterval) {
    const lengthOld = intervals.length;
    for (let i = 0; i < intervals.length; i++) {
        if (intervals[i][0] > newInterval[0]) {
            intervals.splice(i, 0, newInterval);
            break;
        }
    }
    if (lengthOld === intervals.length) {
        intervals.push(newInterval);
    }
    return merge(intervals);
};

var merge = function(intervals) {
    const stack = [intervals[0]];

    for (let i = 1; i < intervals.length; i++) {
        const prev = stack[stack.length - 1];
        const cur = intervals[i];
        if (cur[0] <= prev[1]) { // overlap => merge the two intervals
            prev[1] = Math.max(cur[1], prev[1]);
        } else { // no overlap
            stack.push(cur);
        }
    }
    return stack;
};

Insert newInterval into interval such that it is still sorted by the start time. Next, simply use the merge method from the Merge Intervals problem (but without explicitly sorting intervals).