57. Insert Interval - Hard

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

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].

NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

 

3 cases + edge case at the end
1. if newInterval.start > curInterval.end (or newInterval is null)
   -> curInterval is valid, insert to list, and leave newInterval to compare with next interval
2. if newInterval.end < curInterval.start
   -> both valid, add them to list, and set newInterval to be null, which means has been inserted already
3. other condition, overlap, merge newInterval and curInterval to be a larger interval

 

time = O(n), space = O(1)

class Solution {
    public int[][] insert(int[][] intervals, int[] newInterval) {
        if(intervals.length == 0) {
            return new int[][] {newInterval};
        }
        List<int[]> list = new ArrayList<>();
        for(int i = 0; i < intervals.length; i++) {
            // // non-overlap, current interval valid, leave newInterval to compare with next interval
            if(newInterval == null || newInterval[0] > intervals[i][1]) {
                list.add(intervals[i]);
            } else if(newInterval[1] < intervals[i][0]) {  // non-overlap, both valid
                list.add(newInterval);
                list.add(intervals[i]);
                newInterval = null;     // newInterval inserted, reset to null
            } else {
                newInterval[0] = Math.min(newInterval[0], intervals[i][0]);
                newInterval[1] = Math.max(newInterval[1], intervals[i][1]);
            }
        }
        if(newInterval != null) {   // if newInterval haven't been inserted after traversal, insert at the end
            list.add(newInterval);
        }
        
        int[][] res = new int[list.size()][2];
        for(int i = 0; i < list.size(); i++) {
            res[i] = list.get(i);
        }
        return res;
    }
}