Leetcode Study Day 25

Insert Interval

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

In this solution, we iterate through the intervals. If the current interval’s end is smaller than the new interval’s start, we push the current interval to the answer vector. If the current interval’s start is larger than the new interval’s end, we push the new interval to the answer vector and update the new interval to the current interval. If the new interval overlaps with the current interval, we update the new interval’s start to the minimum of the two intervals’ start and update the new interval’s end to the maximum of the two intervals’ end. After the iteration, we push the new interval to the answer vector.

class Solution {
public:
    vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
        int n = intervals.size();
        vector<vector<int>> ans;
        for (int i = 0; i < n; i ++){
            if (intervals[i][1] < newInterval[0]){
                ans.push_back(intervals[i]);
            }
            else if (intervals[i][0] > newInterval[1]){
                ans.push_back(newInterval);
                newInterval = intervals[i];
            }
            else if (newInterval[1] >= intervals[i][0] || newInterval[0] <= intervals[i][1]){
                newInterval[0] = min (newInterval[0], intervals[i][0]);
                newInterval[1] = max (newInterval[1], intervals[i][1]);
            }
        }
        ans.push_back(newInterval);
        return ans;
    }
};