57. Insert Interval

1 Description #

source: https://leetcode.com/problems/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 <= 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

2 Solution #

#include <algorithm>
class Solution {
public:
    vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
      // Space complexity: O(N), N is the size of intervals
      // Time complexity: O(N)
	std::vector<std::vector<int>> merged;
	for(auto iter = intervals.begin(); iter != intervals.end(); iter++){
	    auto interval = *iter;
	    //  the newInterval is less than the current interval, since the intervals is in ascending order, therefore there is no chance to have overlapping any more, just return
	    if(newInterval[1] < interval[0]){
		merged.push_back(newInterval);
		std::copy(iter, intervals.end(), std::back_inserter(merged));
		return merged;
	    }else if(interval[1] < newInterval[0]){
	    // There is no overlapping here since the left value of newInterval is greater than the right value of interval.
		merged.push_back(interval);
	    }else{
		// Create a new `newInterval` with maximium range.
		newInterval = {std::min(newInterval[0], interval[0]), std::max(newInterval[1], interval[1])};
	    }
	}

	merged.push_back(newInterval);

	return merged;
    }
};