README_EN.md

57. Insert Interval

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Description

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

Solutions

Python3

class Solution:
    def insert(
        self, intervals: List[List[int]], newInterval: List[int]
    ) -> List[List[int]]:
        def merge(intervals):
            intervals.sort()
            ans = []
            st, ed = intervals[0]
            for s, e in intervals[1:]:
                if ed < s:
                    ans.append([st, ed])
                    st, ed = s, e
                else:
                    ed = max(ed, e)
            ans.append([st, ed])
            return ans

        intervals.append(newInterval)
        return merge(intervals)

Java

class Solution {
    public int[][] insert(int[][] intervals, int[] newInterval) {
        List<int[]> list = new LinkedList<>();
        int i = 0;
        while ((i < intervals.length) && (intervals[i][1] < newInterval[0]))
            list.add(intervals[i++]);
        while ((i < intervals.length) && (intervals[i][0] <= newInterval[1])) {
            newInterval[0] = Math.min(intervals[i][0], newInterval[0]);
            newInterval[1] = Math.max(intervals[i][1], newInterval[1]);
            i++;
        }
        list.add(newInterval);
        while (i < intervals.length) list.add(intervals[i++]);
        return list.toArray(new int[list.size()][]);
    }
}

C++

class Solution {
public:
    vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
        intervals.push_back(newInterval);
        return merge(intervals);
    }

    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        sort(intervals.begin(), intervals.end());
        int st = intervals[0][0], ed = intervals[0][1];
        vector<vector<int>> ans;
        for (int i = 1; i < intervals.size(); ++i) {
            int s = intervals[i][0], e = intervals[i][1];
            if (ed < s) {
                ans.push_back({st, ed});
                st = s, ed = e;
            } else
                ed = max(ed, e);
        }
        ans.push_back({st, ed});
        return ans;
    }
};

Go

func insert(intervals [][]int, newInterval []int) [][]int {
	intervals = append(intervals, newInterval)
	return merge(intervals)
}

func merge(intervals [][]int) [][]int {
	sort.Slice(intervals, func(i, j int) bool {
		return intervals[i][0] < intervals[j][0]
	})
	st, ed := intervals[0][0], intervals[0][1]
	var ans [][]int
	for _, e := range intervals[1:] {
		if ed < e[0] {
			ans = append(ans, []int{st, ed})
			st, ed = e[0], e[1]
		} else if ed < e[1] {
			ed = e[1]
		}
	}
	ans = append(ans, []int{st, ed})
	return ans
}

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