Insert Interval – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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

C++ Insert Interval LeetCode Solution

class Solution {
public:
    vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
        vector<Interval> res;
        int index = 0;
        while(index < intervals.size() && intervals[index].end < newInterval.start){
            res.push_back(intervals[index++]);
        }
        while(index < intervals.size() && intervals[index].start <= newInterval.end){
            newInterval.start = min(newInterval.start, intervals[index].start);
            newInterval.end = max(newInterval.end, intervals[index].end);
            index++;
        }
        res.push_back(newInterval);
        while(index < intervals.size()){
            res.push_back(intervals[index++]);
        }
        return res;
    }
};

Java Insert Interval LeetCode Solution

public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
    List<Interval> result = new ArrayList<Interval>();
    for (Interval i : intervals) {
        if (newInterval == null || i.end < newInterval.start)
            result.add(i);
        else if (i.start > newInterval.end) {
            result.add(newInterval);
            result.add(i);
            newInterval = null;
        } else {
            newInterval.start = Math.min(newInterval.start, i.start);
            newInterval.end = Math.max(newInterval.end, i.end);
        }
    }
    if (newInterval != null)
        result.add(newInterval);
    return result;
}
 

Python 3 Insert Interval LeetCode Solution

def insert(self, intervals, newInterval):
    s, e = newInterval.start, newInterval.end
    left = [i for i in intervals if i.end < s]
    right = [i for i in intervals if i.start > e]
    if left + right != intervals:
        s = min(s, intervals[len(left)].start)
        e = max(e, intervals[~len(right)].end)
    return left + [Interval(s, e)] + right

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