Insert Interval LeetCode Solution

Problem – Insert Interval LeetCode Solution

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].


  • 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

Insert Interval LeetCode Solution in Java

public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
    List<Interval> result = new LinkedList<>();
    int i = 0;
    // add all the intervals ending before newInterval starts
    while (i < intervals.size() && intervals.get(i).end < newInterval.start)
    // merge all overlapping intervals to one considering newInterval
    while (i < intervals.size() && intervals.get(i).start <= newInterval.end) {
        newInterval = new Interval( // we could mutate newInterval here also
                Math.min(newInterval.start, intervals.get(i).start),
                Math.max(newInterval.end, intervals.get(i).end));
    result.add(newInterval); // add the union of intervals we got
    // add all the rest
    while (i < intervals.size()) result.add(intervals.get(i++)); 
    return result;

Insert Interval LeetCode Solution in Python

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

Insert Interval LeetCode Solution in C++

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