Insert Interval LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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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

Insert Interval Solutions

Time: O(n)
Space: O(n)

C++

class Solution {
 public:
  vector<vector<int>> insert(vector<vector<int>>& intervals,
                             vector<int>& newInterval) {
    const int n = intervals.size();
    vector<vector<int>> ans;
    int i = 0;

    while (i < n && intervals[i][1] < newInterval[0])
      ans.push_back(intervals[i++]);

    // merge overlapping intervals
    while (i < n && intervals[i][0] <= newInterval[1]) {
      newInterval[0] = min(newInterval[0], intervals[i][0]);
      newInterval[1] = max(newInterval[1], intervals[i][1]);
      ++i;
    }

    ans.push_back(newInterval);

    while (i < n)
      ans.push_back(intervals[i++]);

    return ans;
  }
};

Java

 class Solution {
  public int[][] insert(int[][] intervals, int[] newInterval) {
    final int n = intervals.length;
    List<int[]> ans = new ArrayList<>();
    int i = 0;

    while (i < n && intervals[i][1] < newInterval[0])
      ans.add(intervals[i++]);

    while (i < n && intervals[i][0] <= newInterval[1]) {
      newInterval[0] = Math.min(newInterval[0], intervals[i][0]);
      newInterval[1] = Math.max(newInterval[1], intervals[i][1]);
      ++i;
    }

    ans.add(newInterval);

    while (i < n)
      ans.add(intervals[i++]);

    return ans.toArray(new int[ans.size()][]);
  }
}

Python

class Solution:
  def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
    n = len(intervals)
    ans = []
    i = 0

    while i < n and intervals[i][1] < newInterval[0]:
      ans.append(intervals[i])
      i += 1

    while i < n and intervals[i][0] <= newInterval[1]:
      newInterval[0] = min(newInterval[0], intervals[i][0])
      newInterval[1] = max(newInterval[1], intervals[i][1])
      i += 1

    ans.append(newInterval)

    while i < n:
      ans.append(intervals[i])
      i += 1

    return ans

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