Insert Interval Leetcode Solution

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In this post, we are going to solve the Insert Interval Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

Insert Interval Leetcode Solution
Insert Interval Leetcode Solution

Problem

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

Now, let’s see the leetcode solution of Insert Interval Leetcode Solution.

Insert Interval Leetcode Solution in 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

Insert Interval Leetcode Solution in CPP

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;
  }
};

Insert Interval Leetcode Solution in 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()][]);
  }
}

Note: This problem Insert Interval is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.

NEXT: Length of Last Word Leetcode Solution

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