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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Insert Interval in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemInsert Interval– LeetCode Problem

Insert Interval– LeetCode Problem

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
Insert Interval– LeetCode Solutions
Insert Interval in C++:
class Solution {
 public:
  vector> insert(vector>& intervals,
                             vector& newInterval) {
    const int n = intervals.size();

    vector> 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 in Java:
class Solution {
  public int[][] insert(int[][] intervals, int[] newInterval) {
    final int n = intervals.length;

    List 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()][]);
  }
}
Insert Interval 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

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