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Link for the Problem – Insert 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 bystarti
in ascending order.newInterval.length == 2
0 <= start <= end <= 105
Insert Interval– LeetCode Solutions
Insert Interval in 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; } };
Insert Interval 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()][]); } }
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