Merge Intervals
You are given N number of intervals, where each interval contains two integers denoting the start time and the end time for the interval.

The task is to merge all the overlapping intervals and return the list of merged intervals sorted by increasing order of their start time.

Two intervals [A,B] and [C,D] are said to be overlapping with each other if there is at least one integer that is covered by both of them.

For example:

For the given 5 intervals - [1, 4], [3, 5], [6, 8], [10, 12], [8, 9]. Since intervals [1, 4] and [3, 5] overlap with each other, we will merge them into a single interval as [1, 5]. Similarly, [6, 8] and [8, 9] overlap, merge them into [6,9]. Interval [10, 12] does not overlap with any interval. Final List after merging overlapping intervals: [1, 5], [6, 9], [10, 12].

Input Format:

The first line of input contains an integer N, the number of intervals. The second line of input contains N integers, i.e. all the start times of the N intervals. The third line of input contains N integers, i.e. all the end times of the N intervals.

Output Format:

Print S lines, each contains two single space-separated integers A, and B, where S is the size of the merged array of intervals, 'A' is the start time of an interval and 'B' is the end time of the same interval.

Note

You do not need to print anything, it has already been taken care of. Just implement the given function.

Constraints:

1

Question

Merge Intervals

You are given N number of intervals, where each interval contains two integers denoting the start time and the end time for the interval.

The task is to merge all the overlapping intervals and return the list of merged intervals sorted by increasing order of their start time.

Two intervals [A,B] and [C,D] are said to be overlapping with each other if there is at least one integer that is covered by both of them.

For example:

For the given 5 intervals - [1, 4], [3, 5], [6, 8], [10, 12], [8, 9].

Since intervals [1, 4] and [3, 5] overlap with each other, we will merge them into a single interval as [1, 5].

Similarly, [6, 8] and [8, 9] overlap, merge them into [6,9].

Interval [10, 12] does not overlap with any interval.

Final List after merging overlapping intervals: [1, 5], [6, 9], [10, 12].

Input Format:

The first line of input contains an integer N, the number of intervals.

The second line of input contains N integers, i.e. all the start times of the N intervals.

The third line of input contains N integers, i.e. all the end times of the N intervals.

Output Format:

Print S lines, each contains two single space-separated integers A, and B, where S is the size of the merged array of intervals, 'A' is the start time of an interval and 'B' is the end time of the same interval.

Note

You do not need to print anything, it has already been taken care of. Just implement the given function.

Constraints:

1 <= N <= 10^5 
0 <= START, FINISH <= 10^9

Time Limit: 1sec

Accepted Answer

Practise greedy problems

Accepted Answer

Brute Force

In the brute force approach, mark each interval as non visited. Now for each non-visited interval, while there exists an overlapping interval with the current interval we will merge both intervals, update the current interval with the largest of both intervals, and mark them visited.

Note: Two intervals will be considered to be overlapping if the start time of one interval is less than or equal to the finish time of another interval, and greater than or equal to the start time of that interval.

Space Complexity: O(n)Explanation:

O(N), where N is the number of intervals.

In the worst case, we will be using an extra array of size N to mark intervals visited.

Time Complexity: O(n^2)Explanation:

O(N^2), where N is the number of intervals.

In the worst case, for each interval, we will be checking all other intervals.

Accepted Answer

Sorting

In this approach, we will first sort the intervals by non-decreasing order of their start time. Then we will add the first interval to our resultant list. Now, for each of the next intervals, we will check whether the current interval is overlapping with the last interval in our resultant list. If it is overlapping then we will update the finish time of the last interval in the list by the maximum of the finish time of both overlapping intervals. Otherwise, we will add the current interval to the list.

Space Complexity: O(1)Explanation:

O(1),

In the worst case, we will be using constant extra space.

Time Complexity: O(nlogn)Explanation:

O(N * log(N)), where N is the number of intervals.

In the worst case, we will be sorting the intervals(N * log(N)) and for each adjacent interval, we will be merging the overlapping intervals O(N).