## Merge overlapping intervals

Given N intervals S = {E1,E2,…..En} with each Ei has start time si and end time ei. Some of these intervals are overlapping. The problem statement is to merge these overlapping intervals.

Ei and Ej overlap when start time of Ej i.e sj is less than end time of Ei i.e ei.

For example:

```Input:
[(1,3),(2,4),(5,8), (6,9)]
Output:
[(1, 4),(5,9)]
Explantion:
Interval (1,3) and (2,4) and interval (5,8) and (6,9) overlap.
```

## Merge overlapping intervals solution

As we always do, first try to come up with brute force solution, given enough time and space and money, how would you solve this?
The natural course is to take ith interval and compare start time of all jth intervals with end time of ith, if the start time of jth interval is less than the end time of ith event, then you can merge intervals. What should be end time for merged interval then?  It should be a maximum of end times of two merged intervals.

What will be the time complexity of this approach? We are not using any additional space, however, the worst-case time complexity is O(n2). Can we do better?

What are two times we are comparing in brute force solution? It’s the start time of one interval with the end time of another. If we arrange input in a specific order, can we reduce processing some entries?

If we sort all intervals based on their start time, si < si+1< si+2. Also, interval is always forward looking, ei > si, ei+1 > si+1 and so on.

If si is greater ei-1, then si+1 will be greater than ei-1, so no need to compare si+1 with ei-1, that is no need to go beyond immediate previous interval for any interval Ei.

If si is less than ei-1, update ei-1 with maximum of ei-1 and ei and move to Ei+1.

Notice that we need last interval Ei-1 to decide if to merge new interval into previous one or keep it as standalone. A stack is the best data structure to use. The algorithm will look like:

1. Consider interval Ei.
2. If stack is empty, push Ei to stack.
3. If stack is not empty, then pop interval at top of stack call it Ei-1.
4. Compare si, start time of Ei with ei-1, end time of Ei-1.
5. If si less than ei-1, update ei-1 as max(ei-1, ei), as in maximum of end times of two intervals and push back Ei-1on to stack.
6. Else push Ei on to stack.
7. Continue till all events are considered.
8. At the end of processing, stack will contain all merged interval.

Let’s take an example and see how this algorithm works. We have following intervals and we have to merge overlapping intervals.

Find the maximum of end times of two intervals and update the previous interval with that end time and push it back on to stack.

At this point, when there is no more interval remaining, the stack contains all merged overlapping intervals.

### Merge intervals Java implementation

```package com.company;

import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.Stack;

/**
* Created by sangar on 8.4.18.
*/
public class OverlappingIntervals {
public  static ArrayList<Interval>
mergeOverlappingIntervals(ArrayList<Interval> intervals){

ArrayList<Interval> mergedIntervals = new ArrayList<>();
Stack<Interval> s = new Stack();

//Sort the ArrayList of interval based on start time.
Collections.sort(intervals, Comparator.comparing(p -> p.getStartTime()));
for(Interval currentInterval : intervals){
if(s.empty())s.push(currentInterval);
else {
Interval previousInterval = s.pop();
if(previousInterval.getEndTime() >
currentInterval.getStartTime()){
/*
If current interval's start time is less than end time of
previous interval, find max of end times of two intervals
and push new interval on to stack.
*/
int endTime = Integer.max(previousInterval.getEndTime(),
currentInterval.getEndTime());
/* Notice that we have created new interval and
did not update the old one
This concept is called as immutability of class
*/
s.push(new Interval(previousInterval.getStartTime(),
endTime));
}
else{
s.push(previousInterval);
s.push(currentInterval);
}
}
}
while(!s.empty()){
mergedIntervals.add(s.pop());
}

return mergedIntervals;
}

public static void main(String[] args) {
ArrayList<Interval> intervals = new ArrayList<>();

intervals.add(new Interval(1,3));
intervals.add(new Interval(2,4));
intervals.add(new Interval(5,8));
intervals.add(new Interval(6,9));
ArrayList<Interval> mergedIntervals
= mergeOverlappingIntervals(intervals);
for (Interval interval : mergedIntervals){
System.out.print("(" + interval.getStartTime() +","
+ interval.getEndTime() + ")");
}
}
}
```
Complexity of algorithm to merge overlapping intervals will be O(nlogn) due to sorting with O(n) extra space for stack and then copying into the list to return also takes O(n) space.

There is another way to implement the same function without using the stack, here we use the fact that ArrayList in Java is implemented using the array as the base and getting an element at a particular index should be O(1) operation. The code looks more or less the same, however, there is no traversal of the stack at the end to create the list to return.

find overlapping intervals

```public List<Interval> mergeOptimized(List<Interval> intervals) {

if(intervals.size() == 0) return intervals;

Collections.sort(intervals,
(Interval a, Interval b) -> a.getStartTime() - b.getStartTime());

List<Interval> mergedIntervals = new ArrayList<Interval>();
for(Interval interval : intervals){

/*If the merged list is empty add the interval to
it or check if the last interval in merged list overlaps

/*Remember the get function on ArrayList is O(1) operation
because Arraylists in Java are backed by arrays */
if(mergedIntervals.isEmpty()
|| mergedIntervals.get(
mergedIntervals.size()-1).getEndTime() <
interval.getStartTime() ){
mergedIntervals.add(interval);
}
else {
int lastEndTime = Math.max(
mergedIntervals.get(mergedIntervals.size()-1)
.getEndTime(),
interval.getEndTime()
);
mergedIntervals.get(mergedIntervals.size()-1)
.setEndTime(lastEndTime);
}
}

return mergedIntervals;
}
```

You can use the above snippet of code to submit for this leetcode problem and it should be accepted.

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