LeetCode 56. Merge Intervals (Algorithm Explained)

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there's not many java creators out there ..you are doing a great job dude ..keep going

adarshmishra

It makes sense once you realize updating the last index of current_interval (line 20), also updates it in the ArrayList. Thanks so much for this!

depshallburn

9:16 - "Literally" took me 1 hour to understand from a code from Leetcode.

shubhamchourasia

6:32 I am a little confused because you just initialized an int as an int* and there was no error, i mean intervals was a 2d array... One of which array was containing only pointers and you initialized that pointer to an int ??

aligohar

You are amazing brother, your coding style and explanation is pretty simple.

pradip

I’m stuck on something. How does current_interval[1] = Math.max(current_end, next_end) in the loop cause a change in the arrayList? If it is because we pass by reference then why was there not change in current_interval in the array list when we did current_interval = interval?

TheIntraFerox

Hi Nick.. Can you make a video on "How the Comparator does the sorting stuffs internally" ??

akashrAnD

Thank you so much Nick.

God Bless Everyone😇

sakshiaggarwal

I think the best explanation u have ever given in your videos
may be i had understood in first attempt

surajgrandhi

In which step are we updating the (second no of interval). after encountering an overlap, we do find largest of two curr_end and next_end, but before only we added it to output_arr. so we do need to update [1, 3] to [1, 6]. Where are we doing that.

priyamshah

I got a method which is faster but uses a little more memory. What I did was initialize two variables called which are set as intervals[0][0] and intervals[0][1].

Then I make a third variable in a for loop which is set to intervals[i][0] which compares its value to the upper interval, if its lower, the upper interval is modified if interval[i][1] is higher than the current upper interval. If intervals[i][0] is higher than the lower and upper bound are pushed to a 2nd array and new bounds, intervals[i][0] (lower) and intervals[i][1] (higher) are created.

Keep doing this until the for loop ends after which we push the final values of "c_val" and "bound" as an array to our 2nd vector and return that vector. You'll need to sort intervals first before assigning the variables.

johnnywilson

1. if(intervals.length<=1){
2 return intervals;
3 }
4 Arrays.sort(intervals, (arr1, arr2)->arr1[0]-arr2[0]);
5 List<int[]> l =new ArrayList<>();
6 int[] current;
7 current = intervals[0];
8 l.add(current);
9 for(int[] interval:intervals){

10 int cbegin = current[0];
11 int cend = current[1];
12 int nbegin = interval[0];
13 int nend = interval[1];
14 if(cend>=nbegin){
15 current[1] = Math.max(cend, nend);
16 }
17 else{
18 current = interval;

19 l.add(current);
20 }
21 }
return l.toArray(new int[l.size()][]);

Could u pls xplain line no 18 current is pass by reference, changes is done automatically to list, but at line 18 current=interval, so previous current in list is *LOST*, because list stores reference of current

iamnoob