now i can see improvement in myself, I solved this problem with exactly same approach ,though it took me two hours😅 and came here to see more optimized solustion
@bingochipspass087 ай бұрын
Great solution! I was confused as to why we were sorting intervals based on the starting value: "We want to find values closer to the start & not farther from it since it'll give us a higher chance of finding overlap..."
@shrirambalaji2915 Жыл бұрын
Thank you so much brother...I just start watching your video 3 min into the video and I code the solution without completely watching the video. Your explanation is on point thank you again
@nikoo28 Жыл бұрын
Cheers man!!
@shaikhanuman801213 күн бұрын
Tqs for clear cut explanation on merge intervals. Tq very much
@spacelovertelugu44spacelov469 ай бұрын
Topnotch best explanation for this problem
@shivaniverma42667 ай бұрын
thank you nikhil you are doing great for creating such beautiful content for us , thats what make you unique fro other teachers your explaination keep doing keep posting , u inspire us
@abhcodeАй бұрын
wonderful explanation!! 🌟I would love to see you sharing some java essentials for coding in leetcode I was not knowing methods like asList and use of comparator . So thats the problem I phased int this video .Hope you will come up with video for this!! Eagerly waiting 👀 Lots of love❤
@prashanthshetty83376 ай бұрын
Thank you very much for the great explanation and the solution to the problem. I got confused with the naming of interval and newInterval variables in the code, which could be used other way.
@nandhakumarkr31473 ай бұрын
Your tutorial is just awesome
@Divya-qt1cf11 ай бұрын
Great explanation 🎉
@subee1288 ай бұрын
Thanks
@kalyanamvenumadhav2245 Жыл бұрын
Please Explain 4Sum Problem so that it can be helpful to all of us.
@nikoo28 Жыл бұрын
Yep..will make a video on it too.
@everyontech2716 Жыл бұрын
using stack class Solution: def merge(self, intervals: List[List[int]]) -> List[List[int]]: # by using sort function, we use lamba to select key, in which x is our key # intervals.sort(key = lambda x:x[0]) # without it fail when intervals = [[1,4],[0,4]] stack = [] for i in range(0,len(intervals)): # if stack not empty if stack and stack[0][1] >= intervals [i][0]: # overlapping condition comes..now update end point stack[0][1] = max(stack[0][1], intervals[i][1]) else: stack.insert(0,intervals[i]) # print(i) return stack
@YasarNaser-mr3if2 ай бұрын
Which platform you use to explain the question in Problem Statements and Description?
@nikoo282 ай бұрын
GoodNotes 6
@unemployedcse3514Ай бұрын
awesome ❤
@plutomessi2111 ай бұрын
Thanks bhaiya
@sachinsoma5757 Жыл бұрын
public int[][] merge(int[][] intervals) { int n = intervals.length; int[][] ans = new int[n][2]; Arrays.sort(intervals, new Comparator() { public int compare(int[] a, int[] b) { return a[0] - b[0]; } }); int ansIx = 0; ans[ansIx][0] = intervals[0][0]; ans[ansIx][1] = intervals[0][1]; for(int i = 1;i
@gyandeepdigra846110 ай бұрын
nice yr
@SuriyaT3001 Жыл бұрын
i think in if(condition) you used newinterval[1] instead of result[1] ... Becoz i see here you dont update the value in result list .. in else codition you updated new interval value ...am i right?..
@nikoo28 Жыл бұрын
check it out...we do update the result list in the else block :)
@daffapradana8557 Жыл бұрын
@@nikoo28 can you explain how do you update the result list in the else block? because i've only seen `result.add(newInterval)` and to my knowledge add is like pushing an element to a list. so am i missing something? or is there something that is wrong to my knowledge? because at the first you've already pushed the [1,3] into the result, and when did the result become [1,6] ? because i don't see any .set() function being called.
@apratimkundu2308 Жыл бұрын
@@daffapradana8557 it's getting updated in the if block where the newInterval[1] gets updated to the max. else block is for the disjoint array