Merge Intervals (LeetCode 56) | Full Solution with diagrams and visuals | Interview Essential

[@arjunjadhav6275]

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

[@bingochipspass08]

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]

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]

Cheers man!!

[@shaikhanuman8012]

Tqs for clear cut explanation on merge intervals. Tq very much

[@spacelovertelugu44spacelov46]

Topnotch best explanation for this problem

[@shivaniverma4266]

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

[@abhcode]

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❤

[@prashanthshetty8337]

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.

[@nandhakumarkr3147]

Your tutorial is just awesome

[@Divya-qt1cf]

Great explanation 🎉

[@subee128]

Thanks

[@kalyanamvenumadhav2245]

Please Explain 4Sum Problem so that it can be helpful to all of us.

[@nikoo28]

Yep..will make a video on it too.

[@everyontech2716]

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-mr3if]

Which platform you use to explain the question in Problem Statements and Description?

[@nikoo28]

GoodNotes 6

[@unemployedcse3514]

awesome ❤

[@plutomessi21]

Thanks bhaiya

[@sachinsoma5757]

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

[@gyandeepdigra8461]

nice yr

[@SuriyaT3001]

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]

check it out...we do update the result list in the else block :)

[@daffapradana8557]

​@@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]

@@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

[@manavgora]

I am putting you over striver

[@onkardheemate1066]

Salute you sir🫡🫡🫡