I paused the video at 20:12 and now going to solve the problem at leetcode myself, thank you so much Tushar, your help for the rest of us who have never graduated from the CS but still want to become a sane programmer is invaluable
Brilliant video and explanation! One minor change to the 1st example:initialization: 1) pick smaller len 2) start = 0, end = len, binary search the len of left size of X e.g with the first example: 1st round lo = 0, hi = 5, mid = lo + (hi - lo) / 2 = 2 2nd round lo = mid + 1 = 3, hi = 5
People like me wouldnt understand this video in one time,you gotta watch it multiple times ,read about the problem solving strategy then you will get the problem. He has explained in a very good manner
Clear explanation. Thanks. A small error, 8:33 value of "end" should be 5 (size of small array). I followed just your example and coded, everything was right but failing always, until I watched your code & realized that end should have initial value of smallest array's length.
@@eskuAdradit0 I think it's because in some cases we can have all the elements in x on the left side. He mentioned that partitionX is "the number of elements partitioned to the left in x." So in those cases, partitionX can be the size of the small array. Consider this example: arrayX: [1, 2, 3, 4] and arrayY: [5, 6, 7, 8]. All the elements in x are smaller than elements in y. Hope this helps.
Tushar, thank you so much! You explained it perfectly! If someone need solution in Python, here it is: class Solution: def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float: if len(nums1) > len(nums2): return self.findMedianSortedArrays(nums2, nums1);
x = len(nums1) y = len(nums2) low = 0 high = x while low
@4:30 - I took a second to understand this : Think of the final sorted array (without duplicates) and draw a median line. We need to find the average of numbers which are immediately to the left and right of that median line. If x2 is immediately to the left of the median line, it has to be greater than y5 (remember final array is sorted). Similarly, if y6 is to the right of the median line it has to be lesser than x3. Hope this helps someone!
The best possible explanation of the problem I've seen so far. I have now clear idea on how median of two sorted array algorithm works using binary search & partitioning arrays. Good stuff!
@@emilyhuang2759 Exactly even i'm not able to figure out why the end should be 5 and not 4 as it is clearly mentioned in the video that first array has 5 elements in total.
Beautiful solution, thank you so much! I can't believe how close I was to the solution, I wished I pushed myself a bit harder, but this was a great educational experience. Thanks again!
Great solution. Not every detail was explained but as everyone else was saying this is probably the best solution on the net and if not it is one of the best ! The missing details can be inferred
Brother, you have done well in explaining this. Thanks! I wanted to abandon this until I saw your video. With what I have known now I can go and write it myself until I pass it without looking at random solutions across the internet. I have subscribed for more.
Jahapana tussi great .........sir i read a lot about this problem solution everywhere but finally came to understand from your video only...you made a very tricky concept so simple to understand
don't think there are words to express my gratitude towards your hard work in making such bloody awesome videos @TusharRoy your videos indeed are class apart & blessing for someone who wants to gain in-depth knowledge on DSA.
*EDIT:* Thank you for the video. I watched it carefully, then coded the solution from memory. And then I compared to your solution, and mine is a little simpler and easier to understand (code is in the first comment). *Initial comment* 14:41 here it is obvious from the data that we must compare the rightmost and leftmost elements to see if the values in the arrays even overlap in range. If they don't we should skip the binary search and just treat them as one continuous sorted array, and find the median in it directly. Thank you for the wonderful videos.
Hi Sir, thanks for your great and detailed explanation...just had watched serveral videos and explanations before I saw your video...and finally I understood how it worked...It is really awsome that you used these examples to illustrate the problem (including the edge case) and in your code you also wrote so many annotations...I really appreciate your effort and great work.
15:24 If there are n element then we can partition that array in n+1 no. of ways. E.G 4 elements then we partition it in. 1) 0 - 4 2) 1 - 3 3) 2 - 2 4) 3 - 1 5) 4 - 0
I got asked this question in an interview .Before this video i ignored this question multiple times due to which i was not able to answer this in interview .After getting rejected from the interview i came here and i feel how can a person come up with such a solution in 45 minute interview and code it properly.But The solution is very nice.
Thank u bro.... Your video always give us very good concept..... Please make a session on josephus problem. This problem is very easy to understand but hard to implement.
Thank you for sharing! This hard problem on Leetcode is understandable by any folk with the basic knowledge of median and binary search! That's really crystal clear explanations!
Tushar Roy - Coding Made Simple Sir, please make some more videos on binary search, and explain the strategy how to apply binary search, and please provide the explanation of the code on the github, the circular binary search problem on the github. Thanks
I paused the video at 20:12 and now going to solve the problem at leetcode myself, thank you so much Tushar, your help for the rest of us who have never graduated from the CS but still want to become a sane programmer is invaluable
Man the way you explained it in the first 5 minutes just clarified everything. Thanks. Great video!
I love when he says "once it hits you" like some kind of drug or something
@Kayden Khalil Lol, that is scam! fake accounts
It's a better than a drug and the dopamine released on solving questions is sickkkk.
It finally hit me I'm so high rn
don't do drugs bro. It's harmful.
Quite a common phrase
Brilliant video and explanation!
One minor change to the 1st example:initialization:
1) pick smaller len
2) start = 0, end = len, binary search the len of left size of X
e.g with the first example:
1st round lo = 0, hi = 5, mid = lo + (hi - lo) / 2 = 2
2nd round lo = mid + 1 = 3, hi = 5
I have seen many other videos for this problem, but yours is the one that made me follow every step clearly. Thank you!
I was plagued by this problem for a very long time, you made it crystal clear, sir. Thank you!
People like me wouldnt understand this video in one time,you gotta watch it multiple times ,read about the problem solving strategy then you will get the problem.
He has explained in a very good manner
Even after 5 years your video is best for this problem.Thank you so much for wonderful Explaination
7 years now, its fresh
I can't stress enough, how clear this video explanation is. Really loving your work, and it's very very helpful.
Damn !!! How did anyone come up with such an Algorithm
Clear explanation. Thanks.
A small error, 8:33 value of "end" should be 5 (size of small array). I followed just your example and coded, everything was right but failing always, until I watched your code & realized that end should have initial value of smallest array's length.
any clear explanation on why would we do it like this? is it the same reasoning behind adding + 1 to the sum of both arrays before dividing by two?
@@eskuAdradit0 I think it's because in some cases we can have all the elements in x on the left side. He mentioned that partitionX is "the number of elements partitioned to the left in x." So in those cases, partitionX can be the size of the small array. Consider this example: arrayX: [1, 2, 3, 4] and arrayY: [5, 6, 7, 8]. All the elements in x are smaller than elements in y. Hope this helps.
Tushar, thank you so much! You explained it perfectly!
If someone need solution in Python, here it is:
class Solution:
def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
if len(nums1) > len(nums2):
return self.findMedianSortedArrays(nums2, nums1);
x = len(nums1)
y = len(nums2)
low = 0
high = x
while low
clean implementation good job
I went through 5 videos before landing here and finally understanding the intuition behind the solution. Thank you!
came to "like" this video only found out that I have "liked" this video 4 years ago when I was looking for my last job. Thank you Tushar.
This is amazing. I was so confused by this problem, you explained it so succinctly. Thanks
This video is definitely the most effective one in providing a clear explanation of the algorithm
@4:30 - I took a second to understand this : Think of the final sorted array (without duplicates) and draw a median line. We need to find the average of numbers which are immediately to the left and right of that median line. If x2 is immediately to the left of the median line, it has to be greater than y5 (remember final array is sorted). Similarly, if y6 is to the right of the median line it has to be lesser than x3. Hope this helps someone!
thanks for the extra intuition :)
Thanks a lot!
The best possible explanation of the problem I've seen so far. I have now clear idea on how median of two sorted array algorithm works using binary search & partitioning arrays. Good stuff!
00:00 Introduction
01:39 Solution
06:19 Example 1
14:35 Example 2
20:11 Code
Thanks buddy😊
7:18 end should be 5 not 4 because we are searching which position we want to cut.
YES! Had the last element of first array been 10 instead of 15, the algorithm would've broken at 13:45. Anyways, the explanation was awesome.
@@prakashtiwari7834 I was wondering why did he take end as 4 and read this comment ! Thanks a lot !
But why not 4? The 5th index is empty....so I would think the end is 4...?
@@emilyhuang2759 Exactly even i'm not able to figure out why the end should be 5 and not 4 as it is clearly mentioned in the video that first array has 5 elements in total.
@@sharmamukul938 it should be the size of the array not index that is to be taken as high element according to code.
this is the best video of all time, it brings happiness. thank you!
after reading multiple explanations, this one finally made me understand how to think.
Thanks for the video.
Very well done detailed explanation. Much better than the LeetCode solution example. Thanks for taking the time to make this!
thanks very much.it took me long time to think the algorithm on leetcode, your explanation is so clear and concise.Very nice
The best explaination with perfect testcases that I found for this question on Internet.
Thank you!
Beautiful solution, thank you so much! I can't believe how close I was to the solution, I wished I pushed myself a bit harder, but this was a great educational experience. Thanks again!
probably one of the best explanations i have seen so far, SUBSCRIBED!
Great solution. Not every detail was explained but as everyone else was saying this is probably the best solution on the net and if not it is one of the best ! The missing details can be inferred
Brother, you have done well in explaining this. Thanks! I wanted to abandon this until I saw your video. With what I have known now I can go and write it myself until I pass it without looking at random solutions across the internet. I have subscribed for more.
Spent a lot of time trying to understand this from leetcode solutions and discussion. And you video explained it in 5 minutes
Jahapana tussi great .........sir i read a lot about this problem solution everywhere but finally came to understand from your video only...you made a very tricky concept so simple to understand
What a precise and correct explanation of the algorithm! Thanks for making and sharing this video, Tushar.
Perfect explanation, initially I though it would be difficult for me. But gradually you made me understand in only one go.
Thanks 😀
Great.
im first-year student from Vietnam, thank you a lot for this helpful video, appreciate
i usually don't comments on videos but your method of explaining made me comment on this...
Superb video boss..
Repeat watching the first 6 minutes until you get it. Totally helps. Thank you so much. You simply put it
this is the best channel for coding hand down
Thank you so much Tushar. This explanation will last for generations.
Tushar Sir, you literally saved my time with crystal clear explanation on this problem.
Perfectly explained... Found most useful among all the available videos on RUclips for median in two sorted arrays.
don't think there are words to express my gratitude towards your hard work in making such bloody awesome videos @TusharRoy
your videos indeed are class apart & blessing for someone who wants to gain in-depth knowledge on DSA.
ruclips.net/video/W-UalzYVEiQ/видео.html
*EDIT:* Thank you for the video. I watched it carefully, then coded the solution from memory. And then I compared to your solution, and mine is a little simpler and easier to understand (code is in the first comment). *Initial comment* 14:41 here it is obvious from the data that we must compare the rightmost and leftmost elements to see if the values in the arrays even overlap in range. If they don't we should skip the binary search and just treat them as one continuous sorted array, and find the median in it directly.
Thank you for the wonderful videos.
A somewhat simpler and easier to follow code:
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
int len1 = nums1.length;
int len2 = nums2.length;
int res = 0;
if(len2 < len1)
return findMedianSortedArrays(nums2, nums1);
int partitionX = len1 / 2;
int low = 0;
int high = len1;
while(low 0 ? nums1[partitionX - 1] : Integer.MIN_VALUE;
int num1R = partitionX < len1 ? nums1[partitionX] : Integer.MAX_VALUE;
int num2L = partitionY > 0 ? nums2[partitionY - 1] : Integer.MIN_VALUE;
int num2R = partitionY < len2 ? nums2[partitionY] : Integer.MAX_VALUE;
if(num1L = num2L) { // found correct comb
if((len1 + len2) % 2 != 0)
return Math.max(num1L, num2L);
else
return 1.0*(Math.max(num1L, num2L) + Math.min(num1R, num2R)) / 2.0;
} else if(num1R < num2L) { // need to move to the right
low = ++partitionX;
} else
high = --partitionX;
}
return Double.MIN_VALUE;
}
In the first example, the end value at the start should be 5 as we can have a partition when all the 5 elements of array X will be on the left side.
So elegant and simple. One of the best tutorial ever! Keep it up.
Most Beautiful explanation and solution. My deepest gratitude to all your videos.
Very Nice Explanation. One of the best videos of Tushar
I was so confuse how to solve this without taking care of corner cases as extra, your explanation was awesome. Thanks
Beautiful lovely brilliantly handled edge cases
this is the best explan I ve ever seen
step by step with details,
thank you
Excellent !! Nothing more one can do to explain this problem :)
Thanks for sharing it buddy !!
You are a star, keep up the good work, cheers !!
I can never figure it out without this video.
noob
great sir, got it in single explanation, I looked all over for over 4 hr
+achyut aarjunneupane nice
Crystal clear explanation , Bro. You are helping a lot of people, Please Keep up the good Work.
Hi Sir, thanks for your great and detailed explanation...just had watched serveral videos and explanations before I saw your video...and finally I understood how it worked...It is really awsome that you used these examples to illustrate the problem (including the edge case) and in your code you also wrote so many annotations...I really appreciate your effort and great work.
Best Channel for Data Structure && Algorithms !
Because of you I am going to solve this problem otherwise I thought I will skip it, thank you
After watching 2 times it quite easy to understand. Keep up the good work Tushar.
15:24 If there are n element then we can partition that array in n+1 no. of ways.
E.G 4 elements then we partition it in.
1) 0 - 4
2) 1 - 3
3) 2 - 2
4) 3 - 1
5) 4 - 0
0-4 and 4-0 results in same arrays
Thanks Tushar.Your explanation helped a lot to understand the depth of logic and algorithm behind this question.
These videos are great equivalent to full computer science degree content
My boy Tushar Roy! Earned yourself a sub! Gonna help my algo course so much
Thank you very much for your great explanation. I learn a lot from you, not only the code but the way to explain the problem.
Thank you for the wonderful video. I was stuck for a long time until you made it co clear to me.
Your channel solves all my doubt
Very good explanation. Thank you Tushar Roy.
Thank you so much. Your explanation is the best. Only with you I understood how to solve that problem.
this one of the best explanation on youtube, awesome!
Thanks, good explanation. I was looking for different resources, but it was the best so far :)
Best explanation present on the Internet!
Thanks Bro for the excellent explanation! It is crystal clear now how the logic works for both odd/even scenarios. Appreciate your time. Cheers!
Simple and elegant explanation. Hats off.
I got asked this question in an interview .Before this video i ignored this question multiple times due to which i was not able to answer this in interview .After getting rejected from the interview i came here and i feel how can a person come up with such a solution in 45 minute interview and code it properly.But The solution is very nice.
ruclips.net/video/W-UalzYVEiQ/видео.html
Finally, I understood the crux of this problem. Thanks for the enlightenment.
Simply Amazing !!! This algorithm was really helpful and you gave good examples to solve it. Especially the one in which you covered the corner cases.
Hats off sir...I am really overwhelmed by the way you explained this algorithm and made it so easy and simple...(y) Thank you so much...:-)
You save my life.
Just watch for 8 minutes and come up with a solution by myself!
Very well explained, the code is very clear and precise. Thanks
Such a great explanation! Love how you simplify things. Thanks a lot!!!
Hey Tushar,
You make complicated problem look so simple and easy. Awesome job.
The best explanation that I ever heard
Legendary. Very crisp and clear approach. Thank you very very much!
You are making my life a lot easier! May God bless you
Thank u bro.... Your video always give us very good concept.....
Please make a session on josephus problem. This problem is very easy to understand but hard to implement.
Tushar Sir tumi Bhogowan, Love from West Bengal
Thanks Tushar for the detailed explanation.
very good explanation of the concept here, keep make kind of this videos... thank you
bro tushar please make a video on how to prepare for giant companies like apple,amazon etc. also thanks for the amazing video
I will try.
What do you think this entire channel is for?
Thank you for sharing! This hard problem on Leetcode is understandable by any folk with the basic knowledge of median and binary search! That's really crystal clear explanations!
Tushar Roy - Coding Made Simple
Sir, please make some more videos on binary search, and explain the strategy how to apply binary search, and please provide the explanation of the code on the github, the circular binary search problem on the github.
Thanks
@@83rossb Lmao!
very neat and excellent explanation
Thanks
Might totally be a noob question:
You are using:
"lo = 0, hi = x" with "while lo
same doubt
You truly are making coding simple !
Geniusly explained the algo 🙏🏻👏🏻
Best explanation with examples. Thx a lot.
give this man a medal
Tushar you are awesome! I have watched few video about this question, yours is the most intuitive one
that was so cool explanation sir !! handsdown🙌🙌
Simply the best explanation.
Such an elegant explanation !!
wow there's nothing more than clearer than this! thanks :) you just saved my time
such an incredibly good video. amazing explanation
One of the best explaination!!. Great work. Keep making more video :)