Sort colors | Leetcode #75 

Thank you, you explained it very well, quick n simple. Others are wasting time for around 20 min.

I think you have to use 'else if' instead of if everywhere. Because for this ex 2,0,2,1,1,0 the output will come as 001212 Hope this will help you, or else if I am wrong , Please explain...Thank you

It was the best video on entire youtube. I love your simplest explanation.

Very succinctly explained, I remember solving this problem in O(n) but solution was very messy, this is just clean and simple!. Thanks!

Amazing explanation in such a less time👍👌

Great video! Concise and clear!

One idea can be (though not optimal )...iterating through array & calculating occurrences of 0s 1s 2s in 3 variables and placing them (0 1 2 ) accordingly into original array...

Count Sort will also do this in O(3N) ie. Linear time O(N) but it's mentioned to do it in one-pass and 3N is actually 3 pass

You teach really good, I guess it should be else if instead of the if in the latter two conditions.

very good, helped me a day before my interview. thanks :)

simple and precise! thank you.

sir why in third step we havnot did m++ also? . is it because after swapping arr[h] with arr[m]... the current arr[m] can be 0 or 1 ... if its 1 we'll move forward else if its 0 we'll have to swap it with arr[l]; ??

Thanks found your explanation intutive !

Another (simpler) approach would be counting zeros and ones. Then mutating the array accordingly, it'll also take O(n) time & constant space.

this code does not run for multiple test cases on geeksforgeeks platform.

Nicely explained brother!!

Nice explaination. We can solve this just counting number of 1's,2's and 0's .

Nice straight forward explanation.

I thought of this solution but missed m

sir can we use this technique whenever we required sorting in a problem,suppose there list of no in aaray,we have to sort it,then by using direct sorting can we use this process

Thanks for the video, there are some cases which won't work as the terminating conditions are incorrect. Within the loop, we are incrementing pointers which may go out of bounds or may lead to inconsistent results (as Pratik24 mentioned), below will work. while(m

I think we don't need to swap if we are getting 1 because if we fix two element i.e 0 and 2 then 1 will be automatically fixed in between , so we can reduce work

WHICH SOFTWARE DID YOU USE

Nice explanation...It helped alot. Thank you

Can we achieve the solution using low = 0 , mid = n-1 and high = n -1 ?

Thanks! Your solutions are really well explained and simple to understand.

Nice solution :) keep going

Very nicely explained . good job

The easiest explanation I came across.

Hey , but you need to explain , how this idea comes into your mind , like how you analyse the problem and find this approach ?

Thank you buddy! u made it very clear ! we love u !

nice and fast explanation . liked it bhai.

Nice explanation! Just a quick note, I think the code should be if( ... ) else if ( ... ) else if ( ... ) instead the if (...) if (...) if (...). I'm doing the "Sort Colors" problem from leetcode and I needed to change that

Very nice explanation sir.

you give the best explanation FR

What software do you use?

Awesome

if you want a solution with complexity O(n) then you can also do like this... 1st find all 2's and put it in stack then find all 1's and put it in stack then find all 0's and put it in stack then pop all elements... your output is sorted... here we got (3n) traversal hence equivalent to O(n)... yeah but yours O(n) is better than mine O(n)

Nice explanation sir Tq ❤

This can be solved also with counting sort. And it can also be done in place just using 3 variables instead of array as usually

Why not simply count 0s, 1s and 2s and then fill the array accordingly? I guess it will take two passes as opposed to one so the solution mentioned in the video is more efficient (though might not be scalable when instead of 2s, we have a higher max limit).

Thanks bro !! Ur punjabi fan

At a time only one condition should be executed, so need to use else if / switch cases. Btw explanation was good.

very well explained😁

this is better with no confusion Keep three counter c0 to count 0s, c1 to count 1s and c2 to count 2s Traverse through the array and increase the count of c0 is the element is 0,increase the count of c1 is the element is 1 and increase the count of c2 is the element is 2 Now again traverse the array and replace first c0 elements with 0, next c1 elements with 1 and next c2 elements with 2.

What is the difference if we write len(nums) or len(nums)-1? In 2nd it indicates starting index is 0 then what about len(nums) ?

Awesome explanation

Thanks for the explanation buddy

So well explained 😄😄

Awesome!

Counting sort can also solve this in O(n) since range of nums[i] is constant :)

If we knows that their is only 0,1,2 element. then why not just count frq. Of those 3 and fill the array it's also takes O(n) time and O(1) space.

there are some changes in the solution correct solution : if(a[m] ==2) --h; in video : if(a[m] ==2) h--;

Very Nice explanation sir..

Awesome ❤️

perfect 👌

Op explanation 🤯🤯🤯🤯😀

Awesome explaination, hey you have done a lot of good interview preparation videos. Its a reuqest can u make a repository type something where you can put your videos topicwise ( Graph, Tree,...) . It will help during preparing on a topic to check all imoprtant problems on that topic..

Nice Explanation.

Thanks a lot

Nice explanation sir.......

Thanks for sharing

Why you didn't run the code in java...and where I can get code

Just keep count of number of 0,1 and 2...and fill the array with 0 ,1 and 2 Respectively

We can also use hashing

nice explanation

How do you come up with approaches, is it because of practice ?

short and effective one

Really helpful

what is the intuition behind this?

love youuu broo

Leetcode 75 Sort colors

Maintaining count of 0,1,2's and putting that many back in the array is far simpler

how can we modify it if we want a stable sort?

nice explanation .. but it's not in-place .. the ordering of 2's are changed ... does anybody know any in-place way of doing it .

i have a same solution with same approach but a test case of size 65754 gives an error in gfg idk why and the solution is working fine for the element of small array

This will not pass all the test cases - for eg. 1,0,1,2,2,1. as you are just incrementing when its 1, the resulting array will not be correctly sorted.

what is this algorithm called?

why don’t we just count 0s ,1s and 2s.. Generate new arr and return O(2N)

ALL THREE if statement doesn't work beacause 3 if condition can be true simultaneously.

USE ELSE IF , other wise the output will be wrong , for such like cases 2 2 1 1 2 1 0 0 1 2 2

This solution will not work if we do not have 1 in the list, because m will never increased.

i think in while condition the condition should be while(mid while(mid

Why not simply count the number of 0's 1's and 2's and then fill the array?

not working for 2,0,1

Thanks

Why not just count the 0s 1 2s and place them in array?

Much easier solution-Make a count array with elements (0,0,0) where 0th index is 0 count 1st index is 1st count 2nd index 2nd count.Traverse the given array in question once and keep updating the 0 array then rewrite the question array with appropriate counts of 0,1,2

Understood.

it can b solved using multiset in logn time

instead of a[m]==2 is should be else{} other wise some case will fail for 2 2 1 1 2 1 0 0 1 2 2 and many more

class Solution { public void sortColors(int[] nums) { int i,j,k; int len=nums.length; i=0; j=0; k=len-1; while(j

send this code please

//those who are getting any problem with the mentioned code on video can run this code: int left = 0, right = a.length -1; for(int i=0;i

Dutch algorithm

this explaination coould be more better and clear, next time while making video ensure that you will work on that

U should have explained the intuition. Instead, You just dry run the solution. Not much beneficial.

Can we name this method as Indian Flag Method??

Java Code (Beats 100 %) class Solution { public void sortColors(int[] nums) { int n = nums.length ; int[] arr = new int[3] ; int element = 0 ; for(int i = 0 ; i < n ; i++){ element = nums[i] ; arr[element]++ ; } int count = 0 ; int k = 0 ; for(int i = 0 ; i 0){ nums[k] = i ; k++ ; count-- ; } } } }

two line using Java lib function , but not O(n) :| Arrays.sort(arr); System.out.println(Arrays.toString(arr).replaceAll("\\]","").replaceAll(",",""). replaceAll("\\[",""));