Sort Colors (LeetCode 75) | Dutch National Flag Problem | Full Solution with Visuals and Animations

[@jameswalt2199]

YOU MADE IT THE SIMPLEST TO UNDERSTAND WITHOUT COMPLICATING THINGS!!!!! HAT'S OFF TO YOU!

[@Azeem_Idrisi]

Best solution I've seen for this problem. This channel is so underrated.

[@unanimous8510]

Man that’s the best explanation. I saw the coded solution for this problem which is same as yours but couldn’t wrap my head around it. Now I got it! Thank you!

[@LokeshSharma-hm5jz]

i dont know why i developed a fear for this problem. You made it very easy. Thanks.

[@nikoo28]

I was once in the same boat as you my friend. :)

[@RikitaRoy-m2n]

@@nikoo28 🙏🙏

[@azharsofi2854]

your teaching is next level

[@shabanlukyamuzi4012]

Best explanation on KZbin for the problem

[@sithutun688]

best explanation Sir! I didn't watch the coding part before solving the problem. So I implemented it using if-else statements.After solving, I watched the last part. I found out you wrote it using switch cases. Understanding both versions make me confident in coding

[@omdongare2005]

Your the best bro. The problem seems so easy with the way you explain it. Thanks again. Also this is my solution in python based on your approac

[@Aspiringactor02]

This is better than strivers i watched him then i watched you had a strongn feeling this would be good🎉

[@shwetayadav4244]

your explanations are really amazing. In fact, best so far :) Please make more videos :)

[@sarthakgadge5223]

Thanks man this helped me a lot, loved your energy throughout the problem.

[@moezzzz9341]

Your the best bro. The problem seems so easy with the way you explain it. Thanks again. Also this is my solution in python based on your approach class Solution: def sortColors(self, nums: List[int]) -> None: """ Do not return anything, modify nums in-place instead. """ start = 0 middle = 0 end = len(nums ) - 1 while middle

[@naol80]

wow. such a simple and wonderful solution

[@AnkitkumarDubey-e4r]

VERY NICE EXPLANATION WITH CLARITY . THANKS BHAIYA .

[@RAKESHBABUNAIDU]

Underrated!

[@omsudhamsh.h]

Picture perfect mate! Thanks.

[@mayankbadika3101]

very good explanation! keep up the good work

[@pranavshekhar2048]

Great explaination! But I noticed one thing, in the first example, first swap is wrong as middle was at 0, so swap between start and middle should take place and start and middle should move and not between middle and end i.e 0 and 2 what you did because middle was not at 2.

[@anayatk.007]

Thank you for providing such fantastic content

[@captainstring3820]

really very good explanation. you made it so easy

[@apex095]

elegant solution. could you please add more questions on variety of topics like Stacks and Queues etc on your website ?

[@neeeelll_]

Beautifully explained. could u please come up with topic based important questions

[@arnavkukreti2009]

you explain in a very good manner thank you

[@nikoo28]

So nice of you

[@dylanjara201]

Perfect explained. Ty sir!!

[@technicalguy.]

Thank you soo much ❤❤

[@jerronjones470]

Yay, someone I can understand

[@vishwash7416]

GREAT VIDEO

[@sumitraj9312]

Thank you, ploblem solved. love you 😁

[@BroskiXP]

Great explanation, good work

[@sarthakgupta096]

Thankyou sir your way of teaching is amazing

[@nikoo28]

It's my pleasure

[@iceyyeah]

Thanks

[@subee128]

Thank you

[@Harrish30]

Perfect!! Thanks

[@everyontech2716]

great explanation

[@RishitMohanty]

sir, I actually used in-built sort function, in leetcode ie. sort(nums.begin(),nums.end()), and it said said, u beat 100% users with c++. Can we do this or not???

[@nikoo28]

You can, but your interviewer and ask you to solve it without sorting.

[@sourabhrathore9702]

Thanks Brother❤

[@shenth27]

why don't we just loop through the entire array and count the 0's and 1's in seperate variables, then loop the array again and replace number of 0, 1, 2 in that order.

[@DohaZilaoui-zq1gx]

Its just of solution of 10000000000solutions that u should keep it in your mind

[@nikoo28]

There are multiple ways to approach the problem. You want to do in the fastest way possible. It cannot get faster than a single scan

[@unemployedcse3514]

but interviewer won't be impressed by this approach 😂😂

[@porksonline325]

By this approach I am getting errors for 100000 input of integers in array

[@bhavukgarg3619]

@shenth27 interviewer can modify the problem saying the attribute color is part of a card class like class Card { int color; String suit; Int value; } Now if you have colection of cards with 3 type of colours you cannot simply count and write in same array as other attributes suit and value can not be copied

[@SubhajitDas-mt7sn]

c++ class Solution { public: void sortColors(vector& nums) { int start = 0; int middle = 0; int end = nums.size() - 1; while (middle

[@dineshkm7832]

you are super good

[@KarthikC-ju4fx]

❤

[@nusarathaveliwala]

why we are not incrementing mid when it is arr[mid] is 2?

[@nikoo28]

we swap it out, and put the 2 at the end. so we don't increment it. we need to see what came after the swap.

[@bhargavinaik8145]

@@nikoo28 I solved it on leetcode, i knew it would fail for some testcases and it did, I kept trying to understand what was the pattern of those test cases, your answer made understand that exact point. Thanks :)

[@parthsachan1130]

OP

[@officialdreamplayz]

i used selection sort

[@CPS_XI]

💯❤

[@m-bk4um]

good

[@iiju8212]

Bhai kaafi mast samjhaya.

[@Harshal-dk2kr]

bhai whai is vailya

[@Paadai-Kutti]

more easier!!! class Solution { public static void sortColors(int[] nums) { int zero = 0; int one = 0; int two = 0; for (int i: nums) { if(i==0){ zero++; } else if (i==1) { one++; }else { two++; } } for (int i = 0; i < nums.length ; i++) { if(zero!=0){ zero--; nums[i]= 0; } else if (one!=0) { one--; nums[i]= 1; }else if(two!=0) { two--; nums[i]= 2; } } } }

[@kidoo1567]

But complexity?

[@nenuanenenuane6645]

// Java code to sort an array of integers // with the help of single loop import java.util.*; class Geeks_For_Geeks { // Function for Sorting the array // using a single loop public static int[] sortArrays(int[] arr) { // Finding the length of array 'arr' int length = arr.length; // Sorting using a single loop for (int j = 0; j < length - 1; j++) { // Checking the condition for two // simultaneous elements of the array if (arr[j] > arr[j + 1]) { // Swapping the elements. int temp = arr[j]; arr[j] = arr[j + 1]; arr[j + 1] = temp; // updating the value of j = -1 // so after getting updated for j++ // in the loop it becomes 0 and // the loop begins from the start. j = -1; } } } Bro can we use this as well plz let me know

[@ankitraj4179]

Java Solution (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-- ; } } } }

[@hwval-zw4hy]

Why not just count zeroes and ones and refill the array in place? 😂

[@nikoo28]

You will need to iterate over the array twice. First to count all the different 0 and 1s. Next iteration will be to actually fill all the elements. In the approach I discuss, we just do a single scan of the array.

[@hwval-zw4hy]

@@nikoo28 I like your solution. One pass is good. Though counting involves same big O complexity and simpler approach. I also think your solution fits better definition of in-place. E.g. if these were objects, not integers to sort: mine solution wouldn't be acceptable.

[@xyzsince]

Too difficult to understand what he says because of his accent.

[@nikoo28]

which part were you unable to follow?

[@dineshkm7832]

Best solution I've seen for this problem. This channel is so underrated.

[@vinoths7140]

Great Explanation, Thank You.