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!
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
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
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.
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.
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 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 :)
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; } } } }
// 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
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-- ; } } } }
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.
@@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.
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Best solution I've seen for this problem. This channel is so underrated.
YOU MADE IT THE SIMPLEST TO UNDERSTAND WITHOUT COMPLICATING THINGS!!!!! HAT'S OFF TO YOU!
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!
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
your teaching is next level
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
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
your explanations are really amazing. In fact, best so far :) Please make more videos :)
i dont know why i developed a fear for this problem. You made it very easy. Thanks.
I was once in the same boat as you my friend. :)
@@nikoo28 🙏🙏
Thanks man this helped me a lot, loved your energy throughout the problem.
This is better than strivers i watched him then i watched you had a strongn feeling this would be good🎉
Best explanation on TH-cam for the problem
elegant solution. could you please add more questions on variety of topics like Stacks and Queues etc on your website ?
Beautifully explained. could u please come up with topic based important questions
VERY NICE EXPLANATION WITH CLARITY . THANKS BHAIYA .
Great explanation, good work
very good explanation! keep up the good work
Thank you for providing such fantastic content
Underrated!
Great Explanation, Thank You.
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.
Picture perfect mate!
Thanks.
Perfect explained. Ty sir!!
you explain in a very good manner
thank you
So nice of you
Thank you, ploblem solved. love you
😁
Thankyou sir your way of teaching is amazing
It's my pleasure
Thank you soo much ❤❤
Thank you
Thanks
Yay, someone I can understand
great explanation
Perfect!! Thanks
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.
Its just of solution of 10000000000solutions that u should keep it in your mind
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
but interviewer won't be impressed by this approach 😂😂
c++
class Solution {
public:
void sortColors(vector& nums) {
int start = 0;
int middle = 0;
int end = nums.size() - 1;
while (middle
i used selection sort
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???
You can, but your interviewer and ask you to solve it without sorting.
❤
why we are not incrementing mid when it is arr[mid] is 2?
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.
@@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 :)
💯❤
Bhai kaafi mast samjhaya.
good
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;
}
}
}
}
But complexity?
// 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
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-- ;
}
}
}
}
Why not just count zeroes and ones and refill the array in place? 😂
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.
@@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.
Too difficult to understand what he says because of his accent.
which part were you unable to follow?