Move Zeroes (LeetCode 283) | Full Solution explained with examples | Study Algorithms

Given an array of positive integers with some zeroes. You need to move all the zeroes to the end without changing the relative order of non-zero elements. At the first glance it feels that you will need to swap and move around a lot of elements to arrive at an efficient solution. This video explores a unique way of solving this problem and once you understand, it feels so easy and obvious. Watch the complete video with my final thoughts and a dry-run of code in JAVA.

Chapters:
00:00 - Intro
01:12 - Problem statement and description
02:50 - Straight forward approach to move zeroes
04:15 - Space Efficient Solution In-place
06:56 - Dry-run of code
09:22 - Final Thoughts

📚 Links to topics I talk about in the video:

📖 Reference Books:

🎥 My Recording Gear:

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#leetcode #programming #interview

[HarshPatel-husr]

I honestly love your way of teaching, you would make an AMAZING professor

[reactninja]

Hands Down this is the best soln i have seen so far thanks! keep on going.

[avengersshorts]

3:33 I Completed solving this problem in leetcode thank you ❤

[saichaithrik]

Your way of teaching is one the best i have ever seen thank you sir

[souravsarkarofficials]

Bhaiya apki samjhane ki tarika sach mei pura alag hei ...koi nhi rok sakta apko DSA mei 💪💪💪💪

[supershinobi]

you have made my coding journey very easy and understandable

[ramielwan]

Man I swear I asked chat gpt to explain it to me. I read the leetcode solution, and watched 2 other videos but none of them made sense to me. I wanted to understand how to approach it and not just copy and paste the solution. When you explained it, I was like damn that's easy 😅

[kashikasinghal]

thank you for teaching the best and efficient approach. before watching this video i was struggling with the approach as mine took 336ms and the other videos were complex to understand, although it was only 5 min long but i couldn't get a loop;) nut this approach solved it in only 7ms...like this is crazy!

[ahmar]

On point and quick to grasp explanation. Keep posting

[TechnicalMasterAman]

under if condition, we can just do swap(nums[i], nums[insertPos])

[Vasoya_Jaydeep]

This will take O(n + number of zeros) Time

Here is one more solution with O(n) Time

void moveZeroes(vector<int>& nums) {

int Pi =0, Zi = 0;

while(Pi < nums.size()) {
if (nums[Zi] == 0)
{
while((Pi < nums.size()) && !nums[Pi]) {
Pi++;
}

if (Pi >= nums.size())
break;

int tmp = nums[Zi];
nums[Zi] = nums[Pi];
nums[Pi] = tmp;
}

Zi++;
Pi++;
}
}
/*End*/

[UjjwalRaj-CSE-]

I am watching your video first time, you are great

[TejwantDataLab]

Thank you sir, really good explaination

[vis]

Two pointer approach is also better one

[alexivuckovich]

Loving all the videos so far, but with this one it seems incorrect to say it's O(n) becuase the array is iterated over only once. I think it's actually of O(n) because it's simplified from O(2n) in the worst case that the array is all zeros. What about adding a line a code before inserttPosition is incremented that checks if i > insertPosition and just set the i-th element to zero then?

[karthik-varma-]

nikhil awaiting for the sheet that you said will release!!!

[tufanpondu]

Sir you make me bound to subscribe to your channel and give like.😍😍😍😍