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.
Actual Problem on LeetCode: leetcode.com/problems/move-ze...
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:
Brute Force Solution: • Brute Force algorithms...
What is Big O?: • Big O Notation Simplif...
Time Complexity: • What is the Time Compl...
More LeetCode problems: • Leetcode Solutions
📘 A text based explanation is available at: studyalgorithms.com
Code on Github: github.com/nikoo28/java-solut...
Test-cases on Github: github.com/nikoo28/java-solut...
📖 Reference Books:
Starting Learn to Code: amzn.to/36pU0JO
Favorite book to understand algorithms: amzn.to/39w3YLS
Favorite book for data structures: amzn.to/3oAVBTk
Get started for interview preparation: amzn.to/39ysbkJ
🔗 To see more videos like this, you can show your support on: www.buymeacoffee.com/studyalg...
🎥 My Recording Gear:
Recording Light: amzn.to/3pAqh8O
Microphone: amzn.to/2MCX7qU
Recording Camera: amzn.to/3alg9Ky
Tablet to sketch and draw: amzn.to/3pM6Bi4
Surface Pen: amzn.to/3pv6tTs
Laptop to edit videos: amzn.to/2LYpMqn
💻 Get Social 💻
Follow on Facebook at: / studyalgos
Follow on Twitter at: / studyalgorithms
Follow on Tumblr at: / studyalgos
Subscribe to RSS feeds: studyalgorithms.com/feed/
Join fan mail: eepurl.com/g9Dadv
#leetcode #programming #interview

[@reactninja6567]

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

[@supershinobi7892]

you have made my coding journey very easy and understandable

[@abhinavd2]

one of the best teacher ever studied😍

[@saichaithrik7134]

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

[@deeptidip9864]

Such an easy explanation💫

[@HarshPatel-hu5sr]

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

[@nikoo28]

Wow, thank you!

[@LuckyAndDad]

Thank you brother

[@UjjwalRaj-CSE-]

I am watching your video first time, you are great

[@viveknamdev2605]

Nice Explanation by you

[@pulastyadas3351]

You are really awesome❤

[@ramielwan48]

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 😅

[@nikoo28]

That is so nice of you. :)

[@dewanshanand1395]

Amazing !!

[@TechnicalMasterAman]

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

[@Sakeer89]

Very easy solution

[@ajayupadhyay4964]

you are a really great teacher !!

[@nikoo28]

Wow, thank you!

[@khaleabhishek5023]

amazing keep going

[@shubhamkumar-um2xi]

Great brother

[@shubhamkumar-um2xi]

Very nice

[@sammedpatil3907]

Very Nice Explanation

[@nikoo28]

Thanks and welcome

[@karthik-varma-1579]

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

[@nikoo28]

Will release it soon

[@tufanpondu9511]

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

[@nikoo28]

glad to have you as my audience :)

[@alexivuckovich8005]

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?

[@nikoo28]

O(2n) simplifies to O(n)

[@Vasoya_Jaydeep]

This will take O(n + number of zeros) Time Here is one more solution with O(n) Time void moveZeroes(vector& 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*/

[@rajdeepdas8393]

Just what the fuck happened......