Unique Paths III -(Apple, Amazon, Microsoft, Intuit) : Explanation ➕ Live Coding

This is our 2nd Video on our Backtracking Playlist.
In this video we will try to solve a very Popular and a classical Backtracking Problem "Unique Paths III".
Share your learnings on LinkedIn, Twitter (X), Instagram, Facebook(Meta) with hashtag hashtag#codestorywithmik & feel free to tag me.
We will do live coding after explanation and see if we are able to pass all the test cases.
Problem Name : Unique Paths III
Company Tags : Apple, Amazon(on-site), Meta, Microsoft, Salesforce
My solutions on Github : github.com/MAZHARMIK/Intervie...
Leetcode Link : leetcode.com/problems/unique-...
My GitHub Repo for interview preparation : github.com/MAZHARMIK/Intervie...
Subscribe to my channel : / @codestorywithmik
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#coding #helpajobseeker #easyrecipes #leetcode #leetcodequestionandanswers #leetcodesolution #leetcodedailychallenge #leetcodequestions #leetcodechallenge #hindi #india #hindiexplanation #hindiexplained #easyexplaination #interview #interviewtips #interviewpreparation #interview_ds_algo #hinglish

[@prateekrath5098]

Absolute Gold explanation as always. Happy new year brother, may god bless you with all the wealth, happiness and love. I hope you grow very big in the coming year, which I'm sure you will. Keep making these great explanations❤

[@codestorywithMIK]

Thank you so much Prateek. You made my day. Wish you also all the happiness and success in life ❤️❤️❤️

[@rode_atharva]

sir graph ke questions kiye the to ye ho gaya, lekin appka video pura dekha to dimag me fix ho gaya ab.

[@rajeshsinha290]

You explained the approach really well. I didn't even feel that it is a hard level question.

[@souravjoshi2293]

Dope explanation man. Thanks a lot. No longer a Hard Problem. Happy New year bro Hope to be consistent like you in coming year

[@ajiteshawasthi7311]

Today first time I showed courage to solve hard problem without seeing solution , took me bit of time but I was able to solve it. Its all because of you brother. Thank you for your videos. You really are a gem.

[@aakashtiwari8884]

Been following your channel since last month, I finally received my first ever Leetcode badge today! Thanks for making all these questions feel easy, love your explanations and videos, thanks for doing this. keep posting you surely have miles to go! A very happy new year to you and yours!

[@codestorywithMIK]

Hi Aakash, i am so happy with the news that you got the Leetcode badge. So happy for you. Keep it up and i will also continue it. Feel free to share your badge on LinkedIn or with your colleagues and help others achieve it too. It will strengthen your skill too. Thanks again ❤️❤️❤️

[@EB-ot8uu]

You make Hard to Easy so effortlessly.

[@svdfxd]

Amazing explanation !!

[@GeneralistDev]

holy shit!!! after watching your word search video (I just commented on that about that video few hours back) I was able to solve this question on my own .... totally surprised!!! I just coded casually and submitted in first attempt... can't believe...

[@codestorywithMIK]

Woww. I am so glad. Very well done 👏

[@OIAOa]

enjoying a lot bhaiya❤

[@aryanarora1984]

you deserve a lot moree brother!! your way of explaining is awesome...

[@codestorywithMIK]

Means a lot 😇❤️🙏

[@sunnyvlogs__]

Thanks Bhaiya, just saw the explanation and coded own. At first I was thinking of a solution like:- to calculate all the paths and store its (i,j) of that path in a vector. Then I will check the all the path vectors one by one and if its size is equal to the (size of the matrix(m*n) - given obstacles). then I will increase the count++. But while coding it was getting complicated. So followed this approach. Crystal Clear Explanation✨✨ Happy New Year Aapko. Tokopedia mein referral milega as a fresher😆

[@codestorywithMIK]

Thank you so much Sunny. ❤️❤️❤️ Indeed your earlier approach was also fine but it will get complicated as you mentioned. And currently, there is a hiring freeze unfortunately due to the world wide recession. But once it resumes (soon), i will be glad to refer all of you here. I am pretty sure you all are hard working and will rock wherever you guys go. Mean while, make the most of this time and prepare as much as possible so that when you start giving interviews you are READY 💪💪💪

[@sunnyvlogs__]

@@codestorywithMIKThank you Bhaiya , I will asking once recession is over. Yes Bhaiya preparing and practising regularly. Your videos are so much helpful.

[@varunsheth1003]

Happy New Year !!!!! Nice Explanation

[@codestorywithMIK]

Thank you so much ❤️❤️❤️

[@user-vj3yw5nq7p]

Watching this video in New year 2024, happy new year :)

[@codestorywithMIK]

Happy new year! Thank you for watching 😇🙏❤️

[@girikgarg8]

Nice explanation

[@oqant0424]

bhaiya u r a legit🔥🔥 i mean seriously hope we grow together this upcoming year ... a lots of success is coming in 2023 for us✨✨

[@codestorywithMIK]

Thank you so much ❤️❤️ Yeah, let’s grow together in 2023 Together we all can do it 💪💪💪

[@devsharma181]

nice explanation

[@tauquirahmed1879]

Why the heck watching your explanation is so interesting? I can watch your playlist like a web series....Love from Kolkata.

[@codestorywithMIK]

Wowwww. 😅 Thank you so so much ❤️❤️❤️

[@souravjoshi2293]

True that 😆

[@oqant0424]

literally u r building great future coders! and i am one of them✌✌

[@codestorywithMIK]

We all will rock together soon 🙌💪💪❤️

[@Ramneet04]

Hey how to differentiate wheather we can apply dp in it or not??? Well with question I completed my 300 leetcode questions 💪💪❤ thx a lot you have given amazing Content,like explanation is just out of the box❤.

[@AnandKumar-kz3ls]

Happy New Year bhai hoping to cross 10K in this year

[@codestorywithMIK]

Thanks a lot Anand. ❤️❤️❤️ Means a lot

[@keertilata20]

you are the best!!!!!!!!!!!!!!!

[@codestorywithMIK]

Means a lot 😇🙏

[@thephalcon]

Continue sir

[@pritishpattnaik4674]

Fantastic explanation Wrote the code without seeing void dfs(int i, int j, int count, int nonObstacle, vector& grid, int &ans){ int m = grid.size(); int n = grid[0].size(); //boundary cases if (i < 0 or i >= m or j < 0 or j >= n or grid[i][j] == -1){ return; } if (grid[i][j] == 2){ if (count == nonObstacle){ ans++; } return; } //visit the block grid[i][j] = -1; //call dfs in all 4 directions vector dir = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}}; for (auto x : dir){ int new_i = i + x[0]; int new_j = j + x[1]; dfs(new_i, new_j, count+1, nonObstacle, grid, ans); } //backtrack grid[i][j] = 0; } int uniquePathsIII(vector& grid) { int m = grid.size(); int n = grid[0].size(); int nonObstacle = 0; int start_x, start_y; for (int i=0 ; i

[@codestorywithMIK]

Awesome ❤️❤️❤️

[@piyushacharya7696]

Mera daily comment streak baal baal xutjata.😂😂 On leetcode, I received my first badge. all thanks to you👏

[@codestorywithMIK]

Hi Piyush, i am so happy with the news that you got the Leetcode badge. So happy for you. Keep it up and i will also continue it. Feel free to share your badge on LinkedIn or with your colleagues and help others achieve it too. It will strengthen your skill too. Thanks again ❤️❤️❤️

[@AmandeepSingh-uq3wp]

Please sir kal se leetcode ke contest ke solutions explain karna start kariye. 🙏 Your explanations are really very good. 👍

[@codestorywithMIK]

Hi Amandeep, due to time crunch, it’s in Queue. But soon. Thanks for your appreciation ❤️❤️❤️

[@manishv.8167]

How Amazing You are !!!!!

[@codestorywithMIK]

Thanks a lot Manish ❤️❤️❤️ Stay tuned for video-6 of Graph Concepts today Thanks for watching always ❤️❤️❤️

[@manishv.8167]

@@codestorywithMIK thanks to restart it again

[@oqant0424]

u made a hard ques ssssssssssssooooooooooo easy!!!!!

[@codestorywithMIK]

Thank you so so much ❤️❤️❤️

[@Raj10185]

hii bhaiya one doubt how to check the question is of dp or backtracking since before this problem i am solving unique path 1 and 2 which is a dp problem and this is of backtraking can you give me some keywords so that i look at the question and then decide plz ? bcz once i know its backtraking i done simply by own how to find out which one of backtrcaking and which one of dp

[@udaytewary3809]

Hlo bhai see there is not any category of problem of dp the main thing is DP is used to optimise the recursion approach So if a problem is solvable using recursion then for sure DP will be applicable there And for backtracking part if we have been asked to try out all the possible ans and find the best one backtracking will be applicable

[@dileshchouhan9527]

bhaiya why are we not decrease count as well like when backtrack? i am confuse little bit pls explain :)

[@codestorywithMIK]

Since we pass count+1 and hence value of count is not changed for next iteration of for loop. But if you want it just like backtracking You can change the code like backtracking below - for() { count++ Call recursion count- }

[@svdfxd]

The explanation is awesome. I have a question - in this code, where is the condition to stop counting the paths i.e. at what point do we stop? @codestorywithMIK

[@hyperme1831]

We are checking for starting points and for every starting point we are making dfs calls.. If there is no 1 we will stop

[@hyperme1831]

For suppose if there is only one 1 there will be only 1 recursive call on that 1 internally.. if there are two 1 s in matrix then there will be 2 dfs recursive calls from 2 starting points

[@AnkushKumar-dd1jv]

Aap bihar se ho?

[@YashSinghal]

hny :)

[@codestorywithMIK]

🙌❤️

[@lakshsinghania]

sir can u pls tell where im doing wrong in this and why this code is not working class Solution { public: int f(int i, int j, vector& grid, int di[], int dj[], vector& vis) { int m = grid.size(); int n = grid[0].size(); if(i = n || grid[i][j] == -1 || vis[i][j]) return 0; // Base case if (grid[i][j] == 2) { for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { if (grid[i][j] == 0 && !vis[i][j]) { return 0; } } } return 1; } int paths = 0; for (int k = 0; k < 4; k++) { int new_row = i + di[k]; int new_col = j + dj[k]; if (new_row >= 0 && new_col >= 0 && new_row < m && new_col < n && !vis[new_row][new_col] && grid[new_row][new_col] == 0) { vis[i][j] = 1; paths += f(new_row, new_col, grid, di, dj, vis); vis[i][j] = 0; } } return paths; } int uniquePathsIII(vector& grid) { int m = grid.size(); int n = grid[0].size(); int di[] = {-1, 0, 1, 0}; int dj[] = {0, 1, 0, -1}; // Make the visit matrix vector vis(m, vector(n, 0)); for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { if (grid[i][j] == 1) { return f(i, j, grid, di, dj, vis); } } } return 0; // Return 0 if the ending position is not found. } }; pls sir help me ik backtracking but somewhat fail while implementing