Rotating the Box | Brute Force | Optimal | Leetcode 1861 | codestorywithMIK

[@jeehub041]

Jab se aapke videos dekhne lag gaya hu Consistent rahne lag gaya hu Coding me rum gaya hu Logics sab samajh gaya hu ❤❤🎉🎉 You are my real hero , bhaiya 🙌🙌

[@shikharai-tl4vb]

mik be like "or bss yaha mai pighal gaya hu "

[@codestorywithMIK]

This means a lot ❤️🙏😇

[@DrOnZeR2022]

Bhaiya I solved it by myself it similar to the question rotate image towards right first I rotate the image towards right by transpose and then reverse all the rows and then just add the effect of gravity❤

[@gui-codes]

MIK sir, I feel so good I was able to solve this on my own. My problem solving is improving day by day. Thank you so much

[@KhushnorRahmanMeem]

I never expected to solve it by my own. Bt I did alhamdulillah. Thank you

[@thekindspill]

Thank God I got to know about your channel. I learn everyday from you ❤

[@faizanmohammed7687]

I solved. Observed that transpose and reversing the array on my own. Then while simulating gravity I took a bit help of chatgpt. not completely. I was doing wrong updations on empty cell (`.`). But later I also got to know that simulating gravity would not really require to do rotation first, I could simply move the # to right. So it's new thing to learn.

[@Sarthak2421]

Instead of doing transpose and then reversing it , we can play around the index and it will do the job : int row = box.size(); int col = box[0].size(); vector rotated(col , vector (row)); //for copying elements for(int i = 0 ; i < row ; i++) { for(int j = 0 ; j < col ; j++) { rotated[j][row - i - 1] = box[i][j]; } }

[@gamersgame43]

finally solved the problem alone after 1hr, then saw video came

[@gui-codes]

Same bhai. I took me more than 1 hour but solved it finally.

[@peterfromengland8663]

​@@gui-codes it took me 19 minutes

[@tauheedahmed4073]

Me too bhai took 30 min

[@indiann871]

@@peterfromengland8663 10 min here

[@AmandeepSingh-uq3wp]

I also thought of same optimized approach and then I thought reversing row does not make any sense and it is just increasing the time expense of solution. So I use the gravity simulation logic first and then apply the transpose logic directly without reversing rows. Code (if someone wants to see the code) : class Solution { public: vector rotateTheBox(vector& box) { int n = box.size(); int m = box[0].size(); for (int i = 0; i < n; i++) { int end = m - 1; for (int start = m - 1; start >= 0; start--) { if (box[i][start] == '*') { end = start - 1; } else if (box[i][start] == '#') { box[i][start] = '.'; box[i][end] = '#'; end--; } } } vector ans(m, vector(n, '.')); for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { ans[j][n - i - 1] = box[i][j]; } } return ans; } };

[@ugcwithaddi]

Was waiting 😊. Thank you for the consistency

[@faizanmohammed7687]

I really want to start the jan 1 with good news, but I was thinking how. I am simultaneously developing projects and solving problems. I alotted times. If you have some suggestions, can you reccommend ?

[@tanishqagarg7582]

sir plz make a video what to do as a fresher from 3rd tier cllg . how to make our cv shine within months plz . i get a lot of rejections because my cv nevr gets selected and i think most of us is facing the same issue i trust that u can help us u r genuine thank u for the videos and everything in advance

[@codestorywithMIK]

Hello there ❤️. I understand how challenging it can be to stand out as a fresher. Your CV is the first step in making an impression, and I will definitely work on creating a detailed video to address this. Meanwhile, focus on adding relevant projects, building a strong LinkedIn profile, and highlighting any internships or certifications. Stay consistent, keep learning, and don’t lose hope

[@subasm2160]

No need to do transpose and reverse for the rotation just use res[j][m-i-1] = input [i][j], m = no of col

[@codestorywithMIK]

Indeed that’s a cleaner way. Good one 🙌♥️

[@subasm2160]

@codestorywithMIK Thanks man 🙌😊

[@aws_handles]

Thanks a lot for details

[@jain5184]

mik bhaiya done sir ek news ayega meri or se 1:00

[@codestorywithMIK]

Sure ❤️

[@peterfromengland8663]

Did it on my own 😢 But it seems easy problem

[@tutuimam3381]

Amazing ❤

[@dhirajmahato6500]

Plz upload binary tree concept video bro

[@er.jitu4]

My Approach 👇 class Solution { public: vector rotateTheBox(vector& box) { int m=box.size(); int n=box[0].size(); for(auto &row:box){ int cntEmpty=0; for(int j=n-1;j>=0;j--){ if(row[j]=='.')cntEmpty++; else if(row[j]=='*')cntEmpty=0; else{ if(cntEmpty==0)continue; row[j+cntEmpty]='#'; row[j]='.'; } } } vectorres; for(int col=0;col=0;row--){ temp.push_back(box[row][col]); } res.push_back(temp); } return res; } };

[@rickdutta942]

class Solution { public: vector rotateTheBox(vector& box) { int m = box.size(); // rows int n = box[0].size(); // cols vector ans(n, vector(m)); int j = m-1; // start from end column of ans matrix for (auto& row : box) { int r = n - 1; int l = n - 1; //Arrage the stones in each row of the given box for (l; l >= 0; l--) { if (row[l] == '*' || row[l] == '.' && row[r] != '.') r = l; if (row[l] == '#' && row[r] == '.') { swap(row[l], row[r]); r--; } } //First arranged row goes to last ans column int i = 0; for(char &c: row) { ans[i++][j]= c; } j--; } return ans; } }; TC : O(2(m x n)) SC: O(1) excluding output array

[@invincibleinvincible3215]

Mene to stack ka use karke isko solve kiya I don't know mene itna kyu socha.

[@codestorywithMIK]

Nothing wrong in that. Glad to see yet another approach. Feel free to share the approach with others. ❤️

[@anupamtiwary4265]

//In this I am shifting the stones first then rotating //this is very similar to move zeros to end void moveStones(vector& row) { int n = row.size(); int empty = n - 1; for (int i = n - 1; i >= 0; --i) { if (row[i] == '*') { // move the empty pointer if u get a wall empty = i - 1; } else if (row[i] == '#') { // move stone to the nearest empty spot swap(row[i], row[empty]); empty--; } // do nothing if it's empty space } } vector rotateTheBox(vector& box) { int m = box.size(),n=box[0].size(); for(auto &row : box){ moveStones(row); } vectorans(n,vector(m)); for(int i=0;i

[@harshlalawat]

Done with only single iteration class Solution { public: vector rotateTheBox(vector& box) { vector ans(box[0].size(), vector(box.size(), '.')); for(int i=box.size()-1; i>=0; --i){ for(int j=box[0].size()-1, count=j; j>=0; --j){ if(box[i][j] == '#') ans[count--][box.size()-i-1] = '#'; else if(box[i][j] == '*') count = j, ans[count--][box.size()-i-1] = '*'; } } return ans; } };

[@dayashankarlakhotia4943]

public char[][]rotateTheBox(char[][]box){ int m=box.length,n=box[0].lenght; char[][]ans=new char[n][m]; for(int i=0;i=0;--j){ ans[j][m-i-1]='.'; if(box[i][j]!='.'){ k=box[i][j]=='*'?j:k; ans[k--][m-i-1]=box[i][j]; } } return ans; } 🎉❤

[@KunalRajput-r2q5m]

I exracted each row and did Insertion sort with slight modification and then copied to the resultant matrix. Time Complexity (m * n) public void sort(char[]arr){ for(int i=1;i=0;j--){ if(arr[j]=='*')break; if(arr[j]=='#' && arr[j+1]=='.'){ arr[j+1]='#'; arr[j]='.'; } } }

[@dayashankarlakhotia4943]

First 🎉❤

[@AryanVats603]

sir,my solution vector rotateTheBox(vector& box) { int m=box.size(); int n=box[0].size(); vector rotated(n, vector(m)); for(int i=m-1;i>=0;i--){ for(int j=n-1;j>=0;j--){ if(box[i][j]=='#'){ int x=j+1; while(x

[@sarthakawasthi5686]

can someone please tell the error in my code !! Logic - Shifting stones to the right and then rotating the box class Solution { public: vector rotateTheBox(vector& box) { int m = box.size(); int n = box[0].size(); for(int i =0;i

[@rickdutta942]

Hard to understand what are you trying to do. If you can't figure the solution on your own, try taking solution from solution section and before you implement fully understand that code. Don't worry this kind of things happen with everyone.

[@sarthakawasthi5686]

@@rickdutta942 i am just shifting the stones to voids untill blockages is encountered and then after everything is done , i m rotating the box/ matrix

[@asifulhaquemridul7489]

I was able to solve this using stack. class Solution: def rotateTheBox(self, box: List[List[str]]) -> List[List[str]]: for i in range(len(box)): stack = [] for j in range(len(box[i])-1, -1, -1): if not stack and box[i][j] == ".": stack.append((i,j)) if stack and box[i][j] == "*": stack.pop() if stack and box[i][j] == "#": pos_i, pos_j = stack.pop() box[pos_i][pos_j] = "#" box[i][j] = "." stack.append((pos_i,pos_j-1)) res = [] for col in range(len(box[0])): new_col = [] for row in range(len(box)-1, -1, -1): new_col.append(box[row][col]) res.append(new_col) return res