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i would suggest to type the code instead, like u did yesterday, it may take more time, but it's better...

can you solve this question without using hashmap

bro don't make such wasteless videos if you don't completely understand what really the problem is also gfg is fu** up with this problem

I requested to explain quetion fully. I don't know where i missed I didn't understand the question also

Did you even understand the P.S. ? Please don't do this for views by copying the code from comments and explaining it.

Keep it up brother

Hangup (SIGHUP) Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: Index 41 out of bounds for length 41 at GFG.main(GFG.java:18) I am getting this, can you please check yours, I have reported this issue, even with your code this is coming for a very big test case

Hi u were stammering a lot can u pls speak in hindi

great explanation bhaiya

Kal se kuch n kuch problem aari hai geeksforgeeks mai code bhe sahi hota hai, fir bhe problm aari hai..

what if there are more than two elements of frequency more than n/3 ?

they changed the question now, TC etc and my marks are reduced now as well, 0 marks as it's saying I viewed the solution before successful submit jabki kal sab sahi se submit ho gaya tha

thanks man, you cleared my doubt

yeah mera bhi same issue tha, weird question hai, and output expected output sab sahi tha,

nice explanation sir ji

Thank you so much brother

nice

Thank you!

Yr english ni boli jati to ku ghaseetni h

awesome explanation!

poor explanation skills. Approach is good but explanation is worst!

sir the solution code is not opening.can u please add another link.

sir pls use java language

Can't we do this ???: void sort012(vector<int>& arr) { sort(arr.begin(),arr.end()); }

the code in description is not opening . can u please recheck it.

Do you solve POTD everyday?

very well explained brother!! Thanks!!

why this code is giving segmentation fault? class Solution { public: pair<int, bool> solve(vector<vector<pair<int, bool>>>& dp, vector<vector<int>>& grid, int i, int j) { if (i < 0 || j < 0 || i >= grid.size() || j >= grid[0].size()) return {INT_MAX, false}; if (dp[i][j].first != -1) return dp[i][j]; int res = INT_MAX; bool flag = false; auto p = solve(dp, grid, i + 1, j); if (p.second) { res = min(res, p.first); flag = true; } p = solve(dp, grid, i - 1, j); if (p.second) { res = min(res, p.first); flag = true; } p = solve(dp, grid, i, j - 1); if (p.second) { res = min(res, p.first); flag = true; } p = solve(dp, grid, i, j + 1); if (p.second) { res = min(res, p.first); flag = true; } if (!flag) { dp[i][j] = {INT_MAX, false}; } else { dp[i][j] = {grid[i][j] + res, true}; } return dp[i][j]; } int minimumCostPath(vector<vector<int>>& grid) { int n = grid.size(); int m = grid[0].size(); vector<vector<pair<int, bool>>> dp(n, vector<pair<int, bool>>(m, {-1, false})); dp[0][0] = {grid[0][0], true}; auto p = solve(dp, grid, n - 1, m - 1); return p.second ? p.first : -1; } };

Sir please java me baniye pl😢ease please please please please please please please please please please please please

why not dp my code not working on case 1114 class Solution { //Function to return the minimum cost to react at bottom //right cell from top left cell. public int minimumCostPath(int[][] grid) { // Code here int row = grid.length; int col = grid[0].length; for(int j =1;j<col;j++) grid[0][j] = grid[0][j]+grid[0][j-1]; for(int i =1;i<row;i++) grid[i][0] = grid[i][0]+grid[i-1][0]; for(int i=1;i<row;i++){ for(int j =1;j<col;j++){ grid[i][j] = grid[i][j] + Math.min(grid[i][j-1],grid[i-1][j]); } } return grid[row-1][col-1]; } }

Thank you. I coded myself after understanding your approach.

class Solution{ void convertToString(vector<vector<char>> &a,vector<vector<int>> &ans,int n){ //storing col vector<int> temp; for(int i=0;i<n;i++){ for(int j=0;j<n;j++){ if(a[i][j]=='Q') temp.push_back(j+1); } } ans.push_back(temp); } bool isSafe(int row,int col,int n,vector<vector<char>> &a){ //same col for(int i=0;i<row;i++){ if(a[i][col]=='Q') return false; } //top left for(int i=row-1, j=col-1;i>=0 && j>=0;i--, j--){ if(a[i][j]=='Q') return false; } //top right for(int i=row-1, j=col+1;i>=0 && j<n;i--, j++){ if(a[i][j]=='Q') return false; } return true; } void solve(int i,int n,vector<vector<char>> &a,vector<vector<int>> &ans){ if(i==n){ convertToString(a,ans,n); return; } for(int j=0;j<n;j++){ if(isSafe(i,j,n,a)){ a[i][j]='Q'; solve(i+1,n,a,ans); a[i][j]='.'; } } } public: vector<vector<int>> nQueen(int n) { vector<vector<char>> a(n, vector<char>(n,'.')); vector<vector<int>> ans; solve(0,n,a,ans); return ans; } }; thank you !

The code in Java is also same

Batsman can also take 5 runs by running between the wicket on last ball?

tbh knowledge or Approach was Good but explanation or Intuition is not good keep it up👍

Amazing approach bro .. Keep it up , thanks for explanation

solution ka link provide kar diya Kro

Nice

thankyou so much sir

Keep Hard working bro your time will also come

Bhaiya continue ur videos in English only.. Thank you

Nice

just apply the patter binary search on answers , this is exactly pased on same pattern check my code but it needss few modifications ,plzz correct me class Solution { public: bool check (int mid,int n,int k){ string h = to_string(mid); if(h.size()>n || h.size()<n){ return false; } if(h.size()==n){ if(mid%k>=0){ string r =h; reverse (r.begin(),r.end()); return h==r; } } return false; } string largestPalindrome(int n, int k) { int i=1; int j = 1000000; int ans =-1; while(i<=j){ int mid = (i+j)/2; if(check(mid,n,k)){ ans = j; i= mid+1; } else{ j=mid-1; } } string ok = to_string(ans); return ok; } };

good explanation

bhai bahut hi ganda samjha raha hai... time waste mat kar... mera b 5min waste karwa diye tune...

Why do we need to do b[0] - a[0] becuase priority queue naturally stores the data in max heap format.

Why only top 15 values work fine?

In boolean function, how you are returning integer value to a boolean function? plz copy carefully before uploading the video

🤔 hmmm

Love from Deepanshu bro🫶🏻