Lucky Numbers in a Matrix | 2 Approaches | Dry Run | Leetcode 1380 | codestorywithMIK

[@biswarupacharjya2258]

The Best thing of this video is You expln Approach 1 to understand the logic but also expln the Approach 2 to optimized this problem this proves your dedication level for us . thank you

[@codestorywithMIK]

❤️❤️🙏🙏

[@abhirajsingh3071]

Loved it, did not expected that someone will explain this so smoothly, thanks

[@gui-codes]

I solved it on my own. Thanks a lot mik for helping me being consistent.

[@ganeshjaggineni4097]

NICE SUPER EXCELLENT MOTIVATED

[@jeehub041]

Sir aapke explanations gajab ha, aap Great ho agar kisi ko coding me interest na ho aur voh ek baar aapka teaching delh le 🙌🙌

[@nupurkuhite4967]

Sir, Thanks jo aap ka 2nd approach hai vo que padhane ke baad mujhe bhi ye idea aya tha lekin mene shrif rowMin ka Max value nikala tha, colMax ka Min nahi islye answer wrong aya tha. Thanks to resolve my doubt.

[@anuragsingh2556]

sir i have one doubt i wish you reply to this comment when we add answer in our result you used this condition if (matrix[row][col] == rowMin.get(row) && matrix[row][col] == colMax.get(col)) { result.add(matrix[row][col]); } but i used if(matrix[i][j] == min.get(i) && min.get(i) == max.get(j)) ans.add(min.get(i)); } how these both condition are different

[@RohitKumar-dz8dh]

Thanks 😊

[@ryuu5768]

class Solution { public: pairfindmin(vector& matrix,int &i){ int mini=INT_MAX; int index=-1; for(int j=0;jmatrix[i][j]){ index=j; mini=matrix[i][j]; } } return {mini,index}; } bool ismaxincol(vector& matrix,int maxi,int j){ for(int i=0;i

[@_A__Mohit]

Please segment tree videos sir please 😢 🙏🙏🙏🙏

[@nesh398]

Without using 2 seperate arrays use MAP class Solution { public: vector luckyNumbers(vector& matrix) { int n = matrix.size(); int m = matrix[0].size(); unordered_map holder; vector ans; for (int i = 0; i < n; i++) { int mini = INT_MAX; for (int j = 0; j < m; j++) { mini = min(mini, matrix[i][j]); } holder[mini]++; } for (int i = 0; i < m; i++) { int maxi = INT_MIN; for (int j = 0; j < n; j++) { maxi = max(maxi, matrix[j][i]); } holder[maxi]++; } for (auto elem : holder) { if (elem.second == 2) { return {elem.first}; } } return 0; } };

[@codestorywithMIK]

Very nice ❤️

[@gauravbanerjee2898]

Thanks a lot bhaiya ❤❤

[@thefinalfit]

Done on my own 😊

[@utkarshsahay9908]

class Solution { public: vector luckyNumbers (vector& matrix) { unordered_map mp; vector ans; for(int i = 0;i < matrix.size(); i++){ int mini = *min_element(begin(matrix[i]),end(matrix[i])); int idx = min_element(begin(matrix[i]),end(matrix[i])) - begin(matrix[i]); mp[mini] = idx; } for(auto it : mp){ int ele = it.first; int col = it.second; int maxi = INT_MIN; for(int row = 0;row < matrix.size(); row++){ maxi = max(maxi,matrix[row][col]); } if(maxi == ele) ans.push_back(ele); } return ans; } };❤❤

[@YashMalav-kh1ov]

first viewer...🙂

[@reyazahmed4855]

Dfs approach for number of good leaves yesterday problem please upload

[@codestorywithMIK]

Yes, planning this soon by tomorrow

[@codestorywithMIK]

Coming today only 🙂❤️

[@reyazahmed4855]

​@@codestorywithMIK Thanks buddy. Hope you doing great. Awesome channel

[@j_ourney25]

Can we write in if condition if(rowmin[row] == colmax[col]{ }

[@codestorywithMIK]

Definitely yes 👍🏻

[@nawazthezaifre8870]

Array + Matrix = dimag ghum jata hai konsa loop kha chlega solution dekh lo samjh aa jata hai nhi to solve nhi hote is type ke question.

[@singhshek58]

vector luckyNumbers (vector& matrix) { int n=matrix.size(); int m=matrix[0].size(); unordered_setst; for(int i=0;i

[@ryuu5768]

class Solution { public: pairfindmin(vector& matrix,int &i){ int mini=INT_MAX; int index=-1; for(int j=0;jmatrix[i][j]){ index=j; mini=matrix[i][j]; } } return {mini,index}; } bool ismaxincol(vector& matrix,int maxi,int j){ for(int i=0;i

[@TheBusaSinger]

❤❤

[@maniBrave]

java wale kitne h wsese

[@aggarwalsachin4854]

ye contradiction wala logic kaise aya bro dimaag m???

[@codestorywithMIK]

Hi there, It was a hit and trial initially. I also realised for any example i was always getting one Lucky number. And hence came to this. Hope it helped 😊❤️🙏

[@shabananoor9423]

❤❤❤