Search a 2D matrix (LeetCode 74) | Full solution with visual examples and proof

[@KeerthiKothapalli-g1k]

Your teaching was just amazing .Please continue these Problem Solving sessions

[@conquerorsanthosh]

Very clearly explained. Keep it up brother

[@S8naK]

Thank you so much ❤

[@mohammedharoon1167]

saved my day nice explanation 👍

[@nikoo28]

Glad it helped!

[@pragyanborthakur1333]

Great explanation brother

[@ssss22144]

good annaa

[@ChristForAll-143]

CAN YOU PLEASE CREATE A PLAYLIST FOR INTERVAL PROBLEMS - I DON'T FIND ANY U TUBER DOES THAT

[@nikoo28]

once I have a bunch of problems on intervals, I will put them in a playlist for sure

[@ChristForAll-143]

@@nikoo28 Thanks a lot brother, if you want any questions will share here?

[@AlexTopGT]

can anyone explain why this solution is not O(log m) + O(log n)?

[@sharaabsingh]

This is the law of logarithms.
Log(A) + Log(B) = log(A*B)

[@nikoo28]

Perfect answer

[@ruchiswish]

Say you have 8 in place of 14 and you are searching for 9 … can you explain the algorithm?

[@khushaldodeja8267]

getting time limit exceeded

[@khushaldodeja8267]

Code I have written
class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
int i = 0, j = matrix.length-1;
int mid;
int potentialRow=-1;
while(imatrix[mid+1][0]){
i = mid+1;
}
else if(target

[@nikoo28]

look at the code on github, it is available in video description

[@pravinrathod1676]

var searchMatrix = function(matrix, target) {
if (!matrix.length || !matrix[0].length) return false;
let row = 0;
let col = matrix[0].length - 1;
while (col >= 0 && row target) col--;
else if (matrix[row][col] < target) row++;
}
return false;
};

[@kolawoleabdulrahman]

cool look. great video. lets get started😆

[@nikoo28]

hahaha... 😄

[@codinghero2001]

Your teaching was just amazing .Please continue these Problem Solving sessions