Find Maximum 1's Row | GFG Solution | Searching and Sorting

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[@sujan_kumar_mitra]

GFG wants an O(N+M) solution. This solution is O(N*log(M)).
The O(N*log(M)) implementation in C++ is giving AC but in Java; it's is giving TLE

[@arshadmanxoori07]

public static int findRow(int[][] arr) {
int i = 0;
int j = arr[0].length - 1;

int ans = i;
while(i < arr.length && j >= 0){
if(arr[i][j] == 1){
ans = i;
j--;
}else{
i++;
}
}
return ans;
}

[@divyanshuchaudhari3257]

Thanks..but there is a much more optimized solution with O(m+n) time complexity.

[@danishrockz1]

this is O(n +m ) optimized approach tested of GFG > int rowWithMax1s(int matrix[][], int n, int m) {
int i = 0, j = m - 1, maxRow = - 1;
while(i < n && j >= 0)
{
if(matrix[i][j] == 1)
{
maxRow = i;
j--;
}
else
i++;
}
return maxRow;
}

[@arshadmanxoori07]

@@IamLucky0007 ya

[@mickyman753]

in our O(m+n) solution ,we will do something similar like count 1s question earlier in playlist ,if we get 1 then we can discard check on our column as max count wouln' be larger than our curr 1s count ,we move col-- ,if encounter a 0 then there's a possibility then row++ wali row mai 1 ho so we move down , although thisoltuion works in O(n+m) ,but when we have only 1 row in matrix and that too a large one with first one at very first ,we move O(n) but binarysearch sol finds it in O(logn)
public static int findRow(int[][]mat) {
int row = 0;
int col = mat[0].length-1;
int res = 0;
while(row=0){
if(mat[row][col]==1){
res = row;
col--;
}
else
row++;
}
return res;
}

[@Pepcoding]

You can consume the same content with doubt support on nados.pepcoding.com

[@sunnykakrani7830]

TLE de raha h ye to n*logn me b pass nhi ho rha h esa kyu madam ? esa kyu ??

[@danishrockz1]

check my above comment

[@sujan_kumar_mitra]

@@danishrockz1 Can't see your comment.

[@saniashaikh23b82]

maam plzz discuss optimised approach O

[@danishrockz1]

check my above comment for optimized sol