Which row has the most 1s in a 0-1 matrix with all 1s "on the left"? - algorithm

Question

Problem
Each row of an n x n matrix consists of 1's and 0's such that in any row, all 1's come before any 0's. Find row containing most no of 1's in O(n).
Example
1 1 1 1 1 0 <- Contains maximum number of 1s, return index 1
1 1 1 0 0 0
1 0 0 0 0 0
1 1 1 1 0 0
1 1 1 1 0 0
1 1 0 0 0 0
I found this question in my algorithms book. The best I could do took O(n logn) time.
How to do this in O(n)?

Answer 1

Start at 1,1.
If the cell contains 1, you're on the longest row so far; write it down and go right.
If the cell contains 0, go down.
If the cell is out of bounds, you're done.

Answer 2

You can do it in O(N) as follows:
Start at A[i][j] with i=j=0.
1, keep moving to the right by doing j++
A[i][j] =
0, move down to the next row by doing i++
When you reach the last row or the last column, the value of j will be the answer.
Pseudo code:
Let R be number of rows
Let C be number of columns
Let i = 0
Let j = 0
Let max1Row = 0
while ( i<R && j<C )
if ( matrix[i][j] == 1 )
j++
max1Row = i
else
i++
end-while
print "Max 1's = j"
print "Row number with max 1's = max1Row"

Answer 3

Start with the first row. Keep the row R that has the most numbers of 1s and the index i of the last 1 of R. in each iteration compare the current row with the row R on the index i. if the current row has a 0 on position i, the row R is still the answer.
Otherwise, return the index of the current row. Now we just have to find the last 1 of the current row. Iterate from index i up to the last 1 of the current row, set R to this row and i to this new index.
i
|
v
R-> 1 1 1 1 1 0
|
v 1 1 1 0 0 0 (Compare ith index of this row)
1 0 0 0 0 0 Repeat
1 1 1 1 0 0 "
1 1 1 1 0 0 "
1 1 0 0 0 0 "

Answer 4

Some C code to do this.
int n = 6;
int maxones = 0, maxrow = -1, row = 0, col = 0;
while(row < n) {
while(col < n && matrix[row][col] == 1) col++;
if(col == n) return row;
if(col > maxones){
maxrow = row;
maxones = col;
}
row++;
}

Answer 5

int [] getMax1withRow(int [][] matrix){
int [] result=new int[2];
int rows=matrix.length;
int cols=matrix[0].length;
int i=0, j=0;
int max_row=0;// This will show row with maximum 1. Intialing pointing to 0th row.
int max_one=0;// max one
while(i< rows){
while(matrix[i][j]==1){
j++;
}
if(j==n){
result[0]=n;
result[1]=i;
return result;
}
if(j>max_one){
max_one=j;
max_row=i;
}
j=0;// Again start from the first column
i++;// increase row number
}
result[0]=max_one;
result[1]=max_row;
return result;
}
Time complexity => O(row+col), In worse case If every row has n-1 one except last row which have n 1s then we have be travel till last row.