leftmost_column_with_at_least_a_one 

 1class BinaryMatrix:
 2    def get(self, row, col):
 3        pass
 4
 5    def dimensions(self):
 6        pass
 7
 8
 9# @leet start
10class Solution:
11    def leftMostColumnWithOne(self, binaryMatrix: "BinaryMatrix") -> int:
12        """
13        This question asks us to find the leftmost column that has at least
14        a one, and each row of the matrix is sorted in non-decreasing order.
15        For this, we can start out at the top right and then move to the bottom
16        left.
17
18        We know that if the curr_row, curr_col is a 1, we can always go left
19        and if it's a 0, we go down.
20        We then return the previous column that we saw.
21        This runs in $O(M+N)$ time.
22        """
23        m, n = binaryMatrix.dimensions()
24        curr_row, curr_col = 0, n - 1
25
26        while curr_row < m and curr_col >= 0:
27            if binaryMatrix.get(curr_row, curr_col) == 0:
28                curr_row += 1
29            else:
30                curr_col -= 1
31        return curr_col + 1 if curr_col != n - 1 else -1
32
33
34# @leet end
35
36
37def test():
38    assert 2 + 2 == 4
class BinaryMatrix: 
2class BinaryMatrix:
3    def get(self, row, col):
4        pass
5
6    def dimensions(self):
7        pass
def get(self, row, col): 
3    def get(self, row, col):
4        pass
def dimensions(self): 
6    def dimensions(self):
7        pass
class Solution: 
11class Solution:
12    def leftMostColumnWithOne(self, binaryMatrix: "BinaryMatrix") -> int:
13        """
14        This question asks us to find the leftmost column that has at least
15        a one, and each row of the matrix is sorted in non-decreasing order.
16        For this, we can start out at the top right and then move to the bottom
17        left.
18
19        We know that if the curr_row, curr_col is a 1, we can always go left
20        and if it's a 0, we go down.
21        We then return the previous column that we saw.
22        This runs in $O(M+N)$ time.
23        """
24        m, n = binaryMatrix.dimensions()
25        curr_row, curr_col = 0, n - 1
26
27        while curr_row < m and curr_col >= 0:
28            if binaryMatrix.get(curr_row, curr_col) == 0:
29                curr_row += 1
30            else:
31                curr_col -= 1
32        return curr_col + 1 if curr_col != n - 1 else -1
def leftMostColumnWithOne( self, binaryMatrix: BinaryMatrix) -> int: 
12    def leftMostColumnWithOne(self, binaryMatrix: "BinaryMatrix") -> int:
13        """
14        This question asks us to find the leftmost column that has at least
15        a one, and each row of the matrix is sorted in non-decreasing order.
16        For this, we can start out at the top right and then move to the bottom
17        left.
18
19        We know that if the curr_row, curr_col is a 1, we can always go left
20        and if it's a 0, we go down.
21        We then return the previous column that we saw.
22        This runs in $O(M+N)$ time.
23        """
24        m, n = binaryMatrix.dimensions()
25        curr_row, curr_col = 0, n - 1
26
27        while curr_row < m and curr_col >= 0:
28            if binaryMatrix.get(curr_row, curr_col) == 0:
29                curr_row += 1
30            else:
31                curr_col -= 1
32        return curr_col + 1 if curr_col != n - 1 else -1

This question asks us to find the leftmost column that has at least a one, and each row of the matrix is sorted in non-decreasing order. For this, we can start out at the top right and then move to the bottom left.

We know that if the curr_row, curr_col is a 1, we can always go left and if it's a 0, we go down. We then return the previous column that we saw. This runs in $O(M+N)$ time.

def test(): 
38def test():
39    assert 2 + 2 == 4