Difficulty:
Medium
Points: 4
Given a matrix of size n x m with 0’s and 1’s , you enter the matrix at cell (0,0) in left to right direction. Whenever you encounter a 0 you retain it in the same direction , else if you encounter a 1 you have to change the direction to the right of the current direction and change that 1 value to 0 , you have to find out from which index you will leave the matrix at the end.
Example 1:
Input:
n = 3, m = 3**
** matrix = {{0, 1, 0},
{0, 1, 1},
{0, 0, 0}}
Output:
{1, 0}
Explanation:
Enter the matrix at (0, 0)
-> then move towards (0, 1) -> 1 is encountered
-> turn right towards (1, 1) -> again 1 is encountered
-> turn right again towards (1, 0)
-> now, the boundary of matrix will be crossed ->hence, exit point reached at 1, 0..
Example 2:
Input:
n = 1, m = 2**
** matrix = {{0, 0}}
Output:
{0, 1}
Explanation:
Enter the matrix at cell (0, 0).
Since the cell contains a 0, we continue moving in the same direction.
We reach cell (0, 1), which also contains a 0. So, we continue moving in the same direction, we exit the matrix from cell (0, 1).
Your Task:
You don't need to read or print anything. Your task is to complete the function FindExitPoint() which takes the matrix as an input parameter and returns a list containing the exit point.
Expected Time Complexity: O(n * m) where n = number of rows and m = number of columns.
Expected Space Complexity: O(1)
Constraints:
1 <= n, m <= 100
Samsung
Matrix Data Structures