Unique Path problem with Robot Traversal in python
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Unique Path problem with Robot Traversal in python
0
$begingroup$
I got this problem during a mock interview, and I would like to get code review for the backtracking solution.
Robot Paths
from Leetcode https://leetcode.com/problems/unique-paths/description/
Prompt: Given a matrix of zeroes, determine how many unique paths exist
from the top left corner to the bottom right corner
#
Input: An Array of Array of Integers (matrix)
Output: Integer
#
# Example: matrix = [[0,0,0,0],
# [0,0,0,0],
# [0,0,0,0]]
#
# robotPaths(matrix) = 38
#
#
# matrix = [[0,0,0],
# [0,0,0]]
#
# robotPaths(matrix) = 4
# Note: From any point, you can travel in the four cardinal directions. I decided to do backtracking approach to solve this problem.
# (north, south, east, west). A path is valid as long as it travels
# from the top left corner to the bottom right corner, does not go
# off of the matrix, and does not travel back on itself
def robot_paths(matrix):
num_of_rows = len(matrix)
num_of_cols = len(matrix[0])
def traverse(row, col):
nonlocal num_of_rows
nonlocal num_of_cols
# is row and col inside the boundaries
if row < 0 or row >= num_of_rows or col < 0 or col >= num_of_cols:
return 0
# has row, col already been visited
if matrix[row][col] == 1:
return 0
# is row, col the destination?
if row == num_of_rows - 1 and col == num_of_cols - 1:
return 1
# mark coordinate as visited
matrix[row][col] = 1
# initialize sum of total unique paths to end from that coordinate
s = traverse(row, col + 1) + traverse(row + 1, col) + traverse(row - 1, col) + traverse(row, col - 1)
# backtrack; mark coordinate as unvisited so it can be
# used in another path
matrix[row][col] = 0
return s
return traverse(0, 0)
backtracking
asked 8 mins ago
NinjaGNinjaG
825632
$endgroup$
add a comment |
0
$begingroup$
I got this problem during a mock interview, and I would like to get code review for the backtracking solution.
Robot Paths
from Leetcode https://leetcode.com/problems/unique-paths/description/
Prompt: Given a matrix of zeroes, determine how many unique paths exist
from the top left corner to the bottom right corner
#
Input: An Array of Array of Integers (matrix)
Output: Integer
#
# Example: matrix = [[0,0,0,0],
# [0,0,0,0],
# [0,0,0,0]]
#
# robotPaths(matrix) = 38
#
#
# matrix = [[0,0,0],
# [0,0,0]]
#
# robotPaths(matrix) = 4
# Note: From any point, you can travel in the four cardinal directions. I decided to do backtracking approach to solve this problem.
# (north, south, east, west). A path is valid as long as it travels
# from the top left corner to the bottom right corner, does not go
# off of the matrix, and does not travel back on itself
def robot_paths(matrix):
num_of_rows = len(matrix)
num_of_cols = len(matrix[0])
def traverse(row, col):
nonlocal num_of_rows
nonlocal num_of_cols
# is row and col inside the boundaries
if row < 0 or row >= num_of_rows or col < 0 or col >= num_of_cols:
return 0
# has row, col already been visited
if matrix[row][col] == 1:
return 0
# is row, col the destination?
if row == num_of_rows - 1 and col == num_of_cols - 1:
return 1
# mark coordinate as visited
matrix[row][col] = 1
# initialize sum of total unique paths to end from that coordinate
s = traverse(row, col + 1) + traverse(row + 1, col) + traverse(row - 1, col) + traverse(row, col - 1)
# backtrack; mark coordinate as unvisited so it can be
# used in another path
matrix[row][col] = 0
return s
return traverse(0, 0)
backtracking
asked 8 mins ago
NinjaGNinjaG
825632
$endgroup$
add a comment |
0
0
0
$begingroup$
I got this problem during a mock interview, and I would like to get code review for the backtracking solution.
Robot Paths
from Leetcode https://leetcode.com/problems/unique-paths/description/
Prompt: Given a matrix of zeroes, determine how many unique paths exist
from the top left corner to the bottom right corner
#
Input: An Array of Array of Integers (matrix)
Output: Integer
#
# Example: matrix = [[0,0,0,0],
# [0,0,0,0],
# [0,0,0,0]]
#
# robotPaths(matrix) = 38
#
#
# matrix = [[0,0,0],
# [0,0,0]]
#
# robotPaths(matrix) = 4
# Note: From any point, you can travel in the four cardinal directions. I decided to do backtracking approach to solve this problem.
# (north, south, east, west). A path is valid as long as it travels
# from the top left corner to the bottom right corner, does not go
# off of the matrix, and does not travel back on itself
def robot_paths(matrix):
num_of_rows = len(matrix)
num_of_cols = len(matrix[0])
def traverse(row, col):
nonlocal num_of_rows
nonlocal num_of_cols
# is row and col inside the boundaries
if row < 0 or row >= num_of_rows or col < 0 or col >= num_of_cols:
return 0
# has row, col already been visited
if matrix[row][col] == 1:
return 0
# is row, col the destination?
if row == num_of_rows - 1 and col == num_of_cols - 1:
return 1
# mark coordinate as visited
matrix[row][col] = 1
# initialize sum of total unique paths to end from that coordinate
s = traverse(row, col + 1) + traverse(row + 1, col) + traverse(row - 1, col) + traverse(row, col - 1)
# backtrack; mark coordinate as unvisited so it can be
# used in another path
matrix[row][col] = 0
return s
return traverse(0, 0)
backtracking
asked 8 mins ago
NinjaGNinjaG
825632
$endgroup$
I got this problem during a mock interview, and I would like to get code review for the backtracking solution.
Robot Paths
from Leetcode https://leetcode.com/problems/unique-paths/description/
Prompt: Given a matrix of zeroes, determine how many unique paths exist
from the top left corner to the bottom right corner
#
Input: An Array of Array of Integers (matrix)
Output: Integer
#
# Example: matrix = [[0,0,0,0],
# [0,0,0,0],
# [0,0,0,0]]
#
# robotPaths(matrix) = 38
#
#
# matrix = [[0,0,0],
# [0,0,0]]
#
# robotPaths(matrix) = 4
# Note: From any point, you can travel in the four cardinal directions. I decided to do backtracking approach to solve this problem.
# (north, south, east, west). A path is valid as long as it travels
# from the top left corner to the bottom right corner, does not go
# off of the matrix, and does not travel back on itself
def robot_paths(matrix):
num_of_rows = len(matrix)
num_of_cols = len(matrix[0])
def traverse(row, col):
nonlocal num_of_rows
nonlocal num_of_cols
# is row and col inside the boundaries
if row < 0 or row >= num_of_rows or col < 0 or col >= num_of_cols:
return 0
# has row, col already been visited
if matrix[row][col] == 1:
return 0
# is row, col the destination?
if row == num_of_rows - 1 and col == num_of_cols - 1:
return 1
# mark coordinate as visited
matrix[row][col] = 1
# initialize sum of total unique paths to end from that coordinate
s = traverse(row, col + 1) + traverse(row + 1, col) + traverse(row - 1, col) + traverse(row, col - 1)
# backtrack; mark coordinate as unvisited so it can be
# used in another path
matrix[row][col] = 0
return s
return traverse(0, 0)
backtracking
backtracking
asked 8 mins ago
NinjaGNinjaG
825632
asked 8 mins ago
NinjaGNinjaG
825632
asked 8 mins ago
NinjaGNinjaG
825632
asked 8 mins ago
NinjaGNinjaG
825632
asked 8 mins ago
NinjaGNinjaG
825632
825632
add a comment |
add a comment |
0
active
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