Date: 01 September, 2023
Model: codellama-34b-instruct
Temperature: 0.2
Passing runs: 0
Total runs: 10
Success rate: 0%
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""The above is an incomplete Python code fragment. Return the complete and correct code with no additional text.
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""n = len(grid)
val = n * n + 1
for i in range(n):
for j in range(n):
if grid[i][j] == 1:
temp = []
if i != 0:
temp.append(grid[i - 1][j])
if j != 0:
temp.append(grid[i][j - 1])
if i != n - 1:
temp.append(grid[i + 1][j])
if j != n - 1:
temp.append(grid[i][j + 1])
val = min(temp)
ans = []
for i in range(k):
if i % 2 == 0:
ans.append(1)
else:
ans.append(val)
return ansdef check(candidate):
# Check some simple cases
print
assert candidate([[1, 2, 3], [4, 5, 6], [7, 8, 9]], 3) == [1, 2, 1]
assert candidate([[5, 9, 3], [4, 1, 6], [7, 8, 2]], 1) == [1]
assert candidate([[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]], 4) == [1, 2, 1, 2]
assert candidate([[6, 4, 13, 10], [5, 7, 12, 1], [3, 16, 11, 15], [8, 14, 9, 2]], 7) == [1, 10, 1, 10, 1, 10, 1]
assert candidate([[8, 14, 9, 2], [6, 4, 13, 15], [5, 7, 1, 12], [3, 10, 11, 16]], 5) == [1, 7, 1, 7, 1]
assert candidate([[11, 8, 7, 2], [5, 16, 14, 4], [9, 3, 15, 6], [12, 13, 10, 1]], 9) == [1, 6, 1, 6, 1, 6, 1, 6, 1]
assert candidate([[12, 13, 10, 1], [9, 3, 15, 6], [5, 16, 14, 4], [11, 8, 7, 2]], 12) == [1, 6, 1, 6, 1, 6, 1, 6, 1, 6, 1, 6]
assert candidate([[2, 7, 4], [3, 1, 5], [6, 8, 9]], 8) == [1, 3, 1, 3, 1, 3, 1, 3]
assert candidate([[6, 1, 5], [3, 8, 9], [2, 7, 4]], 8) == [1, 5, 1, 5, 1, 5, 1, 5]
# Check some edge cases that are easy to work out by hand.
assert candidate([[1, 2], [3, 4]], 10) == [1, 2, 1, 2, 1, 2, 1, 2, 1, 2]
assert candidate([[1, 3], [3, 2]], 10) == [1, 3, 1, 3, 1, 3, 1, 3, 1, 3]Start time: 2023-09-01 02:46:30
Duration: 25.93s
Result: Fail
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
n = len(grid)
m = len(grid[0])
dp = [[float('inf')] * (k+1) for _ in range(n)]
dp[0][0] = 0
for i in range(1, n):
for j in range(1, m):
for l in range(k+1):
if dp[i-1][l] != float('inf'):
dp[i][j] = min(dp[i][j], dp[i-1][l] + grid[i][j])
return [grid[i][j] for i in range(n) for j in range(m) if dp[i][j] == dp[n-1][k]]Start time: 2023-09-01 02:49:26
Duration: 42.36s
Result: Fail
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
# Initialize the minimum path length to infinity
min_path_length = float('inf')
# Initialize the minimum path to None
min_path = None
# Loop through each cell in the grid
for i in range(len(grid)):
for j in range(len(grid[i])):
# If the current cell is the starting cell, set the current path length to 0
if grid[i][j] == 1:
current_path_length = 0
# If the current cell is not the starting cell, set the current path length to the minimum path length so far
else:
current_path_length = min_path_length
# Loop through each neighbor of the current cell
for neighbor in get_neighbors(grid, i, j):
# If the neighbor is not the starting cell and has not been visited before, add it to the current path
if neighbor != 1 and neighbor not in min_path:
current_path_length += 1
min_path.append(neighbor)
# If the current path length is less than the minimum path length, update the minimum path length and minimum path
if current_path_length < min_path_length:
min_path_length = current_path_length
min_path = min_path[:]
# Return the minimum path
return min_path
def get_neighbors(grid, i, j):
# Initialize an empty list to store the neighbors
neighbors = []
# Check if the cell to the north is valid and has not been visited before
if i > 0 and grid[i-1][j] != 1 and grid[i-1][j] not in neighbors:
neighbors.append(grid[i-1][j])
# Check if the cell to the south is valid and has not been visited before
if i < len(grid) - 1 and grid[i+1][j] != 1 and grid[i+1][j] not in neighbors:
neighbors.append(grid[i+1][j])
# Check if the cell to the west is valid and has not been visited before
if j > 0 and grid[i][j-1] != 1 and grid[i][j-1] not in neighbors:
neighbors.append(grid[i][j-1])
# Check if the cell to the east is valid and has not been visited before
if j < len(grid[i]) - 1 and grid[i][j+1] != 1 and grid[i][j+1] not in neighbors:
neighbors.append(grid[i][j+1])
# Return the list of neighbors
return neighborsStart time: 2023-09-01 02:52:25
Duration: 24.71s
Result: Fail
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
n = len(grid)
m = len(grid[0])
dp = [[float('inf')] * (k+1) for _ in range(n)]
dp[0][0] = 0
for i in range(1, n):
for j in range(1, m):
for l in range(k+1):
if dp[i-1][l] != float('inf'):
dp[i][j] = min(dp[i][j], dp[i-1][l] + grid[i][j])
return [grid[i][j] for i in range(n) for j in range(m) if dp[i][j] == dp[n-1][k]]Start time: 2023-09-01 02:55:18
Duration: 46.55s
Result: Fail
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
# Initialize the minimum path length
min_path_len = float('inf')
# Initialize the minimum path
min_path = []
# Loop through each cell in the grid
for i in range(len(grid)):
for j in range(len(grid[i])):
# Check if the current cell is the starting cell
if grid[i][j] == 1:
# Initialize the current path
curr_path = [1]
# Initialize the current path length
curr_path_len = 1
# Recursively find the minimum path
min_path, min_path_len = find_min_path(grid, i, j, k, curr_path, curr_path_len, min_path_len)
return min_path
def find_min_path(grid, i, j, k, curr_path, curr_path_len, min_path_len):
# Base case: If the current path length is greater than the minimum path length, return the minimum path
if curr_path_len > min_path_len:
return min_path, min_path_len
# Base case: If the current path length is equal to the minimum path length, return the minimum path
if curr_path_len == min_path_len:
return curr_path, min_path_len
# Recursive case: If the current cell is not the starting cell, explore its neighbors
if grid[i][j] != 1:
# Loop through each neighbor of the current cell
for neighbor in get_neighbors(grid, i, j):
# Check if the neighbor is not the starting cell and has not been visited before
if neighbor not in curr_path and neighbor != 1:
# Add the neighbor to the current path
curr_path.append(neighbor)
# Increment the current path length
curr_path_len += 1
# Recursively find the minimum path
min_path, min_path_len = find_min_path(grid, neighbor[0], neighbor[1], k, curr_path, curr_path_len, min_path_len)
# Remove the neighbor from the current path
curr_path.pop()
# Decrement the current path length
curr_path_len -= 1
return min_path, min_path_len
def get_neighbors(grid, i, j):
# Initialize the list of neighbors
neighbors = []
# Check if the cell to the north is valid
if i > 0:
neighbors.append((i-1, j))
# Check if the cell to the south is valid
if i < len(grid) - 1:
neighbors.append((i+1, j))
# Check if the cell to the west is valid
if j > 0:
neighbors.append((i, j-1))
# Check if the cell to the east is valid
if j < len(grid[i]) - 1:
neighbors.append((i, j+1))
return neighborsStart time: 2023-09-01 02:58:18
Duration: 34.43s
Result: Fail
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
# Initialize the minimum path length to infinity
min_path_length = float('inf')
# Initialize the minimum path to None
min_path = None
# Loop through each cell in the grid
for i in range(len(grid)):
for j in range(len(grid[i])):
# If the current cell is the starting cell, set the current path length to 0
if grid[i][j] == 1:
current_path_length = 0
# If the current cell is not the starting cell, set the current path length to the minimum path length so far
else:
current_path_length = min_path_length
# Loop through each neighbor of the current cell
for neighbor in get_neighbors(grid, i, j):
# If the neighbor is not the starting cell, add the value of the neighbor to the current path length
if neighbor != 1:
current_path_length += grid[neighbor[0]][neighbor[1]]
# If the current path length is less than the minimum path length, update the minimum path length and the minimum path
if current_path_length < min_path_length:
min_path_length = current_path_length
min_path = [grid[i][j]]
# If the current path length is equal to the minimum path length, add the current cell to the minimum path
elif current_path_length == min_path_length:
min_path.append(grid[i][j])
return min_path
def get_neighbors(grid, i, j):
# Return a list of the neighbors of the current cell
return [(i-1, j), (i+1, j), (i, j-1), (i, j+1)]Start time: 2023-09-01 03:01:22
Duration: 26.66s
Result: Fail
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
n = len(grid)
m = len(grid[0])
visited = set()
queue = deque([(0, 0, 0)])
while queue:
i, j, count = queue.popleft()
if (i, j) in visited:
continue
visited.add((i, j))
if count == k:
return grid[i][j]
for di, dj in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
new_i, new_j = i + di, j + dj
if 0 <= new_i < n and 0 <= new_j < m and (new_i, new_j) not in visited:
queue.append((new_i, new_j, count + 1))
return -1Start time: 2023-09-01 03:04:19
Duration: 32.51s
Result: Fail
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
n = len(grid)
m = len(grid[0])
dp = [[float('inf')] * (k+1) for _ in range(n)]
dp[0][0] = 0
for i in range(1, n):
for j in range(1, m):
for l in range(k+1):
if dp[i-1][l] != float('inf'):
dp[i][j] = min(dp[i][j], dp[i-1][l] + grid[i][j])
return [grid[i][j] for i in range(n) for j in range(m) if dp[i][j] == dp[n-1][k]]Start time: 2023-09-01 03:07:21
Duration: 38.81s
Result: Fail
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
# Initialize the minimum path length to infinity
min_path_length = float('inf')
# Initialize the minimum path to None
min_path = None
# Loop through each cell in the grid
for i in range(len(grid)):
for j in range(len(grid[i])):
# If the current cell is the starting cell, set the current path length to 0
if grid[i][j] == 1:
current_path_length = 0
# If the current cell is not the starting cell, set the current path length to the minimum path length so far
else:
current_path_length = min_path_length
# Loop through each neighbor of the current cell
for neighbor in get_neighbors(grid, i, j):
# If the neighbor is not the starting cell, add the value of the neighbor to the current path length
if neighbor != 1:
current_path_length += grid[neighbor[0]][neighbor[1]]
# If the current path length is less than the minimum path length, update the minimum path length and minimum path
if current_path_length < min_path_length:
min_path_length = current_path_length
min_path = [grid[i][j]]
# If the current path length is equal to the minimum path length, add the current cell to the minimum path
elif current_path_length == min_path_length:
min_path.append(grid[i][j])
# Return the minimum path
return min_path
def get_neighbors(grid, i, j):
# Initialize an empty list to store the neighbors
neighbors = []
# Check if the cell to the north is valid
if i > 0:
neighbors.append((i-1, j))
# Check if the cell to the south is valid
if i < len(grid) - 1:
neighbors.append((i+1, j))
# Check if the cell to the west is valid
if j > 0:
neighbors.append((i, j-1))
# Check if the cell to the east is valid
if j < len(grid[i]) - 1:
neighbors.append((i, j+1))
# Return the list of neighbors
return neighborsStart time: 2023-09-01 03:10:24
Duration: 40.63s
Result: Fail
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
n = len(grid)
m = len(grid[0])
dp = [[0] * (m + 1) for _ in range(n + 1)]
for i in range(1, n + 1):
for j in range(1, m + 1):
dp[i][j] = min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]) + 1
path = []
i = n
j = m
while i > 0 and j > 0:
if dp[i][j] == dp[i - 1][j - 1]:
path.append(grid[i - 1][j - 1])
i -= 1
j -= 1
elif dp[i][j] == dp[i - 1][j]:
path.append(grid[i - 1][j])
i -= 1
else:
path.append(grid[i][j - 1])
j -= 1
return pathStart time: 2023-09-01 03:13:21
Duration: 25.94s
Result: Fail
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
n = len(grid)
m = len(grid[0])
dp = [[0] * (k+1) for _ in range(n)]
for i in range(n):
for j in range(m):
if i == 0 and j == 0:
dp[i][j] = grid[i][j]
elif i == 0:
dp[i][j] = min(dp[i][j-1], grid[i][j])
elif j == 0:
dp[i][j] = min(dp[i-1][j], grid[i][j])
else:
dp[i][j] = min(dp[i-1][j], dp[i][j-1], grid[i][j])
return dp[n-1][m-1]