Date: 18 July, 2023
Model: gpt-3.5-turbo
Temperature: 0.2
Passing runs: 0
Total runs: 10
Success rate: 0%
Tokens: 422
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: 2023-07-18 17:07:16
Duration: 7.41s
Completion tokens: 206
Result: Fail
def minPath(grid, k):
def dfs(i, j, path, visited):
if len(path) == k:
return path
visited.add((i, j))
neighbors = [(i-1, j), (i+1, j), (i, j-1), (i, j+1)]
for ni, nj in neighbors:
if 0 <= ni < len(grid) and 0 <= nj < len(grid[0]) and (ni, nj) not in visited:
path.append(grid[ni][nj])
result = dfs(ni, nj, path, visited)
if result:
return result
path.pop()
visited.remove((i, j))
for i in range(len(grid)):
for j in range(len(grid[0])):
path = [grid[i][j]]
visited = set()
result = dfs(i, j, path, visited)
if result:
return resultStart time: 2023-07-18 17:07:08
Duration: 8.03s
Completion tokens: 190
Result: Fail
def minPath(grid, k):
def dfs(i, j, path, visited):
if len(path) == k:
return path
visited.add((i, j))
neighbors = [(i-1, j), (i+1, j), (i, j-1), (i, j+1)]
for ni, nj in neighbors:
if 0 <= ni < len(grid) and 0 <= nj < len(grid[0]) and (ni, nj) not in visited:
result = dfs(ni, nj, path + [grid[ni][nj]], visited)
if result:
return result
visited.remove((i, j))
for i in range(len(grid)):
for j in range(len(grid[0])):
path = dfs(i, j, [grid[i][j]], set())
if path:
return pathStart time: 2023-07-18 17:06:59
Duration: 8.99s
Completion tokens: 259
Result: Fail
def minPath(grid, k):
def dfs(i, j, path):
if len(path) == k:
return path
path.append(grid[i][j])
neighbors = []
if i > 0:
neighbors.append((i-1, j))
if i < len(grid) - 1:
neighbors.append((i+1, j))
if j > 0:
neighbors.append((i, j-1))
if j < len(grid[0]) - 1:
neighbors.append((i, j+1))
neighbors.sort(key=lambda x: grid[x[0]][x[1]])
for neighbor in neighbors:
ni, nj = neighbor
if (ni, nj) not in visited:
visited.add((ni, nj))
result = dfs(ni, nj, path)
if result:
return result
visited.remove((ni, nj))
path.pop()
visited = set()
for i in range(len(grid)):
for j in range(len(grid[0])):
visited.add((i, j))
result = dfs(i, j, [])
if result:
return result
visited.remove((i, j))
return []Start time: 2023-07-18 17:06:50
Duration: 8.09s
Completion tokens: 176
Result: Fail
def minPath(grid, k):
N = len(grid)
start = (0, 0)
visited = set()
path = []
def dfs(cell, steps):
if steps == k:
return True
x, y = cell
visited.add(cell)
path.append(grid[x][y])
neighbors = [(x-1, y), (x+1, y), (x, y-1), (x, y+1)]
for neighbor in neighbors:
nx, ny = neighbor
if 0 <= nx < N and 0 <= ny < N and neighbor not in visited:
if dfs(neighbor, steps+1):
return True
visited.remove(cell)
path.pop()
return False
dfs(start, 0)
return pathStart time: 2023-07-18 17:06:43
Duration: 7.2s
Completion tokens: 199
Result: Fail
def minPath(grid, k):
N = len(grid)
visited = [[False] * N for _ in range(N)]
path = []
def dfs(i, j, count):
if count == k:
return True
if i < 0 or i >= N or j < 0 or j >= N or visited[i][j]:
return False
visited[i][j] = True
path.append(grid[i][j])
if dfs(i+1, j, count+1) or dfs(i-1, j, count+1) or dfs(i, j+1, count+1) or dfs(i, j-1, count+1):
return True
visited[i][j] = False
path.pop()
return False
for i in range(N):
for j in range(N):
if dfs(i, j, 1):
return pathStart time: 2023-07-18 17:06:36
Duration: 7.19s
Completion tokens: 214
Result: Fail
def minPath(grid, k):
def dfs(i, j, path, visited):
if len(path) == k:
return path
visited.add((i, j))
neighbors = [(i+1, j), (i-1, j), (i, j+1), (i, j-1)]
for ni, nj in neighbors:
if 0 <= ni < len(grid) and 0 <= nj < len(grid[0]) and (ni, nj) not in visited:
path.append(grid[ni][nj])
result = dfs(ni, nj, path, visited)
if result:
return result
path.pop()
visited.remove((i, j))
for i in range(len(grid)):
for j in range(len(grid[0])):
path = [grid[i][j]]
visited = set()
visited.add((i, j))
result = dfs(i, j, path, visited)
if result:
return resultStart time: 2023-07-18 17:06:27
Duration: 8.47s
Completion tokens: 228
Result: Fail
def minPath(grid, k):
n = len(grid)
m = len(grid[0])
visited = [[False] * m for _ in range(n)]
path = []
def dfs(i, j, count):
if count == k:
return True
if i < 0 or i >= n or j < 0 or j >= m or visited[i][j]:
return False
visited[i][j] = True
path.append(grid[i][j])
if dfs(i + 1, j, count + 1):
return True
if dfs(i - 1, j, count + 1):
return True
if dfs(i, j + 1, count + 1):
return True
if dfs(i, j - 1, count + 1):
return True
visited[i][j] = False
path.pop()
return False
for i in range(n):
for j in range(m):
if dfs(i, j, 0):
return path
return pathStart time: 2023-07-18 17:06:18
Duration: 9.69s
Completion tokens: 257
Result: Fail
def minPath(grid, k):
def dfs(i, j, path):
if len(path) == k:
return path
path.append(grid[i][j])
neighbors = []
if i > 0:
neighbors.append((i-1, j))
if i < len(grid) - 1:
neighbors.append((i+1, j))
if j > 0:
neighbors.append((i, j-1))
if j < len(grid[0]) - 1:
neighbors.append((i, j+1))
neighbors.sort(key=lambda x: grid[x[0]][x[1]])
for neighbor in neighbors:
ni, nj = neighbor
if (ni, nj) not in visited:
visited.add((ni, nj))
res = dfs(ni, nj, path)
if res:
return res
visited.remove((ni, nj))
path.pop()
visited = set()
for i in range(len(grid)):
for j in range(len(grid[0])):
visited.add((i, j))
res = dfs(i, j, [])
if res:
return res
visited.remove((i, j))Start time: 2023-07-18 17:06:10
Duration: 7.87s
Completion tokens: 220
Result: Fail
def minPath(grid, k):
def dfs(i, j, path):
if len(path) == k:
return path
path.append(grid[i][j])
neighbors = []
if i > 0:
neighbors.append((i-1, j))
if i < len(grid) - 1:
neighbors.append((i+1, j))
if j > 0:
neighbors.append((i, j-1))
if j < len(grid[0]) - 1:
neighbors.append((i, j+1))
neighbors.sort(key=lambda x: grid[x[0]][x[1]])
for neighbor in neighbors:
ni, nj = neighbor
if grid[ni][nj] not in path:
result = dfs(ni, nj, path)
if result:
return result
path.pop()
for i in range(len(grid)):
for j in range(len(grid[0])):
result = dfs(i, j, [])
if result:
return resultStart time: 2023-07-18 17:06:00
Duration: 9.38s
Completion tokens: 255
Result: Fail
def minPath(grid, k):
n = len(grid)
m = len(grid[0])
visited = [[False] * m for _ in range(n)]
path = []
def dfs(i, j, count):
if count == k:
return True
visited[i][j] = True
path.append(grid[i][j])
if i > 0 and not visited[i-1][j]:
if dfs(i-1, j, count+1):
return True
if i < n-1 and not visited[i+1][j]:
if dfs(i+1, j, count+1):
return True
if j > 0 and not visited[i][j-1]:
if dfs(i, j-1, count+1):
return True
if j < m-1 and not visited[i][j+1]:
if dfs(i, j+1, count+1):
return True
visited[i][j] = False
path.pop()
return False
for i in range(n):
for j in range(m):
if dfs(i, j, 1):
return path