Bingo Check

Challenge Difficulty: Medium | Estimated completion time: ~20 minutes

Create a function that takes a 5x5 2D list and returns True if it has at least one Bingo, and False If it doesn’t.

Examples

bingo_check([
  [45, "x", 31, 74, 87],
  [64, "x", 47, 32, 90],
  [37, "x", 68, 83, 54],
  [67, "x", 98, 39, 44],
  [21, "x", 24, 30, 52]
])
output = True
 
bingo_check([
  ["x", 43, 31, 74, 87],
  [64, "x", 47, 32, 90],
  [37, 65, "x", 83, 54],
  [67, 98, 39, "x", 44],
  [21, 59, 24, 30, "x"]
]) 
output = True
 
bingo_check([
  ["x", "x", "x", "x", "x"],
  [64, 12, 47, 32, 90],
  [37, 16, 68, 83, 54],
  [67, 19, 98, 39, 44],
  [21, 75, 24, 30, 52]
]) 
output = True
 
bingo_check([
  [45, "x", 31, 74, 87],
  [64, 78, 47, "x", 90],
  [37, "x", 68, 83, 54],
  [67, "x", 98, "x", 44],
  [21, "x", 24, 30, 52]
]) 
output = False

Notes

Only check for diagonals, horizontals and verticals.

Solution

Python

from typing import Any def bingo_check(board: list[list[Any]]) -> bool: """Check if a 5x5 bingo board has at least one bingo. Args: board: 5x5 2D list where "x" marks called numbers Returns: True if there is at least one bingo (row, column, or diagonal) """ size = 5 for row_index in range(size): if all(board[row_index][column_index] <span class="text-highlight"> "x" for column_index in range(size)): return True for column_index in range(size): if all(board[row_index][column_index] </span> "x" for row_index in range(size)): return True if all(board[index][index] <span class="text-highlight"> "x" for index in range(size)): return True if all(board[index][size - 1 - index] </span> "x" for index in range(size)): return True return False if __name__ == "__main__": print(bingo_check([ [45, "x", 31, 74, 87], [64, "x", 47, 32, 90], [37, "x", 68, 83, 54], [67, "x", 98, 39, 44], [21, "x", 24, 30, 52] ])) # True (column) print(bingo_check([ ["x", 43, 31, 74, 87], [64, "x", 47, 32, 90], [37, 65, "x", 83, 54], [67, 98, 39, "x", 44], [21, 59, 24, 30, "x"] ])) # True (diagonal)

Output