Here's a recent interview problem. Find the walk from the top left to the bottom right that maximizes the minimum element traveled. I believe I have the brute force solution and the most efficient solution.

find_max_min.py

import sys

def find_max_min_brute(array, x=0, y=0):
    max_x = max_y = len(array) - 1
    current = array[y][x]
    if x == max_x and y == max_y:
        return current
    if x == max_x:
        return min(current, find_max_min_brute(array, x, y + 1))
    if y == max_y:
        return min(current, find_max_min_brute(array, x + 1, y))

    return max(min(current, find_max_min_brute(array, x + 1, y)), \
               min(current, find_max_min_brute(array, x, y + 1)))

def find_max_min_helper(array, row, col):
  max_child = None;
  if col < len(array) - 1:
    max_child = find_max_min_helper(array, row, col + 1);
  if row < len(array) - 1:
    next_row_min = find_max_min_helper(array, row + 1, col)
    if max_child:
      max_child = max(next_row_min, max_child) 
    else:
      max_child = next_row_min;
                        
  return min(array[row][col], max_child or sys.maxsize);


def find_max_min(array):
    if len(array) == 0:
        return None
    return find_max_min_helper(array, 0, 0)


a = [[5, 4], [2, 5]]
b = [[9, 4, 8], [6, 5, 6], [3, 2, 9]]
c = [[9, 2, 7], [6, 5, 8], [8, 9, 10]]
d = [[1]]

print(find_max_min_brute(a))
print(find_max_min_brute(b))
print(find_max_min_brute(c))
print(find_max_min_brute(d))

print(find_max_min(a))
print(find_max_min(b))
print(find_max_min(c))
print(find_max_min(d))