检查整数列表中是否存在升序（Python）

Question

Given a list of lists like this:

[[2, 7], [1, 4], [0, 5, 6]]

How can I check if there is a valid ascending order in this list. For example this would be True because I can form this order:

[2,4,5]

I need an algorithm which can somehow find a valid order for an arbitrarily large list with lists of any size. No integer value will repeat and sublists are sorted.

edit: This is what I have currently tried but it won't scale to larger lists.

allNumbers = [[2, 7], [1, 4], [0, 5, 6]]

smallest = min(allNumbers[0])
largest = max(allNumbers[2])
for n in allNumbers[1]:
    if smallest < n < largest:
        return True

Answer 1

You can solve it by using a greedy algorithm like this.

a = [[2, 7], [1, 4], [0, 5, 6]]
current = min(a[0])

possible = True
for l in a[1:]:
    possible = False
    for i in l:
        if i >= current:
            possible = True
            current = i
            break
    if not possible:
        break
print(possible) #returns True    

Answer 2

For each sublist, find the smallest element that's still larger than the previous smallest. If there's no such element, then there's no ascending path.

from math import inf

def ascend(xss):
    smallest = -inf
    path = []
    for xs in xss:
        smallest = min(x for x in xs if x > smallest)
        path.append(smallest)
    return path

>>> ascend([[2, 7], [1, 4], [0, 5, 6]])
[2, 4, 5]
>>> ascend([[2, 7], [1, 4], [0, 5, 6], [0]])
Traceback (most recent call last):
 ...
ValueError: min() arg is an empty sequence

Answer 3

Here is my answer.

def tell_if_the_lists_are_in_order(lists):
    min_first = min(lists[0])  # for empty input, return True.
    prev_min = min_first
    for sublist in lists[1:]:
        min_num = min(sublist)
        max_num = max(sublist)
        if max_num <= prev_min:
            return False
        prev_min = min(x for x in sublist if x > prev_min)
    return True

Answer 4

Here is my answer:

def IsAscending(list):
    ascend = True
    int a;
    for i in list:
        if i == 0:
            a = list[i]
        else:
            if list[i] >= a:

                a = list[i]
            else:
                ascend = False
    return ascend

Answer 5

Here is the final solution I derived

def validList(chLocations):
     ascendingOrder = []

     smallest = min(chLocations[0])
     maximum = max(chLocations[-1])

     ascendingOrder.append(smallest)
     chLocations = chLocations[1:]

     for i in range(len(chLocations)-1):
         for number in chLocations[i]:
             if number > max(ascendingOrder):
                 ascendingOrder.append(number)
                 break
     ascendingOrder.append(maximum)

     if sorted(ascendingOrder) == ascendingOrder:
          return True
     else:
          return False

>>> validList([[2, 7], [1, 4], [0, 5, 6]]) 
True
>>> validList([[1, 3], [7], [2]])
False

Explanation:

[[2, 7], [1, 4], [0, 5, 6]] --> [2,4,5] == True
[[1, 3], [7], [2]] --> [1,7,2] == False