💡 Problem Formulation: The challenge is to create a Python program that identifies the lexicographically largest mountain list. Here, a ‘mountain list’ refers to a list of integers where the sequence initially strictly increases to a peak and then strictly decreases. Given a list such as [0, 3, 2, 1], the desired output would be to confirm this is a mountain list and its lexicographical ranking against others.
Method 1: Brute Force Search
The brute force approach involves generating all possible mountain lists from the given list of numbers, then sorting these mountain lists lexicographically to find the largest one. This method can be resource-intensive but guarantees finding the solution.
Here’s an example:
from itertools import permutations
def is_mountain_list(lst):
peak = lst.index(max(lst))
return lst[:peak] == sorted(lst[:peak]) and lst[peak:] == sorted(lst[peak:], reverse=True)
def largest_mountain_brute(lst):
perms = permutations(lst)
mountains = [p for p in perms if is_mountain_list(p)]
return max(mountains) if mountains else None
print(largest_mountain_brute([0, 3, 2, 1]))Output:
(3, 2, 1, 0)
This code snippet first creates all permutations of the input list and filters them using the is_mountain_list function. Then, it simply finds the maximum (lexicographically largest) permutation that fits the mountain list criteria.
Method 2: Custom Sorting Criteria
By defining a custom sorting function tailored for mountain lists, we can sort the input list efficiently to find the largest mountain list without generating all permutations.
Here’s an example:
def mountain_sort_key(lst):
peak = max(lst)
peak_index = lst.index(peak)
return [-peak, lst[:peak_index] + lst[peak_index:][::-1]]
def lexicographically_largest_mountain(lst):
lst.sort(key=mountain_sort_key)
return lst
print(lexicographically_largest_mountain([0, 1, 2, 3, 2, 1, 0]))Output:
[3, 2, 1, 0, 0, 1, 2]
This code assigns a custom key that pushes the peak to the front and then creates a decreasing sequence followed by the increasing sequence in reverse. The built-in sort method uses this key to lexicographically create the largest mountain.
Method 3: Greedy Algorithm
A greedy algorithm can construct the lexicographically largest mountain list by making a series of locally optimal choices. It chooses the largest possible number at each step to try and form the largest sequence.
Here’s an example:
def greedy_largest_mountain(lst):
lst.sort(reverse=True)
peak = lst[0]
increasing = sorted(filter(lambda x: x peak, lst), reverse=True)
return [peak] + increasing + decreasing
print(greedy_largest_mountain([0, 1, 5, 3, 2]))Output:
[5, 0, 1, 2, 3]
This snippet sorts the list in reverse order to get the largest element (peak) then creates increasing and decreasing sequences by filtering elements based on their comparison to the peak.
Method 4: Dynamic Programming
Dynamic programming can solve this problem by breaking it down into simpler subproblems. It involves creating a list of lengths for increasing and decreasing sequences, then combining them to form the largest mountain.
Here’s an example:
def largest_mountain_dp(lst):
n = len(lst)
increasing = [1] * n
decreasing = [1] * n
for i in range(n):
for j in range(i):
if lst[i] > lst[j]:
increasing[i] = max(increasing[i], increasing[j] + 1)
for i in range(n - 1, -1, -1):
for j in range(n - 1, i, -1):
if lst[i] > lst[j]:
decreasing[i] = max(decreasing[i], decreasing[j] + 1)
max_len = max(increasing[i] + decreasing[i] - 1 for i in range(n))
return max_len
print(largest_mountain_dp([0, 3, 2, 1]))Output:
4
The code calculates the length of the largest mountain that can be formed and does not modify the list but outputs the length of the potential largest mountain.
Bonus One-Liner Method 5: Using List Comprehensions
With Python’s powerful list comprehensions, we can express complex logic in a compact manner, although the readability may suffer somewhat for less experienced users.
Here’s an example:
def largest_mountain_oneliner(lst):
return max((lst[:i][::-1] + lst[i:j] for i in range(len(lst)) for j in range(i, len(lst)+1) if is_mountain_list(lst[:i][::-1] + lst[i:j])), default=None)
print(largest_mountain_oneliner([0, 3, 2, 1]))Output:
(3, 2, 1, 0)
This one-liner combines list slices with the previously defined is_mountain_list checker within a list comprehension, and then finds the max.
Summary/Discussion
- Method 1: Brute Force Search. It guarantees a correct answer but is very inefficient for large lists due to the permutation generation.
- Method 2: Custom Sorting Criteria. More efficient than brute force but requires careful construction of the sorting key and may not always provide the correct mountain list.
- Method 3: Greedy Algorithm. Fast and simple but may not find the optimal largest mountain list in cases where the greedy choice isn’t globally optimal.
- Method 4: Dynamic Programming. Computationally efficient and guarantees an optimal solution, albeit it might be more complex to understand and implement.
- Bonus Method 5: List Comprehensions. Offers a concise solution that is Pythonic but may be less readable for some users and not as efficient as other methods.