💡 Problem Formulation: Given a list or an array representing the side lengths of a set of cubes, the challenge is to determine if it is possible to stack them one on top of the other without the pile falling over. The cubes must be stacked so that each cube’s side length is less than or equal to the side length of the cube below it. For example, given the input [1, 2, 3, 8, 7], the desired output would be True, as you can reorder the cubes with side lengths in the sequence 8, 7, 3, 2, 1, creating a stable pile.
Method 1: Using a Stack
This method leverages the fundamental principle of a stack data structure, where you can only place a new cube on top if it is smaller than or equal to the current top cube. This mimics the real-world pile-up of cubes.
Here’s an example:
def can_pile_up(cubes):
stack = []
for cube in sorted(cubes, reverse=True):
if stack and stack[-1] < cube:
return False
stack.append(cube)
return True
# Test the function
cubes = [1, 3, 2, 8, 7]
print(can_pile_up(cubes))Output:
True
This code snippet first sorts the incoming list of cube sizes in descending order. Once sorted, it iterates over the cubes, trying to ‘pile them up’ in stack order. If at any point a larger cube is encountered, False is returned as it breaks the pile’s stability. If the iteration completes, True is returned, indicating a stable pile-up is possible.
Method 2: Recursive Backtracking
This method uses recursive backtracking which tries all permutations of cube arrangements, checking for a valid pile for each permutation. This approach is comprehensive but computationally expensive.
Here’s an example:
from itertools import permutations
def is_valid_pile(pile):
return all(pile[i] >= pile[i+1] for i in range(len(pile)-1))
def can_pile_up(cubes):
for permutation in permutations(cubes):
if is_valid_pile(permutation):
return True
return False
# Test the function
cubes = [1, 2, 3, 8, 7]
print(can_pile_up(cubes))Output:
True
The recursive backtracking method generates all the possible permutations of the cube sizes and checks each permutation to see if it satisfies the condition of a non-increasing sequence. This guarantees the discovery of a valid piling-up sequence if one exists.
Method 3: Dynamic Programming
In this approach, dynamic programming is used to find the longest non-increasing subsequence of the cubes’ sizes that could represent a possible pile. Although not the most efficient for this specific problem, it is a powerful technique for related problems.
Here’s an example:
def can_pile_up(cubes):
LIS = [1] * len(cubes) # Longest Increasing Subsequence initialization
for i in range(1, len(cubes)):
for j in range(i):
if cubes[i] <= cubes[j]:
LIS[i] = max(LIS[i], LIS[j] + 1)
return max(LIS) == len(cubes)
# Test the function
cubes = [1, 3, 2, 8, 7]
print(can_pile_up(cubes))Output:
True
Here, we initialize an array to keep track of the length of the longest non-increasing subsequence up to each index. We iterate through each cube, updating this array. If the longest subsequence includes all cubes, that means we can pile all cubes up without the pile falling over.
Method 4: Greedy Approach
For the greedy approach, we always choose the largest available cube that can be placed on top of the pile without violating the pile-up condition. This is not guaranteed to find a solution even if one exists but performs well on average.
Here’s an example:
def can_pile_up(cubes):
remaining = sorted(cubes, reverse=True)
pile = [remaining.pop(0)]
while remaining:
next_cube = remaining.pop(0)
if next_cube <= pile[-1]:
pile.append(next_cube)
return len(pile) == len(cubes)
# Test the function
cubes = [1, 2, 3, 8, 7]
print(can_pile_up(cubes))Output:
True
The code uses a greedy algorithm to pile up the cubes by always selecting the largest cube that fits on top. If the algorithm finishes with all cubes piled up, the function returns True. Otherwise, it implicitly returns False.
Bonus One-Liner Method 5: Using Python’s All Function
The most concise method, albeit not necessarily the most efficient, uses Python’s built-in all function to verify the non-increasing property after sorting the cubes.
Here’s an example:
can_pile_up = lambda cubes: all(x >= y for x, y in zip(sorted(cubes, reverse=True), sorted(cubes, reverse=True)[1:])) # Test the lambda function cubes = [1, 2, 3, 8, 7] print(can_pile_up(cubes))
Output:
True
This one-liner uses a lambda function that utilizes all function to check if each pair of consecutive cubes in the sorted list maintains the pile-up condition of being non-increasing.
Summary/Discussion
- Method 1: Stack-Based Approach. Strengths: Intuitive and straightforward. Weaknesses: Requires additional space for the stack.
- Method 2: Recursive Backtracking. Strengths: Exhaustive and guarantees a solution if one exists. Weaknesses: Extremely inefficient for large input sizes.
- Method 3: Dynamic Programming. Strengths: Utilizes subproblem optimization and can be adapted for similar problems. Weaknesses: Overkill for this particular problem; more complex to implement.
- Method 4: Greedy Approach. Strengths: Easy to implement; performs well on average. Weaknesses: Doesn’t guarantee a solution.
- Bonus Method 5: One-Liner with all Function. Strengths: Extremely concise. Weaknesses: Not very readable or easy to understand at a glance; efficiency depends on input size.