Finding the Maximum Common Stack Height Efficiently
In the world of data structures and algorithms, solving problems efficiently is key. One such problem is determining the maximum possible equal height of three stacks by removing cylinders from the top.
The given Python function, `equalStacks,` provides an optimal solution to this problem using set intersections for quick lookup and comparison.
I will break down the problem and consequently explain the solution straightforwardly.
Further, we optimize it and explore alternative approaches.
Understanding the Problem
We have three stacks of cylinders, each represented by a list of integers.
Each integer represents the height of a cylinder.
The goal is to find the maximum possible height at which all three stacks are equal after removing cylinders from the top.
# Example Input
h1 = [3, 2, 1, 1, 1] # top to bottom
h2 = [4, 3, 2]
h3 = [1, 1, 4, 1]
print(equalStacks(h1, h2, h3))
# Expected Output
5
How the Function Works
def equalStacks(h1, h2, h3):
"""
Given three stacks represented as lists of cylinder heights,
find the maximum possible height where all stacks are equal.
"""
# Compute cumulative heights from bottom to top
def cumulative_heights(heights):
result = []
total = 0
for h in reversed(heights):
total += h
result.append(total)
return set(result) # Convert to set for fast lookup
# Compute sets of possible heights for each stack
s1, s2, s3 = map(cumulative_heights, [h1, h2, h3])
# Find the maximum common height using a set intersection
return max(s1 & s2 & s3) if (s1 & s2 & s3) else 0
Optimized Version with Better Readability
def equalStacks(h1, h2, h3):
"""
Given three stacks of cylinders, find the maximum height at which they can be equal.
Args:
h1(List): List of integers representing stack 1
h2(List): List of integers representing stack 2
h3(List): List of integers representing stack 3
Returns:
The maximum possible equal height after removal
"""
def get_possible_heights(stack):
""" Compute cumulative heights from bottom to top. """
heights = set()
total = 0
for height in reversed(stack):
total += height
heights.add(total)
return heights
# Get sets of cumulative heights
heights1, heights2, heights3 = map(get_possible_heights, [h1, h2, h3])
# Find the maximum common height
common_heights = heights1 & heights2 & heights3
return max(common_heights) if common_heights else 0
Optimizations and Why This Works Best
Set Operations for Fast Lookup: Instead of iterating over lists, using sets enables quick intersection (`&` operation) to find common heights.
Reverse Processing: Instead of summing heights each time, we process the list in reverse to store cumulative sums efficiently.
Early Exit Condition: If there are no common heights, we return `0` immediately.
Let's now explore alternative approaches and their trade-offs.
Using a Min-Heap (Less Efficient)
Another way to solve this would be to continuously remove elements from the tallest stack until all are equal. However, this requires additional sorting and heap operations, making it O(n log n) in complexity.
This approach keeps track of the three stack heights and removes elements from the tallest stack until all stacks have the same height.
import heapq
def equalStacks_min_heap(h1, h2, h3):
# Compute the initial heights of the stacks
height1, height2, height3 = sum(h1), sum(h2), sum(h3)
# Convert lists to stacks (reverse to match stack behaviour)
h1, h2, h3 = h1[::-1], h2[::-1], h3[::-1]
# Use a min-heap to track the smallest height
# Min-Heap - (Python uses max-heap by default, so we negate values)
heap = [-height1, -height2, -height3]
heapq.heapify(heap)
while len(set([-x for x in heap])) > 1: # If all heights are not equal
# Get the tallest stack height
tallest_height = -heapq.heappop(heap)
# Determine which stack is the tallest and remove the top element
if tallest_height == height1:
height1 -= h1.pop()
elif tallest_height == height2:
height2 -= h2.pop()
else:
height3 -= h3.pop()
#Push updated height back to heap
heapq.heappush(heap, -max(height1, height2, height))
return -heap[0] # Return the final equal height
How It Works
Compute the initial heights of all three stacks.
Store heights in a min-heap (Python negates values because heapq is a min-heap).
Repeatedly remove elements from the tallest stack until all stacks have the same height.
Return the common height.
Time Complexity Analysis
Heap operations (push and pop) take O(log 3) = O(1).
Removing elements from stacks takes O(n).
Overall, this runs in O(n log n).
Trade-offs
- Efficient compared to brute force
- Still slower than the O(n) solution with sets
Brute Force (Inefficient)
A naive approach would check all possible heights using nested loops, but this is computationally expensive with O(n^3) complexity.
def equalStacks_brute_force(h1, h2, h3):
#Compute all possible heights for each stack
def all_possible_heights(stack):
heights = set()
total = 0
for h in reversed(stack):
total += h
heights.add(total)
return heights
heights1, heights2, heights3 = all_possible_heights(h1), all_possible_heights(h2), all_possible_heights(h3)
# Brute-force check for all common heights
max_common_height = 0
for h1 in heights1:
for h2 in heights2:
for h3 in heights3:
if h1 == h2 == h3:
max_common_height = max(max_common_height, h1)
return max_common_height
How It Works
Compute all possible heights for each stack.
Use three nested loops to check for a common height.
Return the maximum common height.
Time Complexity Analysis
- Generating possible heights takes O(n).
- Nested loops take O(n³) in the worst case.
- Overall complexity: O(n³) (Extremely inefficient).
Trade-offs
- Easy to implement
- Extremely slow for large inputs
- Not practical beyond small test cases
Why Using Sets Is the Best
- O(n) Complexity: We traverse each list once and use set operations, making it faster.
- Minimal Memory Usage: We store only cumulative heights in sets.
- Scalability: Works well for large inputs due to efficient lookups.
Comparison Table
Approach Time Complexity Space Complexity Efficiency
Optimized (Set Intersection) O(n) O(n) Best
Min-Heap O(n log n) O(1) Good
Brute Force O(n³) O(n) Worst
- The Optimized Set Intersection approach is the best solution.
- The Min-Heap approach is an alternative but slower.
- The Brute Force approach is impractical for large inputs.
What You Should Do Next
If you found this helpful, try modifying the function to work with N stacks instead of just three.
Experiment with other data structures like heaps and queues to compare performance.
For more efficient algorithmic solutions, stay tuned for upcoming deep dives into data structures!