Maximizing Sum Across Multiple Stacks by Popping Elements in Python

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πŸ’‘ Problem Formulation: We aim to design a program in Python that, given multiple stacks of integers, identifies the maximum possible sum we can obtain by popping some elements off these stacks. The task is to equalize the height of the stacks without breaching the maximal sum achievable. For example, given stacks [3, 2, 1, 1, 1], [4, 3, 2], and [1, 1, 4, 1], the maximum sum for equal height stacks is 5.

Method 1: Brute Force Iteration

This method involves iterating over all possible combinations of pops to equalize the stacks and then calculating the maximum sum. The function specification would include iterating through each possible state of the stacks until the heights are equal and recording the maximum sum encountered.

Here’s an example:

def max_sum_stacks(*stacks):
    stack_heights = [sum(stack) for stack in stacks]
    max_sum = 0
    while not all(height == stack_heights[0] for height in stack_heights):
        highest_stack_index = stack_heights.index(max(stack_heights))
        stack_heights[highest_stack_index] -= stacks[highest_stack_index].pop()
        max_sum = max(max_sum, min(stack_heights))
    return max_sum

print(max_sum_stacks([3, 2, 1, 1, 1], [4, 3, 2], [1, 1, 4, 1]))

Output:

5

This code snippet implements the brute-force approach by continuously equating the heights of the stacks. It does so by popping elements from the tallest stack and updating the maximum sum possible. It runs in a loop until all stacks reach an equal height, at which point it returns the maximum sum.

Method 2: Using Heap Data Structure

Method 2 optimizes the process by using a heap data structure to track and remove the largest elements efficiently. It involves pushing and popping from a priority queue to maintain the maximum at all times.

Here’s an example:

import heapq

def max_heap_sum_stacks(*stacks):
    max_heap = []
    for stack in stacks:
        heapq.heappush(max_heap, (-sum(stack), stack))
    max_sum = 0
    while len(set(sum(stack) for _, stack in max_heap)) != 1:
        _, highest_stack = heapq.heappop(max_heap)
        popped_value = highest_stack.pop()
        heapq.heappush(max_heap, (-(sum(highest_stack)), highest_stack))
        max_sum = max(max_sum, -max_heap[0][0])
    return max_sum

print(max_heap_sum_stacks([3, 2, 1, 1, 1], [4, 3, 2], [1, 1, 4, 1]))

Output:

5

This code snippet utilizes a heap to maintain the order of stacks by their sum in descending order. A negative sum is used since Python’s heapq is a min-heap. The algorithm pops the largest element when necessary and recalculates the maximum sum after each operation.

Method 3: Prefix Sum with Hash Map

This method leverages pre-computed prefix sums and a hash map to quickly identify overlaps in the sum of stacks. It’s about optimizing search operations for matching stack heights.

Here’s an example:

def prefix_sum_max(stacks):
    prefix_sums = [list(reversed([sum(stack[:i + 1]) for i in range(len(stack))])) for stack in stacks]
    common_sums = set(prefix_sums[0])
    for sums in prefix_sums[1:]:
        common_sums.intersection_update(sums)
    return max(common_sums)

stacks = [[3, 2, 1, 1, 1], [4, 3, 2], [1, 1, 4, 1]]
print(prefix_sum_max(stacks))

Output:

5

This snippet calculates the prefix sums for each stack in a reversed order to facilitate easy popping and then uses set operations to identify common sums across stacks. The maximum common sum is the desired output, representing the maximum sum obtainable with equal-height stacks.

Method 4: Dynamic Programming

Dynamic Programming (DP) can solve this by breaking it down into smaller subproblems and caching results for re-use. The focus is on finding the maximum sum of substacks with equal height and then building up to the full stacks.

Here’s an example:

def dp_max_sum_stacks(*stacks):
    memo = {}
    def recurse(*indices):
        if any(index == 0 for index in indices): return 0
        if indices in memo: return memo[indices]
        height = min(sum(stacks[i][0:indices[i]]) for i in range(len(stacks)))
        max_sum = 0
        for i in range(len(stacks)):
            if sum(stacks[i][0:indices[i]]) > height:
                next_indices = list(indices)
                next_indices[i] -= 1
                max_sum = max(max_sum, recurse(*next_indices))
        memo[indices] = max_sum
        return max_sum
    return recurse(*[len(stack) for stack in stacks])
print(dp_max_sum_stacks([3, 2, 1, 1, 1], [4, 3, 2], [1, 1, 4, 1]))

Output:

5

This dynamic programming example defines a recursive function to explore different substacks while memoizing results to avoid recalculating known sums. It incrementally builds up towards the maximum sum of all stacks after popping, thereby improving performance compared to an uninformed brute-force approach.

Bonus One-Liner Method 5: Using Python Set Intersections

The one-liner approach emphasizes Pythonic brevity, relying on the language’s set intersection capabilities to find common sums quickly.

Here’s an example:

stacks = [[3, 2, 1, 1, 1], [4, 3, 2], [1, 1, 4, 1]]
max_sum = max(set().intersection(*(set(sum(stack[:i+1]) for i in range(len(stack))) for stack in stacks)), default=0)
print(max_sum)

Output:

5

The one-liner code snippet provides a compact means of obtaining the maximum sum of equal-height stacks. It generates prefix sums for each stack and uses set intersection to retrieve the highest common sum, all in a single statement.

Summary/Discussion

  • Method 1: Brute Force Iteration. Simple to implement but inefficient for large or many stacks. Time-consuming as it checks all possible combinations.
  • Method 2: Using Heap Data Structure. More efficient than brute force, especially for larger data sets. Prioritizes operations on the tallest stack for faster convergence.
  • Method 3: Prefix Sum with Hash Map. Offers quick lookup times and avoids unnecessary operations. Efficient for small to medium-sized datasets with moderate stack sizes.
  • Method 4: Dynamic Programming. Best for cases with overlapping subproblems, and it’s more efficient than brute force, conserving computations. However, it requires more memory and a proper understanding of DP for implementation.
  • Method 5: Python Set Intersections One-Liner. Extremely concise and takes advantage of Python’s powerful data structure operations. However, can be less readable and obscure for those not familiar with Python’s set operations.