Warehouse Packer's Dilemma

Alloha veggies is a new grocery store. As part of stock clearance guidlines of the store, given many piles of fresh product, follow the rules given below to stack the products in orderly manner.

There are total of n piles of products. The number of product in each pile is represented by the array numProducts. Select any subarray from the array numProducts and pickup product from the subarray such that number of product you pick from the ith pile is strictly less than the number of products you pick from the (i+1)th pile for all the indices i of the subarray. Find Maximum number of products that can be picked.

Example - numProducts = [7,4,5,2,6,5] Ans : 12 Subarray choosen (0,2). i.e [7,4,5] From the 0th index pile, we choose only 3 products. So, 3 from 0th pile, 4 from 1st pile, and 5 from 2nd pile. Total 12 products.

Names

Here are some potential names for this problem:

  • Max Incremental Product Picking

  • Greedy Product Stacking

  • Incremental Subarray Product Selection

  • Ordered Subarray Sum Maximization

  • Monotonically Increasing Subarray Aggregation

The key ideas the name aims to convey are:

  • Picking/selecting products in a greedy incremental way

  • Considering all subarrays to maximize the total

  • Following a monotonic ordering rule within subarrays

  • Aggregating products from the subarrays into a total sum

  • Greedily optimizing the total count selected

Some other ideas that capture aspects of the problem:

  • Warehouse Packer’s Dilemma

  • Storekeeper’s Stacking Challenge

  • Stockroom Optimization Puzzle

Overall, a descriptive name like “Greedy Incremental Subarray Product Picking” or “Monotonically Increasing Subarray Aggregation” would accurately characterize the essence of this problem.

Solution

Here is Python code to find the maximum number of products that can be picked following the given rules:

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def max_products(num_products):
  n = len(num_products)

  max_count = 0

  for i in range(n):
    for j in range(i+1, n):
      count = 0
      prev = 0
      for k in range(i, j+1):
        curr = min(num_products[k], prev + 1)
        count += curr
        prev = curr

      max_count = max(max_count, count)

  return max_count

# Example
num_products = [7,4,5,2,6,5] 

print(max_products(num_products)) # 12

The key steps are:

  1. Consider all possible subarrays within num_products
  2. For each subarray, greedily choose products from each pile, picking at most 1 more than the previous pile
  3. Update the max_count based on the maximum products picked from each subarray

This greedy approach ensures we follow the rules by incrementally increasing the picked products per pile within a subarray.