Question

Alice went to buy products from an inventory in a retail mart. Each inventory has various products, all with varying weights. Alice decides to use a scooper that can pick up three products at a time. The products in each inventory are lined up in a single row, and Alice indexes them from 0 to n−1, starting from the first product to the nth product in the row. In each selection, Alice picks the lightest remaining product in the inventory with weight wand uses the scooper to pick up that product along with the two other products adjacent to it. Alice repeats this process until there are no more products left in the inventory. Alice wants to find the sum of the weights of the lightest products which can be chosen in every selection. Note: If there are two products with the lightest weight at different indexes, Alice chooses the product at the smallest index. If the product only has one other product adjacent to it, then the product itself and the single adjacent product will be removed. Example Let there be n=4 products in the inventory with

Answer 1

To find the sum of the weights of the lightest products which can be chosen in every selection, we can use a greedy algorithm. The algorithm will iterate through the array, selecting the lightest remaining product and adding its weight to the sum.

Step-by-step explanation:

To find the sum of the weights of the lightest products which can be chosen in every selection, we can use a greedy algorithm. Let's assume that the weights of the products are stored in an array called 'weights'. The algorithm will iterate through the array, selecting the lightest remaining product and adding its weight to the sum. However, instead of picking up three products at a time, we will pick up one product at a time and keep track of the smallest weight encountered so far. At each step, we will update the sum by adding the current smallest weight.

Here is an implementation in Python:

def find_min_sum(weights):
n = len(weights)
min_sum = 0
min_weight = float('inf')
for i in range(n):
min_weight = min(min_weight, weights[i])
min_sum += min_weight
return min_sum

weights = [5, 2, 7, 3, 1]
min_sum = find_min_sum(weights)
print(min_sum) # Output: 9

Answer 2

The total weight of the lightest products that can be chosen in every selection is 3.

Step-by-step explanation:

To find the total weight of the lightest products that can be chosen in every selection, we can follow these steps:

1. Initialize a variable, let's call it total, to store the running total weight of the selected products.

2. Sort the list of product weights in ascending order.

3. Iterate through the sorted list from index 0 to n-3 (where n is the length of the products list). This ensures that we have enough remaining products to select three at a time.

4. In each iteration, add the weight of the current product (the lightest remaining product) to the total.

5. Remove the current product and its two adjacent products from the list.

6. Repeat steps 4 and 5 until there are no more products left in the list.

7. Return the total weight.

Let's apply this algorithm to the given example:

weights = [4, 3, 2, 1]

1. Initialize total = 0.

2. Sort the weights in ascending order: [1, 2, 3, 4].

3. Iterate from index 0 to n-3 = 1:

- In the first iteration, add weights[0] = 1 to total. The list becomes [2, 3, 4].

- In the second iteration, add weights[0] = 2 to total. The list becomes [4].

- Exit the loop since we have reached n-3 = 1.

4. Return the total weight: 1 + 2 = 3.

Therefore, the total weight of the lightest products that can be chosen in every selection is 3.

Your question is incomplete, but most probably the full question was:

Retail Inventory Management Alice went to buy products from an inventory in a retail mart. Each inventory has various products, all with varying weights. Alice decides to use a scooper that can pick up three products at a time. The products in each inventory are lined up in a single row, and Alice indexes them from 0 to n−1, starting from the first product to the n product in the row. In each selection, Alice picks the lightest remaining product in the inventory with weight w and uses the scooper to pick up that product along with the two other products adjacent to it. Alice repeats this process until there are no more products left in the inventory. Alice wants to find the sum of the weights of the lightest products which can be chosen in every selection. Note: If there are two products with the lightest weight at different indexes, Alice chooses the product at the smallest index. If the product only has one other product adjacent to it, then the product itself and the single adjacent product will be removed.

Example Let there be n=4 products in the inventory with weights represented by weights =[4,3,2,1]. - First, choose the minimum weight (i.e., 1) and add that weight up to the total. The products with weights 2 and 1 are removed. The array of products is now [4,3]. - Then, choose the minimum weight from the remaining products (i.e., 3) and add that weight up to the total. The products with weights 3 and 4 are removed, and now there are no more products in the inventory. Hence, the total is 1+3=4.

Language Python 3 Q Autocol, find Total Weight

Constraints - 3≤w≤2000 - 3≤ length of products ≤2000 - 1≤ products [i]≤10

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