AMAZON Coding Question – Solved

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As an operations engineer at Amazon, you are responsible for organizing the distribution of n different items in the warehouse. The size of each product is provided in an array productSize, where productSize[i] represents the size of the i-th product. You construct a new array called variation, where each element variation[i] is the difference between the largest and smallest product sizes among the first i products. Mathematically, this is defined as: variation[i] = max(productSize[1], productSize[2], ..., productSize[i]) - min(productSize[1], productSize[2], ..., productSize[i]) Your goal is to arrange the products in a way that minimizes the total variation, i.e., the sum of variation[1] + variation[2] + ... + variation[n]. Determine the minimum possible value of this sum after you have reordered the products. Example: n = 3 productSize = [3, 1, 2] By reordering the products as productSize = [2, 3, 1]: - variation[0] = max(2) - min(2) = 2 - 2 = 0 - variation[1] = max(2, 3) - min(2, 3) = 3 - 2 = 1 - variation[2] = max(2, 3, 1) - min(2, 3, 1) = 3 - 1 = 2 The sum is variation[0] + variation[1] + variation[2] = 0 + 1 + 2 = 3. This is the minimum possible total variation after rearranging. Function Description: Complete the function minimizeVariation in the editor below. minimizeVariation has the following parameter(s): int productSize[n]: The size of each product. Returns: int: the minimum possible total variation after rearranging the array productSize. Constraints: 1 ≤ n ≤ 2000 1 ≤ productSize[i] ≤ 10^6 Input Format for Custom Testing: The first line contains an integer, n, the number of elements in productSize. Each of the next n lines contains an integer describing productSize[i]. Sample Case 0: Sample Input: 7 productSize = [4, 5, 4, 6, 2, 1, 1] Sample Output: 4

Asked in: AMAZON

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Solution

from functools import lru_cache
from sys import setrecursionlimit
setrecursionlimit(10 ** 6)
def minimizeVariation(productSize):
// ... rest of solution available after purchase

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Explanation

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To solve this problem, the key challenge is to reorder the products so that the sum of variations is minimized. Each variation[i] depends on the range (max - min) of the first i products in the sequence. The goal is to find an arrangement where these ranges accumulate to the smallest possible total.

Let's break down the problem and the reasoning approach:

1. **Understanding the variation definition:**
variation[i] = max(productSize[1..i]) - min(productSize[1..i])

This means for each prefix of the reordered array, you look at the maximum and minimum product sizes in that prefix and compute their difference. The sum of all these differences for prefixes 1 to n is what you want to minimize.

2. **Effect of reordering on prefix ranges:**
Since the sum includes every prefix, early positions in the array have smaller prefixes (few elements), while later positions have larger prefixes (more elements). The maximum and minimum values in early prefixes are critical because their difference contributes multiple times in the overall sum through the accumulation over prefixes.

3. **Initial observations:**
- If you put large numbers too early, they might increase the max values of early prefixes, potentially increasing variation early.
- Similarly, putting very small numbers early might keep the min low and stable, reducing variation.
- Balancing these extremes is important.

4. **Role of sorting:**
If you sort the array in ascending order, then the max of each prefix i is productSize[i] (since the array is sorted ascending), and the min is productSize[1] for all prefixes. Therefore, variation[i] = productSize[i] - productSize[1].

The sum then becomes:
sum_{i=1 to n} (productSize[i] - productSize[1]) = (sum of all elements) - n * productSize[1]

Since productSize[1] is minimum in sorted order, this sum is influenced strongly by how far other elements are from this minimum.

5. **Is sorting the best?**
While sorting ascending is a natural thought, we want to see if some rearrangement can do better. Placing smaller elements earlier keeps the min stable, but large elements appear later and create large gaps in the later prefixes, possibly increasing variation significantly for those large prefixes.

6. **Exploring the opposite - sorting descending:**
If sorted descending, the max in prefixes is always the first element, and the min decreases as you move forward, affecting variation differently.

7. **Trying a balanced approach:**
Since the variation depends on both max and min values within prefixes, an arrangement that gradually introduces values from both ends (small and large) could keep the prefix ranges low over the entire array.

This idea hints at placing elements to keep the max and min close to each other for as many prefixes as possible, thus minimizing the difference.

8. **Mathematical insight with cumulative ranges:**
Since the sum of variations is the sum over all prefixes, each element’s position affects the total in a weighted way.

In fact, if we think about the contribution of each element to the overall sum:

- Larger elements increase the max in prefixes that include them.
- Smaller elements lower the min in prefixes that include them.

The position of each element determines how many prefixes include it.

9. **Dynamic programming and combinatorial solutions:**
With constraints (n ≤ 2000), a dynamic programming approach can be considered. The problem can be viewed as choosing the order of elements to minimize the cumulative variation.

One way to think is:

- For each prefix, you must keep track of the minimal possible sum of variations up to that prefix, considering the elements placed so far.
- You might explore the possibility of placing elements from either end of the sorted list to the prefix to control the range (max-min).

This approach resembles interval DP or constructing sequences optimally.

10. **Practical strategy - sorting and greedy intuition:**
Sort the array ascending. Then, attempt to build the sequence from left to right by choosing elements either from the smallest or the largest available remaining elements to minimize the prefix range.

By doing this, you can keep the variation within prefixes low and avoid large jumps in max or min values.

11. **Summary and key points for solving:**

- Sort the array to understand the range of values.
- Use a two-pointer approach or DP to decide the order of placing smallest and largest remaining elements to minimize prefix variations.
- Remember that placing the smallest element first sets the minimal baseline for min values in prefixes.
- Try to place elements in a way that avoids drastic increases in max or decreases in min values early in the sequence.
- Calculate the running max and min for prefixes efficiently to sum variations.
- The final result is the minimal total sum of prefix variations after rearranging.

12. **Edge cases:**

- All elements equal: variation is zero no matter the order.
- Large differences: careful placement can prevent sudden large jumps in prefix range.
- Small arrays (n=1 or 2): trivial or direct checking.

By following these conceptual steps—understanding how max and min evolve with prefix expansions, how sorting affects prefix ranges, and employing dynamic or greedy strategies to reorder—you can approach the problem systematically and devise an efficient solution to minimize the total variation sum.
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