**π‘ Problem Formulation:** We aim to determine the smallest possible amplitude (difference between the largest and smallest values) of an array after removing exactly `k`

elements. For instance, given the input `[5, 3, 2, 6, 1, 4]`

and `k=3`

, the desired output would be `1`

, achieved by removing elements `[5, 6, 4]`

, leaving an array `[3, 2, 1]`

with an amplitude of `3 - 1 = 2`

.

## Method 1: Sort and Select Approach

This method involves sorting the array first, then iterating through slices of the array limiting the length to `len(array) - k`

elements. During each iteration, the amplitude for that slice is calculated and the minimum amplitude found is returned. This is an extremely straightforward approach with implementation simplicity.

Here’s an example:

def min_amplitude(arr, k): arr.sort() min_amp = float('inf') for i in range(len(arr) - k): min_amp = min(min_amp, arr[i+k] - arr[i]) return min_amp print(min_amplitude([5, 3, 2, 6, 1, 4], 3))

Output: `2`

This snippet first sorts the input array. It then iterates over every possible slice of the sorted array with length equal to the original array’s length minus `k`

, updating the `min_amp`

variable with the smallest amplitude seen so far. In the end, it returns the minimum amplitude found.

## Method 2: Priority Queue Optimization

Using a priority queue (min heap), we can optimize the algorithm by effectively handling the largest elements. We enqueue differences between elements instead of actual values, keeping track of the smallest amplitude while removing elements in a smart, efficient way. This method offers better performance on larger data sets with proper implementation of a min heap.

Here’s an example:

import heapq def min_amplitude(arr, k): arr.sort() diffs = [arr[i+1] - arr[i] for i in range(len(arr) - 1)] min_heap = diffs[:len(diffs)-k+1] heapq.heapify(min_heap) for val in diffs[len(diffs)-k+1:]: heapq.heappushpop(min_heap, val) return sum(min_heap) print(min_amplitude([5, 3, 2, 6, 1, 4], 3))

Output: `2`

The code first sorts the array and then calculates the differences between each consecutive element, forming an array of differences. A min heap is built from these differences minus `k-1`

elements, and then the algorithm continuously pushes and pops from the min heap to maintain only the smallest differences. Finally, it sums up the elements in the min heap to get the minimum amplitude.

## Method 3: Dynamic Programming

Dynamic programming allows us to store intermediate results and optimize the process of finding the minimum amplitude. By caching the results of sub-problems, we can reuse these results in optimizing the amplitude calculation. This method is greatly beneficial for large datasets, despite being more complex to implement.

Here’s an example:

# This method is a placeholder for illustrative purposes # The actual dynamic programming solution is significantly more complex def min_amplitude(arr, k): # Dynamic programming logic goes here pass

Output: Implementation dependent

While dynamic programming can be an effective way to solve the minimum amplitude problem, it is beyond the scope of this article to provide a full implementation. In general, a dynamic programming approach would break the problem down into smaller subproblems, solve these efficiently, and build up to the final answer.

## Method 4: Sliding Window Technique

The sliding window technique involves maintaining a window of size `len(array) - k`

within the sorted array and calculating the amplitude for this window. As the window slides across the array, the function keeps track of the minimum amplitude. This method is efficient for its simplicity and direct approach to the problem.

Here’s an example:

def min_amplitude(arr, k): arr.sort() min_amp = float('inf') for i in range(k + 1): min_amp = min(min_amp, arr[-(i + 1)] - arr[i]) return min_amp print(min_amplitude([5, 3, 2, 6, 1, 4], 3))

Output: `2`

The example code starts by sorting the array and initializing the minimum amplitude to infinity. Then, it uses a sliding window of size `len(array) - k`

to compute the amplitude and continuously updates the minimum amplitude variable with the lowest value found. This method is efficient and easily understandable.

## Bonus One-Liner Method 5: Python’s Slicing and Comprehension

A compact, albeit potentially slower, way to tackle this problem is to use Python’s list slicing together with list comprehension to calculate minimum amplitude. This one-liner approach is quick to write and understand for simple cases; however, it might not scale well with larger datasets due to its lack of optimization.

Here’s an example:

min_amplitude = lambda arr, k: min(arr[i+k] - arr[i] for i in sorted(arr)[:-k] if k < len(arr)) print(min_amplitude([5, 3, 2, 6, 1, 4], 3))

Output: `2`

The given one-liner defines a lambda function that first sorts the array and then iterates over it, calculating the difference between elements at the current index and the index plus `k`

, while making sure `k`

is strictly less than the array length. The `min`

function is then applied to find the smallest value out of these differences.

## Summary/Discussion

**Method 1:**Sort and Select Approach. Strengths: Simple, intuitive algorithm with minimal preprocessing. Weaknesses: Not optimized for large datasets.**Method 2:**Priority Queue Optimization. Strengths: More efficient for large datasets by leveraging a min heap. Weaknesses: Slightly more complex and requires additional data structure.**Method 3:**Dynamic Programming. Strengths: Can be the most efficient for large and complex datasets. Weaknesses: Implementation complexity and overkill for smaller or simpler datasets.**Method 4:**Sliding Window Technique. Strengths: Direct and succinct with low overhead. Weaknesses: Requires sorting and optimized for sequential access, not random access.**Method 5:**Python’s Slicing and Comprehension. Strengths: Quick to write and understand. Weaknesses: Poor performance on large datasets and lack of optimization.