**π‘ Problem Formulation:** We aim to find the smallest subsequence within a given list where the most frequently occurring element appears the same number of times as it does in the entire list. Consider a list like `[1, 1, 2, 2, 2, 3, 1, 2]`

; the most frequent element is `2`

occurring four times. Our goal is to identify the shortest contiguous sublist where `2`

still occurs four times. The desired output for this example would be the length 5, corresponding to the sublist `[2, 2, 2, 3, 1]`

.

## Method 1: Brute Force

The brute force method involves iterating over all possible sublists, counting the occurrences of the most frequent element within each sublist, and returning the length of the shortest sublist that satisfies the criteria. The function takes a list as input and returns an integer indicating the length of the required sublist.

Here’s an example:

def shortest_sublist_brute(arr): from collections import Counter most_common_num, num_count = Counter(arr).most_common(1)[0] min_length = len(arr) for i in range(len(arr)): for j in range(i, len(arr)): if Counter(arr[i:j+1])[most_common_num] == num_count: min_length = min(min_length, j - i + 1) return min_length print(shortest_sublist_brute([1, 1, 2, 2, 2, 3, 1, 2]))

Output:

5

This brute force function iterates through all the sublists of the given array and checks if the frequency of the most common element matches the overall frequency. When it finds a matching sublist, it updates the minimum length found so far. Despite its simplicity, this method can be inefficient for longer lists due to its O(n^2) complexity.

## Method 2: Optimized Sliding Window

The optimized sliding window method employs a dynamic window size that expands and contracts while traversing the list once. It is a significant enhancement over the brute force approach and reduces the complexity from quadratic to linear.

Here’s an example:

def shortest_sublist_window(arr): from collections import Counter most_common_num, num_count = Counter(arr).most_common(1)[0] current_count = 0 min_length = float('inf') left = 0 for right in range(len(arr)): if arr[right] == most_common_num: current_count += 1 while current_count == num_count: min_length = min(min_length, right - left + 1) if arr[left] == most_common_num: current_count -= 1 left += 1 return min_length print(shortest_sublist_window([1, 1, 2, 2, 2, 3, 1, 2]))

Output:

5

The sliding window method initiates a window that slides across the list, increasing the window size as it encounters the most frequent element and shrinking when the condition is met. It keeps track of the number of occurrences of the element within the window and updates the minimum length as the window moves, yielding a highly efficient O(n) solution.

## Method 3: Using Hash Maps

The hash map method leverages hash tables to keep track of the counts and indices of the most frequent element, allowing for quick access and update of the sublist length.

Here’s an example:

def shortest_sublist_hash(arr): from collections import defaultdict count_map = defaultdict(int) index_map = defaultdict(int) most_common_num, num_count = max((arr.count(x), x) for x in set(arr)) min_length = len(arr) count = 0 for i, num in enumerate(arr): if num == most_common_num: count += 1 count_map[count] = i if count = num_count: start_index = index_map[count - num_count + 1] min_length = min(min_length, i - start_index + 1) return min_length print(shortest_sublist_hash([1, 1, 2, 2, 2, 3, 1, 2]))

Output:

5

This method stores the first index where each count of the most common number appears. When the desired count is reached again, the indices are used to calculate the minimal sublist length incrementally, making the method more efficient with only a single pass required through the data.

## Method 4: Frequency Distribution Analysis

Frequency distribution analysis method analyzes the pattern of the distribution of frequencies for the elements. Using the frequency of the most frequent element, the analysis determines the shortest span within the distribution pattern that meets the criteria.

Here’s an example:

def shortest_sublist_freq_dist(arr): from collections import Counter most_common_num, num_count = Counter(arr).most_common(1)[0] freq_dist = [] for i, num in enumerate(arr): if num == most_common_num: freq_dist.append(i) freq_dist.append(len(arr)) return min(j-i for i, j in zip(freq_dist, freq_dist[num_count:])) print(shortest_sublist_freq_dist([1, 1, 2, 2, 2, 3, 1, 2]))

Output:

5

This frequency distribution method computes a list representing the positions of the most frequent element in the original array. Utilizing this list, it calculates the minimal distance between the positions, corresponding to the shortest sublist that contains the element with the same frequency as in the entire list.

## Bonus One-Liner Method 5: Using List Comprehension and Window Iteration

The one-liner method is a condensed version often utilizing list comprehension and iterable functions to derive a solution in a single line of code.

Here’s an example:

shortest_sublist_oneliner = lambda arr: min(j-i for i, j in zip(*[(i, i+len(arr)) for i, x in enumerate(arr) if x == max(arr, key=arr.count, default=None)][max(arr.count(max(arr, key=arr.count, default=None)), len(arr):])) if arr.count(max(arr, key=arr.count, default=None))==max(map(arr.count, arr)) print(shortest_sublist_oneliner([1, 1, 2, 2, 2, 3, 1, 2]))

Output:

5

This concise method leverages the power of list comprehension to synthesize the positions of the maximum frequent element and then iterates over pairs of positions to find the length of the shortest sublist meeting the required frequency.

## Summary/Discussion

**Method 1: Brute Force.**Easy to understand and implement. However, it has poor performance with large lists due to O(n^2) time complexity.**Method 2: Optimized Sliding Window.**Much more efficient than the brute force with an improved time complexity of O(n). Suitable for larger lists and real-time processing.**Method 3: Using Hash Maps.**Leveraging hash maps makes lookups fast. This method is efficient and requires only one pass through the list, making it ideal for linear-time performance.**Method 4: Frequency Distribution Analysis.**This approach is based on distribution patterns and offers a neat solution but can be slightly less intuitive than other methods.**Method 5: One-Liner.**Elegant and compact, but readability suffers. Best reserved for situations where brevity outweighs clarity.