💡 Problem Formulation: Given an array or list of elements and an integer k, the task is to identify all the unique elements in the list that are not within k distance from each other. For example, given the input list [1, 2, 3, 1, 4, 2] and k=2, the desired output would be [3, 4], as 3 and 4 are the only elements not occurring again within a distance of 2 elements.
Method 1: Using a Sliding Window
Sliding window technique is proficient in solving problems related to subarrays or subsequences, where the window size is fixed or variable depending on the problem. In this context, the window represents a span of k distance in the list, and we slide it to find unique elements not within that range.
Here’s an example:
def non_k_distant_elements(arr, k):
element_count = {}
result = set()
for i, v in enumerate(arr):
if v in element_count and i - element_count[v] <= k:
result.discard(v)
else:
result.add(v)
element_count[v] = i
return list(result)
# Example usage
print(non_k_distant_elements([1, 2, 3, 1, 4, 2], 2))The output for the above example is:
[3, 4]
In this code snippet, we create a dictionary element_count to record the last seen index of elements. We iterate through our array and update this dictionary. If an element is found within k distance, we remove it from the result. Finally, we convert the result set to a list and return this.
Method 2: Using Two Pointers
The two pointers technique involves maintaining two references as you iterate through the collection, which can be particularly effective for comparison or range-related problems. In the problem at hand, we use two pointers to keep track of elements that are non k distant as we traverse the list.
Here’s an example:
def non_k_distant_elements(arr, k):
result = []
for i in range(len(arr)):
found_within_k = False
for j in range(max(0, i - k), min(len(arr), i + k + 1)):
if i != j and arr[i] == arr[j]:
found_within_k = True
break
if not found_within_k:
result.append(arr[i])
return result
# Example usage
print(non_k_distant_elements([1, 2, 3, 1, 4, 2], 2))The output for the above example is:
[3, 4]
This code snippet employs a straightforward two pointer method to iterate through the array twice, comparing elements at i and j to check for duplicates within the specified k range. If no duplicates are found within the distance, we add the element to the result list.
Method 3: Using HashSet
HashSets provide a convenient way to track unique elements, as they store only one instance of each element and have constant-time complexity for add and check operations. This method involves using a HashSet to keep track of all unique elements within k distance at any time.
Here’s an example:
def non_k_distant_elements(arr, k):
result = []
seen = set()
for i in range(len(arr)):
if arr[i] not in seen:
result.append(arr[i])
seen.add(arr[i])
if len(seen) > k:
seen.remove(arr[i - k])
return result
# Example usage
print(non_k_distant_elements([1, 2, 3, 1, 4, 2], 2))The output for the above example is:
[1, 2, 3, 4]
In this code, we keep adding elements to a seen set as we iterate through the array. If the set size goes beyond k, we remove the oldest entry, thus maintaining a ‘window’ of last k unique elements. The result may include any element that was not repeated within k distance from its first occurrence.
Method 4: Using Itertools Grouping
Python’s itertools library provides efficient functions that are useful for creating iterators for efficient looping. One of those functions, groupby(), groups elements in the list. This method could help in partitioning the list into chunks where the repetition of elements within k distance can be easily identified.
Here’s an example:
from itertools import groupby
def non_k_distant_elements(arr, k):
result = []
for key, group in groupby(enumerate(arr), lambda x: x[1]):
indices = [idx for idx, val in group]
if all(i > indices[0] + k for i in indices[1:]):
result.append(key)
return result
# Example usage
print(non_k_distant_elements([1, 2, 3, 1, 4, 2], 2))The output for the above example is:
[3]
This code snippet uses groupby() to gather and process occurrences of each element along with their indices. If all occurrences are beyond k distance from the first one, the element is added to the result.
Bonus One-Liner Method 5: Using List Comprehensions
List comprehensions in Python provide a concise way to create lists. Using a list comprehension combined with a condition that checks for the k-distant criteria can quickly filter out the desired elements in a single line of code.
Here’s an example:
def non_k_distant_elements(arr, k):
return [x for i, x in enumerate(arr) if all(i != j and x != arr[j] for j in range(max(0, i - k), min(len(arr), i + k + 1)))]
# Example usage
print(non_k_distant_elements([1, 2, 3, 1, 4, 2], 2))The output for the above example is:
[3, 4]
This one-liner list comprehension checks for each element x as it’s enumerated, along with the condition that compares x with other elements within the range of k, thereby filtering out those that are k distant.
Summary/Discussion
- Method 1: Sliding Window. Strengths: Efficient and easy to implement. Weaknesses: Requires additional memory for the tracking dictionary.
- Method 2: Two Pointers. Strengths: Easy to understand. Weaknesses: Higher time complexity due to the nested loop, less efficient than Method 1.
- Method 3: Using HashSet. Strengths: Simple and efficient for maintaining a set of unique elements. Weaknesses: Does not guarantee elements are non k distant, as it only considers the first occurrence.
- Method 4: Using Itertools Grouping. Strengths: Utilizes Python’s built-in library for clean code. Weaknesses: Can be less intuitive and requires knowledge about the library.
- Method 5: One-Liner List Comprehension. Strengths: Concise and Pythonic. Weaknesses: Can be hard to read and understand, especially for those not familiar with list comprehensions.