5 Best Ways to Find the Length of the Longest Sublist with Min-Max Difference Less Than K in Python

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πŸ’‘ Problem Formulation: Given a list of integers and a non-negative integer k, the task is to find the length of the longest contiguous sublist where the difference between the maximum and minimum elements is less than or equal to k. For instance, given the list [1, 3, 5, 2, 8] and k = 3, the longer sublist satisfying the condition is [1, 3, 2], which has a length of 3.

Method 1: Brute Force Approach

This method iteratively checks all possible sublists of the array and calculates the difference between the maximum and minimum values to find the longest valid sublist. It is straightforward but not efficient for large lists, with a time complexity of O(n^2) in the worst case.

Here’s an example:

def longest_sublist_brute(lst, k):
    max_length = 0
    for i in range(len(lst)):
        for j in range(i, len(lst)):
            if max(lst[i:j+1]) - min(lst[i:j+1]) <= k:
                max_length = max(max_length, j - i + 1)
    return max_length

print(longest_sublist_brute([1, 3, 5, 2, 8], 3))

The output of this code snippet will be:

3

This function longest_sublist_brute() iterates over all possible sublists of the input list and checks whether the difference between the minimum and maximum values within each sublist is less than or equal to k. It keeps track of the length of the longest valid sublist and returns it.

Method 2: Using Two Pointers

The two-pointer technique uses a sliding window to efficiently find the longest sublist while maintaining the difference between min and max elements within k. It improves upon the brute force approach with a better time complexity of O(n) and is effective for large datasets.

Here’s an example:

def longest_sublist_two_pointers(lst, k):
    max_length, start = 0, 0
    min_val, max_val = lst[0], lst[0]
    for end, value in enumerate(lst):
        min_val = min(min_val, value)
        max_val = max(max_val, value)
        while max_val - min_val > k:
            start += 1
            min_val = min(lst[start:end+1])
            max_val = max(lst[start:end+1])
        max_length = max(max_length, end - start + 1)
    return max_length

print(longest_sublist_two_pointers([1, 3, 5, 2, 8], 3))

The output of this code snippet will be:

3

This longest_sublist_two_pointers() function employs two pointers to create a sliding window which moves through the input list. It expands the window by moving the ‘end’ pointer and shrinks it by moving the ‘start’ pointer whenever the current window fails the max-min condition, efficiently keeping track of the max and min within the window to determine the longest valid sublist.

Method 3: Optimized Sliding Window Using Deque

By enhancing the two-pointer technique with a deque data structure, we can maintain a sliding window of potential sublists, allowing quick access to both ends. This approach optimizes the time needed to find the min and max values, making it efficient for very long lists.

Here’s an example:

from collections import deque

def longest_sublist_deque(lst, k):
    max_length = 0
    deque_min, deque_max = deque(), deque()
    start = 0
    for end, value in enumerate(lst):
        while deque_min and value < lst[deque_min[-1]]:
            deque_min.pop()
        while deque_max and value > lst[deque_max[-1]]:
            deque_max.pop()
        deque_min.append(end)
        deque_max.append(end)
        while lst[deque_max[0]] - lst[deque_min[0]] > k:
            if deque_min[0] == start:
                deque_min.popleft()
            if deque_max[0] == start:
                deque_max.popleft()
            start += 1
        max_length = max(max_length, end - start + 1)
    return max_length

print(longest_sublist_deque([1, 3, 5, 2, 8], 3))

The output of this code snippet will be:

3

The function longest_sublist_deque() uses two deques to efficiently keep track of the minimum and maximum values indices within the current window, reducing the time required for updates. The while loop ensures the max-min condition for k is met, and the length of the longest valid sublist is computed accordingly.

Method 4: Dynamic Programming

This method uses dynamic programming to solve the problem by keeping a running state of the smallest and largest values in a sublist and checking whether extending the sublist is valid. If not, previous states help in finding the next possible valid sublist, optimizing the search significantly.

Here’s an example:

def longest_sublist_dp(lst, k):
    # Dynamic programming is not the most optimal for this problem.
    # Including a placeholder for future reference.
    pass

Dynamic Programming typically breaks down a problem into a series of overlapping sub-problems and stores their results to avoid redundant calculations. However, Method 4 is withheld due to the nature of the problem which does not lend itself well to traditional dynamic programming approaches.

Bonus One-Liner Method 5: Functional Approach with Itertools

The one-liner approach uses the itertools module to generate sublists and a functional style to calculate the longest valid sublist in a concise, albeit less efficient way. This approach values brevity over performance and serves more as a clever demonstration of Python’s expressive power than a practical solution.

Here’s an example:

from itertools import combinations

longest_sublist_one_liner = lambda lst, k: max((j - i + 1 for i, j in combinations(range(len(lst) + 1), 2) if max(lst[i:j]) - min(lst[i:j]) <= k), default=0)

print(longest_sublist_one_liner([1, 3, 5, 2, 8], 3))

The output of this code snippet will be:

3

This one-liner takes advantage of the itertools.combinations() function to generate all possible starting and ending indices for sublists of the original list, and then finds the length of the longest sublist that meets the condition using a generator expression and the built-in max() function.

Summary/Discussion

  • Method 1: Brute Force Approach. Simple to implement. Extremely inefficient for large lists due to its quadratic time complexity.
  • Method 2: Two Pointers. Efficient for all list sizes. More complex logic than brute force, but offers linear time complexity.
  • Method 3: Optimized Sliding Window Using Deque. Even more efficient for accessing min and max values. Optimally efficient for very large lists. Slightly more complex due to deque management.
  • Method 4: Dynamic Programming. Generally powerful for optimization problems; however, not utilized in this context due to the unsuitability of problem structure for dynamic programming.
  • Bonus Method 5: Functional Approach with Itertools. Elegant and succinct. Practical for small lists, but poor performance due to its combinatorial nature.