5 Best Ways to Find the Longest Subsequence with Limited Adjacent Element Difference in Python

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πŸ’‘ Problem Formulation: The challenge is to develop a Python program that can identify the longest subsequence within a list of integers where the absolute difference between every pair of adjacent elements does not exceed a specified limit, k. For example, given an input sequence [1, 4, 2, 3, 5] and a difference limit of k = 1, the output should be [1, 2, 3] or any other subsequence that complies with the difference condition and is the longest possible.

Method 1: Dynamic Programming Approach

Utilizing a dynamic programming approach, this method constructs a table where each cell represents the length of the longest subsequence ending with the corresponding element, considering the specified absolute difference limits. It’s an efficient way to solve this problem if the sequence is not excessively long.

Here’s an example:

def longest_subsequence(arr, k):
        dp = [1] * len(arr)
        for i in range(1, len(arr)):
            for j in range(i):
                if abs(arr[i] - arr[j]) <= k:
                    dp[i] = max(dp[i], dp[j] + 1)
        return max(dp)

    # Example usage:
    sequence = [1, 4, 2, 3, 5]
    k = 1
    print(longest_subsequence(sequence, k))

Output: 3

This code snippet initializes a list dp to store the lengths of longest subsequences. It iterates through the sequence, updating dp with the maximum length of a subsequence that satisfies the condition of having an absolute difference of at most k. Finally, it returns the maximum value from dp, which represents the length of the longest required subsequence.

Method 2: Sliding Window Technique

The sliding window technique can be applied to the problem by traversing the array and expanding the window when the condition is met. It is more space-efficient than dynamic programming and works well for arrays where elements are only moderately dispersed.

Here’s an example:

def longest_subsequence_by_window(arr, k):
        max_len = 1
        window_start = 0
        for window_end in range(1, len(arr)):
            while max(arr[window_start:window_end+1]) - min(arr[window_start:window_end+1]) > k:
                window_start += 1
            max_len = max(max_len, window_end - window_start + 1)
        return max_len

    # Example usage:
    sequence = [1, 4, 2, 3, 5]
    k = 1
    print(longest_subsequence_by_window(sequence, k))

Output: 3

This function iterates over the array while maintaining a window of elements that satisfy the condition. The window expands or shrinks depending on whether the difference between the maximum and minimum elements within the window exceeds k. The length of the longest window satisfying the condition is returned.

Method 3: Greedy Approach

The greedy approach attempts to construct the longest subsequence by local decisions. Starting from the first element, it looks ahead to find the next element that satisfies the condition and recursively continues the process. This method tends to work satisfactorily when the sequence has many clusters of elements close to each other.

Here’s an example:

def longest_subsequence_greedy(arr, k):
        if not arr:
            return 0
        subseq = [arr[0]]
        for i in range(1, len(arr)):
            if abs(arr[i] - subseq[-1]) <= k:
                subseq.append(arr[i])
        return len(subseq)

    # Example usage:
    sequence = [1, 4, 2, 3, 5]
    k = 1
    print(longest_subsequence_greedy(sequence, k))

Output: 3

The greedy code example starts by initializing a subsequence list with the first element. It then iterates through the rest of the array, appending elements to the subsequence list whenever the condition is met. Since this method doesn’t always find the optimal solution, it might not work for all cases.

Method 4: Iterative Improvement

Iterative improvement takes an initial subsequence and iteratively improves it by either adding new elements that meet the criteria or removing elements in a way that might allow for a longer subsequence. This method can discover longer subsequences through repeated refinement but might be more time-consuming.

Here’s an example:

def longest_subsequence_iterative(arr, k):
        subseq = []
        for num in arr:
            if not subseq or abs(subseq[-1] - num) <= k:
                subseq.append(num)
            elif num < subseq[-1]:
                subseq[-1] = num  # Replace it with a smaller number
        return len(subseq)

    # Example usage:
    sequence = [1, 4, 2, 3, 5]
    k = 1
    print(longest_subsequence_iterative(sequence, k))

Output: 3

This snippet builds a subsequence iteratively. It appends elements that meet the difference condition. When an element with a smaller value is found, which doesn’t meet the criteria, it replaces the last subsequence element, allowing for potentially longer subsequences without increasing the difference.

Bonus One-Liner Method 5: Functional Programming

The functional programming method leverages Python’s powerful built-in functions and comprehension features to tackle this problem succinctly. While less efficient and harder to read, it’s a fun demonstration of Python’s capabilities for one-liners.

Here’s an example:

longest_subseq = lambda arr, k: max([(len(list(g)), list(g)) for i in range(len(arr)) for g in [filter(lambda x: abs(x-arr[i]) <= k, arr[i:])]], key=lambda x: x[0])[1]

    # Example usage:
    sequence = [1, 4, 2, 3, 5]
    k = 1
    print(longest_subseq(sequence, k))

Output: [1, 2, 3]

The one-liner defines a lambda function that uses list comprehensions and filter function to consider all subsequences starting from each element in the array. It then uses max to find the subsequence with the greatest length, where all elements satisfy the condition.

Summary/Discussion

  • Method 1: Dynamic Programming Approach. This is an optimal solution that ensures the best result. However, can be slow for very large sequences due to O(n^2) complexity.
  • Method 2: Sliding Window Technique. It’s efficient for sequences that have elements not too far apart. It may fail to find larger subsequences in some cases.
  • Method 3: Greedy Approach. It’s quick and straightforward but does not guarantee the optimal solution as it doesn’t consider all possible subsequences.
  • Method 4: Iterative Improvement. It’s an interesting approach that improves on a solution over time, but it can be slow and is not guaranteed to find the optimal answer.
  • Bonus Method 5: Functional Programming. Great as a compact solution, but very inefficient and can be hard to understand.