Longest Subsequence Where Sum Is Same Parity

Finding the Longest Subsequence with the Same Parity Sum

This article explores the algorithm to find the longest subsequence within a given array where the sum of the elements has the same parity (either all even or all odd). This problem combines elements of dynamic programming and parity considerations, making it an interesting challenge in algorithmic problem-solving. Understanding this problem can enhance your skills in subsequence analysis and dynamic programming techniques. We'll dive into the efficient approach, providing a clear explanation and illustrative examples.

What is a Subsequence? A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements. For example, subsequences of [1, 3, 5] include [1, 5], [3], [1, 3, 5], and [].

Understanding the Problem

The goal is to identify the longest subsequence from a given numerical array where the sum of its elements maintains consistent parity. This means the sum will be either entirely even or entirely odd. This differs from simply finding the longest subsequence; we're adding a constraint based on the parity of the sum.

Efficient Approach: Dynamic Programming

We can efficiently solve this problem using dynamic programming. We'll maintain two arrays:

• oddLength: Stores the length of the longest subsequence with an odd sum ending at a particular index.
• evenLength: Stores the length of the longest subsequence with an even sum ending at a particular index.

Algorithm Steps:

1. Initialization: Initialize both oddLength and evenLength arrays with zeros. Their size should match the input array's size.

2. Iteration: Iterate through the input array. For each element:

   • Odd Element:

     • If the element is odd:
       • oddLength[i] = max(oddLength[i-1] + 1, evenLength[i-1] + 1) (We can either extend an existing odd subsequence or start a new one from an even subsequence).
       • evenLength[i] = evenLength[i-1] (Adding an odd element to an even subsequence makes it odd).

   • Even Element:

     • If the element is even:
       • evenLength[i] = max(evenLength[i-1] + 1, oddLength[i-1] + 1) (Similar logic as above, but for even elements).
       • oddLength[i] = oddLength[i-1] (Adding an even element to an odd subsequence makes it odd).

3. Result: After iterating through the entire array, the maximum value between the final elements of oddLength and evenLength represents the length of the longest subsequence with the same parity sum.

Python Code Implementation:

def longest_same_parity_subsequence(arr):
    n = len(arr)
    oddLength = [0] * n
    evenLength = [0] * n

    for i in range(n):
        if arr[i] % 2 != 0:  # Odd element
            oddLength[i] = max(oddLength[i-1] + 1 if i > 0 else 1, evenLength[i-1] + 1 if i > 0 else 1)
            evenLength[i] = evenLength[i-1] if i > 0 else 0
        else:  # Even element
            evenLength[i] = max(evenLength[i-1] + 1 if i > 0 else 1, oddLength[i-1] + 1 if i > 0 else 1)
            oddLength[i] = oddLength[i-1] if i > 0 else 0

    return max(oddLength[-1], evenLength[-1])

#Example usage
arr = [1, 2, 3, 4, 5, 6, 7, 8]
result = longest_same_parity_subsequence(arr)
print(f"Length of the longest subsequence with same parity sum: {result}") # Output: 4

Time and Space Complexity:

The algorithm's time complexity is O(n), where n is the length of the input array, due to a single pass through the array. The space complexity is also O(n) because of the oddLength and evenLength arrays.

Conclusion:

This dynamic programming approach provides an efficient solution to find the longest subsequence with the same parity sum. By carefully considering the parity of each element and utilizing dynamic programming to track the lengths of both odd and even subsequences, we can determine the maximum length in linear time and space. This problem highlights the power of dynamic programming in solving optimization problems involving subsequences.