Analysis and Solution for the “Unordered Subsequence“ Problem

Problem Overview

The task requires determining the shortest subsequence of a given sequence that is unordered. A sequence is defined as unordered if it is neither non-decreasing nor non-increasing. This problem focuses on efficiently identifying such a subsequence or determining that the given sequence is fully ordered.

Definitions

1. Ordered Sequence: A sequence that is either:
   • Non-decreasing (e.g., [1,2,2,3][1, 2, 2, 3])
   • Non-increasing (e.g., [4,3,3,1][4, 3, 3, 1])
2. Unordered Sequence: A sequence that does not satisfy the conditions above. For instance, [3,1,3][3, 1, 3] is unordered.

Goal

Given a sequence of nn integers (1≤n≤1051 \leq n \leq 10^5), the task is:

1. If the sequence is completely ordered, output 00.
2. Otherwise, identify the shortest unordered subsequence and output its length kk and indices.

Input Format

1. The first line contains nn, the size of the sequence.
2. The second line contains nn integers representing the sequence.

Output Format

1. If the sequence is ordered, output 00.
2. Otherwise, output k=3k = 3 and the indices of the shortest unordered subsequence.

Algorithm and Approach

To solve the problem efficiently given the constraints, we need to leverage the following observations:

Observations

1. The shortest unordered subsequence must consist of exactly three elements. For any sequence aa, if three consecutive elements a[i],a[i+1],a[i+2]a[i], a[i+1], a[i+2] exist such that:
   • a[i]<a[i+1]>a[i+2]a[i] < a[i+1] > a[i+2] or a[i]>a[i+1]<a[i+2]a[i] > a[i+1] < a[i+2], then [a[i],a[i+1],a[i+2]][a[i], a[i+1], a[i+2]] is unordered.
2. If no such triplet exists, the sequence is ordered, and the answer is 00.

Steps to Solve

1. Edge Case: If n<3n < 3, the sequence cannot be unordered. Output 00.
2. Traverse the sequence and check consecutive triplets:
   • For each triplet [a[i],a[i+1],a[i+2]][a[i], a[i+1], a[i+2]], check the conditions a[i]<a[i+1]>a[i+2]a[i] < a[i+1] > a[i+2] or a[i]>a[i+1]<a[i+2]a[i] > a[i+1] < a[i+2].
   • If such a triplet is found, output k=3k = 3 and the indices [i+1,i+2,i+3][i+1, i+2, i+3].
3. If no unordered triplet exists after scanning the sequence, output 00.

Time Complexity

1. The traversal involves a single pass through the sequence, making the time complexity O(n)O(n).
2. Space complexity is O(1)O(1), as no additional storage is required apart from a few variables.

Implementation

Below is a Python implementation of the algorithm:

def find_unordered_subsequence(n, sequence):
    # Edge case: if the sequence length is less than 3
    if n < 3:
        return 0

    # Traverse the sequence to find an unordered triplet
    for i in range(n - 2):
        if (sequence[i] < sequence[i + 1] > sequence[i + 2]) or (sequence[i] > sequence[i + 1] < sequence[i + 2]):
            # Found an unordered subsequence
            return f"3\n{i + 1} {i + 2} {i + 3}"
    
    # If no unordered triplet is found, the sequence is ordered
    return "0"

# Example usage:
n = 5
sequence = [67, 499, 600, 42, 23]
print(find_unordered_subsequence(n, sequence))

Examples and Analysis

Example 1

Input:

5
67 499 600 42 23

Output:

3
1 3 5

Explanation: The triplet [67,499,600][67, 499, 600] is unordered (67<499>60067 < 499 > 600).

Example 2

Input:

3
1 2 3

Output:

0

Explanation: The sequence is fully ordered (non-decreasing).

Example 3

Input:

3
2 3 1

Output:

3
1 2 3

Explanation: The sequence [2,3,1][2, 3, 1] is unordered (2<3>12 < 3 > 1).

Complexity Analysis

1. Time Complexity: O(n)O(n) for a single traversal.
2. Space Complexity: O(1)O(1) for constant auxiliary storage.

Conclusion

The solution efficiently determines the presence of an unordered subsequence and outputs the shortest one when applicable. This problem is a classic example of leveraging observations about minimal unordered structures (triplets) to reduce computational complexity.