Longest Subsequence With Decreasing Adjacent Difference

This problem involves finding the length of the longest subsequence where the absolute differences between consecutive elements form a non-increasing sequence. It requires a dynamic programming approach where the state transition helps in tracking the subsequences' valid configurations based on their difference patterns.

Problem Statement

You are given an array of integers nums. Your task is to find the length of the longest subsequence seq of nums, such that the absolute differences between consecutive elements form a non-increasing sequence of integers.

In other words, for a subsequence seq0, seq1, seq2, ..., seqm of nums, the absolute differences |seq1 - seq0| >= |seq2 - seq1| >= ... >= |seqm - seqm - 1| must hold. Return the length of such a subsequence.

Examples

Example 1

Input: nums = [16,6,3]

Output: 3

The longest subsequence is [16, 6, 3] with the absolute adjacent differences [10, 3] .

Example 2

Input: nums = [6,5,3,4,2,1]

Output: 4

The longest subsequence is [6, 4, 2, 1] with the absolute adjacent differences [2, 2, 1] .

Example 3

Input: nums = [10,20,10,19,10,20]

Output: 5

The longest subsequence is [10, 20, 10, 19, 10] with the absolute adjacent differences [10, 10, 9, 9] .

Constraints

• 2 <= nums.length <= 104
• 1 <= nums[i] <= 300

Solution Approach

State Transition Dynamic Programming

This problem is best solved using dynamic programming with state transitions. We define a state where each element of the array has a value representing the longest subsequence ending at that index, given that the absolute differences between consecutive elements are non-increasing. By processing the array and updating this state, we can find the solution efficiently.

Iterative Comparison

Iterate through the array, and for each element, compare it with previous elements to check if adding it to a subsequence will maintain the non-increasing absolute difference condition. If so, update the corresponding state. The process ensures the longest subsequence is found by progressively validating all potential subsequences.

Optimizing Space Complexity

To optimize space complexity, store only the necessary values that affect future calculations. Instead of maintaining a full DP table, track just the lengths of subsequences that are relevant to each comparison, reducing the required memory overhead.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity depends on the final approach, but a typical solution would run in O(n^2), where n is the length of the input array. Space complexity can be reduced to O(n) if we optimize the storage of subsequence lengths.

What Interviewers Usually Probe

• Ability to apply dynamic programming effectively.
• Understanding of the relationship between adjacent differences and subsequence patterns.
• Skill in optimizing space complexity while maintaining correct results.

Common Pitfalls or Variants

Common pitfalls

• Forgetting to properly handle the non-increasing condition for differences.
• Incorrectly updating the subsequence lengths during the iteration.
• Over-complicating the solution and not reducing the space complexity.

Follow-up variants

• Problem with different constraints on the array length or value range.
• Variant where the subsequence must have an exact difference pattern rather than a non-increasing one.
• Challenge where the input array contains negative numbers or zero, requiring adjustments to the difference calculation.

FAQ

What is the primary approach for solving 'Longest Subsequence With Decreasing Adjacent Difference'?

The primary approach is using dynamic programming to track the longest subsequence where absolute adjacent differences are non-increasing.

What is the time complexity of the solution?

The time complexity is typically O(n^2), where n is the length of the input array, though it can vary depending on the approach.

Can this problem be solved with a greedy approach?

A greedy approach is not effective for this problem since it requires careful tracking of subsequences and their differences, which is best handled by dynamic programming.

How do I optimize the space complexity for this problem?

Space complexity can be optimized by tracking only the necessary subsequence lengths and not maintaining a full DP table.

What are common mistakes when solving 'Longest Subsequence With Decreasing Adjacent Difference'?

Common mistakes include not correctly handling the non-increasing difference condition, updating subsequences incorrectly, and failing to optimize space complexity.