What is the time complexity of Longest Subsequence With Decreasing Adjacent Difference?

The optimal solution for Longest Subsequence With Decreasing Adjacent Difference runs in O(n²) time and uses O(n) space. The nested loop checks pairs, leading to O(n²), while the DP array requires O(n) space.

What is the brute force solution for Longest Subsequence With Decreasing Adjacent Difference?

The brute force approach for Longest Subsequence With Decreasing Adjacent Difference is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible subsequences and check their differences. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Longest Subsequence With Decreasing Adjacent Difference?

Longest Subsequence With Decreasing Adjacent Difference uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Longest Subsequence With Decreasing Adjacent Difference?

Explain your thought process clearly. Consider edge cases and constraints. Practice similar DP problems.

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

Forgetting to check the non-increasing condition. Not initializing the DP array correctly.

#3409

Longest Subsequence With Decreasing Adjacent Difference

Medium

ArrayDynamic ProgrammingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n²)Space O(n)

We can use dynamic programming to track the longest subsequence while considering the differences between elements.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP array where dp[i] stores the longest subsequence ending at index i.
2Step 2: For each pair of indices (i, j), calculate the absolute difference and update dp[i] based on valid previous indices.
3Step 3: Return the maximum value in the DP array.

solution.py13 lines

1def longestDecreasingSubsequence(nums):
2    n = len(nums)
3    dp = [1] * n
4    max_len = 1
5    for i in range(n):
6        for j in range(i):
7            diff1 = abs(nums[i] - nums[j])
8            if dp[j] > 1:
9                diff2 = abs(nums[j] - nums[j-1])
10                if diff1 >= diff2:
11                    dp[i] = max(dp[i], dp[j] + 1)
12                    max_len = max(max_len, dp[i])
13    return max_len

ℹ

Complexity note: The nested loop checks pairs, leading to O(n²), while the DP array requires O(n) space.

1The absolute differences must be non-increasing.
2Dynamic programming is effective for subsequence problems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.