3409. Longest Subsequence With Decreasing Adjacent Difference LeetCode Solution

In this guide, you will get 3409. Longest Subsequence With Decreasing Adjacent Difference LeetCode Solution with the best time and space complexity. The solution to Longest Subsequence With Decreasing Adjacent Difference problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Longest Subsequence With Decreasing Adjacent Difference solution in C++
4. Longest Subsequence With Decreasing Adjacent Difference solution in Java
5. Longest Subsequence With Decreasing Adjacent Difference solution in Python
6. Additional Resources

Problem Statement of Longest Subsequence With Decreasing Adjacent Difference

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, |seq1 – seq0| >= |seq2 – seq1| >= … >= |seqm – seqm – 1|.
Return the length of such a subsequence.

Example 1:

Input: nums = [16,6,3]
Output: 3
Explanation:
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
Explanation:
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
Explanation:
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

Complexity Analysis

• Time Complexity:
• Space Complexity:

3409. Longest Subsequence With Decreasing Adjacent Difference LeetCode Solution in C++

class Solution {
 public:
  int longestSubsequence(vector<int>& nums) {
    const int mx = ranges::max(nums);
    // dp[num][diff] := the length of the longest subsequence ending in `num`
    // s.t. the last absolute difference between consecutive elements is `diff`
    vector<vector<int>> dp(mx + 1, vector<int>(mx + 1));

    for (const int num : nums) {
      for (int prev = 1; prev <= mx; ++prev) {
        const int diff = abs(num - prev);
        dp[num][diff] = max(dp[num][diff], dp[prev][diff] + 1);
      }
      // dp[num][diff] := max(dp[num][j]), where j >= diff
      for (int j = mx - 1; j >= 0; --j)
        dp[num][j] = max(dp[num][j], dp[num][j + 1]);
    }

    return ranges::max_element(dp, ranges::less{}, [](const vector<int>& row) {
      return row[0];
    })->at(0);
  }
};
/* code provided by PROGIEZ */

3409. Longest Subsequence With Decreasing Adjacent Difference LeetCode Solution in Java

class Solution {
  public int longestSubsequence(int[] nums) {
    final int mx = Arrays.stream(nums).max().getAsInt();
    // dp[num][diff] := the length of the longest subsequence ending in `num`
    // s.t. the last absolute difference between consecutive elements is `diff`
    int[][] dp = new int[mx + 1][mx + 1];

    for (final int num : nums) {
      for (int prev = 1; prev <= mx; ++prev) {
        final int diff = Math.abs(num - prev);
        dp[num][diff] = Math.max(dp[num][diff], dp[prev][diff] + 1);
      }
      // dp[num][diff] := max(dp[num][j]), where j >= diff
      for (int j = mx - 1; j >= 0; --j)
        dp[num][j] = Math.max(dp[num][j], dp[num][j + 1]);
    }

    return Arrays.stream(dp).mapToInt(row -> row[0]).max().getAsInt();
  }
}
// code provided by PROGIEZ

3409. Longest Subsequence With Decreasing Adjacent Difference LeetCode Solution in Python

class Solution:
  def longestSubsequence(self, nums: list[int]) -> int:
    mx = max(nums)
    # dp[num][diff] := the length of the longest subsequence ending in `num`
    # s.t. the last absolute difference between consecutive elements is `diff`
    dp = [[0] * (mx + 1) for _ in range(mx + 1)]

    for num in nums:
      for prev in range(1, mx + 1):
        diff = abs(num - prev)
        dp[num][diff] = max(dp[num][diff], dp[prev][diff] + 1)
      # dp[num][diff] := max(dp[num][j]) for j >= diff
      for j in range(mx - 1, -1, -1):
        dp[num][j] = max(dp[num][j], dp[num][j + 1])

    return max(map(max, dp))
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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