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

The optimal solution has a time complexity of O(n*V).

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

The space complexity is O(V*D).

What data structures are used to solve Longest Subsequence With Decreasing Adjacent Difference?

This problem uses Array, Dynamic Programming. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Longest Subsequence With Decreasing Adjacent Difference?

Insight 1: Dynamic programming is suitable to solve this problem since the optimal solution for a larger sequence can be built from optimal solutions of smaller subsequences. Insight 2: The key to forming the dynamic programming state is to consider both the last value added to the subsequence and the last absolute difference calculated. This allows us to verify the non-increasing condition.

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Longest Subsequence With Decreasing Adjacent Difference - LeetCode 3409 Solution

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Longest Subsequence With Decreasing Adjacent Difference - Complete Solution Guide

Longest Subsequence With Decreasing Adjacent Difference is LeetCode problem 3409, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 seq 0 , seq 1 , seq 2 , ..., seq m of nums , |seq 1 - seq 0 | >= |seq 2 - seq 1 | >= ... >= |seq m - seq m - 1 | . Return the length of such a subsequence. Example 1: Input: nums = [16,6,3] Output: 3 Explanation: The longest subsequence is [16

Detailed Explanation

The problem asks us to find the length of the longest subsequence of a given array `nums` such that the absolute differences between consecutive elements in the subsequence form a non-increasing sequence. A subsequence is formed by taking elements from the original array in order, but not necessarily consecutively. The absolute difference between two numbers is the non-negative difference between them. A non-increasing sequence means the numbers are decreasing or staying the same. For example, if nums = [10, 20, 10, 19, 10, 20], one possible subsequence is [10, 20, 10, 19, 10]. The absolute differences between consecutive elements would be |20-10| = 10, |10-20| = 10, |19-10| = 9, |10-19| = 9 which forms the non-increasing sequence [10, 10, 9, 9]. The constraints are that the array length is between 2 and 10000 and that each number is between 1 and 300.

Solution Approach

The provided solutions use dynamic programming to find the length of the longest subsequence. The core idea is to maintain a DP table `dp[val][diff]` that stores the length of the longest subsequence ending with value `val` and last absolute difference `diff`. Additionally, it maintains a `suffix_max_dp` table to store the maximum DP values for each value `val`, indexed by difference. The `max_len_at_val` array stores the maximum subsequence length ending with a particular value regardless of what the last absolute difference was. The solution iterates through each number in `nums`, building the DP table as it goes.

Step-by-Step Algorithm

Step 1: Initialize DP tables: `dp[V_max + 1][D_max + 1]`, `suffix_max_dp[V_max + 1][D_max + 1]`, and `max_len_at_val[V_max + 1]`. Initialize `ans` to 1 (since the minimum subsequence length is 1).
Step 2: Iterate through each number `num` in the input `nums` array. Let `current_val = num`.
Step 3: Create a `candidate_dp_row` to temporarily store the potential new lengths of the subsequence ending at `current_val`. This prevents changes to `dp` from affecting calculations in the current iteration.
Step 4: Iterate through all possible previous values `prev_val` from 1 to `V_max`. If there exists a subsequence ending with `prev_val` (indicated by `max_len_at_val[prev_val] > 0`), calculate the absolute difference `d = abs(current_val - prev_val)`.
Step 5: Find the maximum subsequence length ending at `prev_val` with the difference greater or equal to `d` from `suffix_max_dp[prev_val][d]`. Let `len_to_extend = suffix_max_dp[prev_val][d]`.
Step 6: Calculate the new potential subsequence length: `new_len = 1 + max(1, len_to_extend)`. The `max(1, len_to_extend)` is necessary because `len_to_extend` might be 0, and we still need to add the current value to extend the subsequence.
Step 7: Update `candidate_dp_row[d]` if `new_len` is greater.
Step 8: After iterating through all possible `prev_val`s, update the `dp[current_val][d]` table with the values from the `candidate_dp_row`. Maintain a flag `updated_dp_row` to track if the current DP row has been updated. Update the answer if a longer subsequence is found.
Step 9: If `updated_dp_row` is true, update the `suffix_max_dp[current_val]` table by taking the suffix maximum of the `dp[current_val]` row. The suffix max is needed because the subsequent difference `d` must be less than or equal to the previous difference. Starting from `d=D_max` copy the value from the main `dp` table. Iterate backwards over `d`, taking the max of the current value from `dp` and the suffix max for `d+1`.
Step 10: Update the length of the longest subsequence ending at `current_val` in `max_len_at_val`.
Step 11: After iterating through all numbers in `nums`, return `ans`.

Key Insights

Insight 1: Dynamic programming is suitable to solve this problem since the optimal solution for a larger sequence can be built from optimal solutions of smaller subsequences.
Insight 2: The key to forming the dynamic programming state is to consider both the last value added to the subsequence and the last absolute difference calculated. This allows us to verify the non-increasing condition.
Insight 3: The relatively small constraint on num[i] (1 <= num[i] <= 300) allows us to use the value of the numbers and differences as indices in our DP tables. This greatly simplifies the implementation.

Complexity Analysis

Time Complexity: O(n*V)

Space Complexity: O(V*D)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: Juspay.

Longest Subsequence With Decreasing Adjacent Difference | LeetCode 3409

Medium

Array Dynamic Programming

Time: O(n*V)Space: O(V*D)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int longestSubsequence(int[] nums) {
        int V_max = 300;
        int D_max = 299;
        
        int[][] dp = new int[V_max + 1][D_max + 1];
        int[][] suffix_max_dp = new int[V_max + 1][D_max + 1];
        int[] max_len_at_val = new int[V_max + 1];
        
        int ans = 1;
        
        for (int num : nums) {
            int current_val = num;
            
            int[] candidate_dp_row = new int[D_max + 1];

            for (int prev_val = 1; prev_val <= V_max; prev_val++) {
                if (max_len_at_val[prev_val] > 0) {
                    int d = Math.abs(current_val - prev_val);
                    
                    int len_to_extend = suffix_max_dp[prev_val][d];
                    
                    int new_len = 1 + Math.max(1, len_to_extend);
                    
                    if (candidate_dp_row[d] < new_len) {
                        candidate_dp_row[d] = new_len;
                    }
                }
            }

            boolean updated_dp_row = false;
            for (int d = 0; d <= D_max; d++) {
                if (candidate_dp_row[d] > dp[current_val][d]) {
                    dp[current_val][d] = candidate_dp_row[d];
                    updated_dp_row = true;
                    if (ans < dp[current_val][d]) {
                        ans = dp[current_val][d];
                    }
                }
            }

            if (updated_dp_row) {
                suffix_max_dp[current_val][D_max] = dp[current_val][D_max];
                for (int k = D_max - 1; k >= 0; k--) {
                    suffix_max_dp[current_val][k] = Math.max(suffix_max_dp[current_val][k + 1], dp[current_val][k]);
                }
            }

            if (max_len_at_val[current_val] < 1) {
                 max_len_at_val[current_val] = 1;
            }
                 
            if (updated_dp_row) {
                if (max_len_at_val[current_val] < suffix_max_dp[current_val][0]) {
                    max_len_at_val[current_val] = suffix_max_dp[current_val][0];
                }
            }
                
        }
        
        return ans;
    }
}

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Problem Info

Description

Problem 3409: Longest Subsequence With Decreasing Adjacent Difference - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

Juspay

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3409

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Initialize DP tables: `dp[V_max + 1][D_max + 1]`, `suffix_max_dp[V_max + 1][D_max + 1]`, and `max_len_at_val[V_max + 1]`. Initialize `ans` to 1 (since the minimum subsequence length is 1).
2
Step 2: Iterate through each number `num` in the input `nums` array. Let `current_val = num`.
3
Step 3: Create a `candidate_dp_row` to temporarily store the potential new lengths of the subsequence ending at `current_val`. This prevents changes to `dp` from affecting calculations in the current iteration.
4
Step 4: Iterate through all possible previous values `prev_val` from 1 to `V_max`. If there exists a subsequence ending with `prev_val` (indicated by `max_len_at_val[prev_val] > 0`), calculate the absolute difference `d = abs(current_val - prev_val)`.
5
Step 5: Find the maximum subsequence length ending at `prev_val` with the difference greater or equal to `d` from `suffix_max_dp[prev_val][d]`. Let `len_to_extend = suffix_max_dp[prev_val][d]`.
6
Step 6: Calculate the new potential subsequence length: `new_len = 1 + max(1, len_to_extend)`. The `max(1, len_to_extend)` is necessary because `len_to_extend` might be 0, and we still need to add the current value to extend the subsequence.
7
Step 7: Update `candidate_dp_row[d]` if `new_len` is greater.
8
Step 8: After iterating through all possible `prev_val`s, update the `dp[current_val][d]` table with the values from the `candidate_dp_row`. Maintain a flag `updated_dp_row` to track if the current DP row has been updated. Update the answer if a longer subsequence is found.
9
Step 9: If `updated_dp_row` is true, update the `suffix_max_dp[current_val]` table by taking the suffix maximum of the `dp[current_val]` row. The suffix max is needed because the subsequent difference `d` must be less than or equal to the previous difference. Starting from `d=D_max` copy the value from the main `dp` table. Iterate backwards over `d`, taking the max of the current value from `dp` and the suffix max for `d+1`.
10
Step 10: Update the length of the longest subsequence ending at `current_val` in `max_len_at_val`.
11
Step 11: After iterating through all numbers in `nums`, return `ans`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*V)

💾

Space Complexity

O(V*D)

📝

Why?

The outer loop iterates through `n` elements of the `nums` array. The nested loop iterates through `V` (V_max) values. Thus, the time complexity is O(n*V). The space complexity is dominated by the 2D DP tables, which are of size (V+1) * (D+1). The other arrays take only O(V) space, which is dominated by O(V*D).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursion with Memoization

A recursive approach with memoization could be used. This involves defining a recursive function that explores all possible subsequences. To avoid redundant computations, a memoization table can store the results of previously computed subproblems.

Time: O(n * V)Space: O(n * V)

Trade-offs: The recursive approach with memoization can be more intuitive to understand for some. However, it may have a higher overhead due to function calls and can be harder to optimize compared to the iterative dynamic programming solution. Space usage could also be higher depending on the recursion depth.

✨

Key Insights

1Insight 1: Dynamic programming is suitable to solve this problem since the optimal solution for a larger sequence can be built from optimal solutions of smaller subsequences.
2Insight 2: The key to forming the dynamic programming state is to consider both the last value added to the subsequence and the last absolute difference calculated. This allows us to verify the non-increasing condition.
3Insight 3: The relatively small constraint on num[i] (1 <= num[i] <= 300) allows us to use the value of the numbers and differences as indices in our DP tables. This greatly simplifies the implementation.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: nums = [1, 1, 1, 1]. The algorithm correctly identifies the longest subsequence as the entire array, with a length of 4. All adjacent differences are 0, which is a non-increasing sequence.
Edge case 2: nums = [300, 1]. The algorithm correctly returns 2, with a single difference of 299. This is correctly captured in dp[1][299].
Edge case 3: nums = [16, 6, 3]. This example shows a simple decreasing sequence, and the answer will be 3.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not correctly identifying the non-increasing difference sequence. The difference must be calculated between adjacent terms *in the subsequence*, not necessarily adjacent terms in the original array.

✗Common mistake 2: Incorrectly updating the DP table. It's crucial to update the DP table based on previous values and differences correctly to ensure the non-increasing property is maintained.

✗Common mistake 3: Not handling the case where the differences are the same, leading to incorrect calculations of subsequence length. Since non-increasing means decreasing or staying the same, equal differences are valid.

✗Common mistake 4: Not correctly calculating and applying the suffix max for the given value. Failing to do so would mean not accounting for valid previous sequence differences for building the DP table.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Stock price analysis. Finding patterns in stock prices where the price difference between consecutive time points are decreasing or staying the same.

🏦Real-world scenario 2: Data compression. Identifying subsequences with decreasing differences in data values for efficient compression algorithms.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 17, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Longest Subsequence With Decreasing Adjacent Difference"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?