Longest Arithmetic Subsequence of Given Difference

“Longest Arithmetic Subsequence of Given Difference” is approximately isomorphic to “Longest Increasing Subsequence”.

Reasoning: Both problems focus on finding the longest subsequence within an array that follows a specific pattern. In “Longest Arithmetic Subsequence of Given Difference”, the pattern is a constant difference between adjacent elements. In “Longest Increasing Subsequence”, the pattern is that each element is larger than the previous one.

Another similar problem is “Longest Harmonious Subsequence”.

Reasoning: The “Longest Harmonious Subsequence” problem also revolves around finding the longest subsequence that fits a particular pattern. In this case, the pattern is that the maximum and minimum elements in the subsequence differ by exactly one. It requires a similar approach to tracking and comparing elements in the sequence.

The order from simple to more complex is:

1. Longest Harmonious Subsequence
2. Longest Arithmetic Subsequence of Given Difference
3. Longest Increasing Subsequence

“Longest Harmonious Subsequence” is simpler as it only requires tracking the count of each number and its adjacent number. “Longest Arithmetic Subsequence of Given Difference” is a bit more complex as it needs to keep track of the longest sequence for each value. “Longest Increasing Subsequence” is the most complex as it requires maintaining multiple potential subsequences at the same time.

10 Prerequisite LeetCode Problems

Here are 10 problems to build a strong foundation of the concepts needed to solve the problem 1218. “Longest Arithmetic Subsequence of Given Difference”:

1. Problem 167. Two Sum II - Input array is sorted

   • This problem will help you practice working with sorted arrays and two-pointer technique.

2. Problem 15. 3Sum

   • This problem helps in understanding the pointers and how to manipulate them to get the required output.

3. Problem 217. Contains Duplicate

   • Understanding the use of hashing to check for duplicates.

4. Problem 53. Maximum Subarray

   • This problem can help you understand how to deal with subarray problems using dynamic programming.

5. Problem 300. Longest Increasing Subsequence

   • This problem is about finding the longest increasing subsequence in an array, which will help in understanding subsequence problems.

6. Problem 674. Longest Continuous Increasing Subsequence

   • Understanding how to track the length of a continuous sequence.

7. Problem 128. Longest Consecutive Sequence

   • This problem can help you in understanding how to deal with consecutive sequence problems using hash sets.

8. Problem 646. Maximum Length of Pair Chain

   • This problem will help you practice finding longest sequences in a more complex scenario.

9. Problem 718. Maximum Length of Repeated Subarray

   • This problem is also about subsequences, but here we’re looking for the longest common subarray in two integer arrays.

10. Problem 1143. Longest Common Subsequence

    • This problem will help in understanding the concept of longest common subsequence which is similar to longest arithmetic subsequence.

These cover two-pointers, hashing, dynamic programming, and working with subsequences.

Clarification Questions

Here are some potential clarification questions you might ask about this problem:

1. Can the input array arr contain duplicate elements?
2. What should be returned if there are multiple subsequences with the same maximum length? Should we return the length of just one of them, or all of them?
3. Is the difference always a positive integer, or can it be negative or zero?
4. In the case where difference is zero, does that mean we are looking for the longest subsequence of identical numbers?
5. If the array is empty, should the function return 0?
6. What should be the return value if no arithmetic sequence is found in the array?
7. Is it possible for the difference to be larger than any possible difference in the array? If so, what should be returned in this case?
8. Can the input array arr contain both positive and negative integers?

Always make sure to ask questions that help you understand edge cases or corner cases in your input. This will allow you to write a more robust solution.

Problem Analysis and Key Insights

The problem is asking for the length of the longest arithmetic subsequence in the given array where the difference between adjacent elements in the subsequence is equal to a given difference. This is essentially a dynamic programming problem as it involves finding an optimal solution by making a set of choices, where the choice depends on the choices made before it. Here are some key insights from the problem statement:

1. Subsequence, not Subarray: A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements. This is different from a subarray, where the elements must be contiguous. This means that the elements in our arithmetic subsequence don’t necessarily need to be next to each other in the original array.

2. Arithmetic Sequence: An arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. This constant is called the common difference. In this case, we are given the common difference, so we are looking for a subsequence where the difference between each pair of consecutive elements is equal to the given difference.

3. Longest Subsequence: We’re interested in finding the longest such subsequence. This indicates that there may be multiple valid arithmetic subsequences, but we want the one with the most elements.

4. Dynamic Programming: This problem has both optimal substructure and overlapping subproblems properties, making it a good candidate for a dynamic programming solution. The length of the longest arithmetic subsequence ending at a particular index in the array depends on the longest arithmetic subsequence in the previous part of the array.

5. Array Constraints: The constraints of the problem (1 <= arr.length <= 105 and -104 <= arr[i], difference <= 104) indicate that the array could be large, so an efficient solution will be required. This further points towards a dynamic programming solution.

From these insights, we can start to form an approach to solve the problem.

Problem Boundary

The scope of this problem revolves around the field of dynamic programming and sequence analysis within the realm of computer science and data structures. The problem requires an understanding of arrays, sequences, and the concept of dynamic programming to find an optimal solution.

Specifically, the problem is focused on:

1. Understanding the nature of arithmetic sequences: An arithmetic sequence is a sequence of numbers such that the difference of any two successive members is a constant.

2. Identifying and analyzing subsequences: In the context of this problem, a subsequence is a sequence that can be derived from the array by deleting some or no elements without changing the order of the remaining elements.

3. Applying dynamic programming techniques: Dynamic programming is a method for efficiently solving a broad range of search and optimization problems which exhibit the characteristics of overlappling subproblems and optimal substructure.

4. Managing constraints: There are constraints on the length of the input array and the values it can hold, as well as the difference. The array can be quite large, so an efficient solution is necessary.

5. Optimization: The goal is to identify the longest arithmetic subsequence in the given array with a specified difference, suggesting a need to optimize the subsequence length.

The solution to this problem would have applicability in a number of computational and algorithmic contexts where sequence analysis and dynamic programming techniques are required.

Establishing the boundaries of a problem involves understanding the problem’s constraints and clearly defining the inputs and outputs. Here’s how we can establish the boundary for this problem:

Input:

1. An integer array arr of length n (1 <= n <= 10^5), with each element in the range of -10^4 to 10^4.
2. An integer difference in the range of -10^4 to 10^4.

Output:

1. The output should be an integer that represents the length of the longest arithmetic subsequence in arr where the difference between adjacent elements in the subsequence equals difference.

Functional Constraints/Requirements:

1. A subsequence is a sequence that can be derived from arr by deleting some or no elements without changing the order of the remaining elements.
2. An arithmetic sequence is a sequence of numbers such that the difference of any two successive members is a constant (difference in this case).
3. We need to find the length of the longest subsequence in arr which is an arithmetic sequence.
4. If multiple such subsequences exist, we should return the longest one.
5. If no such subsequence exists, the length would be 0.

Performance Constraints:

1. The large upper limit on the size of arr (10^5 elements) indicates that a naive solution might not be efficient enough. Therefore, we need to design our solution keeping time and space complexity in mind, suggesting that we should aim for an algorithm with a time complexity better than O(n^2).

Setting these boundaries will help in understanding the problem’s requirements clearly and ensuring that the solution addresses all the constraints and edge cases.

Problem Classification

This problem is a combinatorial problem, specifically a variant of the Longest Increasing Subsequence problem. It falls into the domain of Dynamic Programming because it involves finding an optimal solution by breaking the problem down into simpler, overlapping subproblems.

‘What’ Components:

1. We are given an integer array, arr.
2. We are also given an integer, difference.
3. We need to find a subsequence in arr such that it is an arithmetic sequence and the difference between adjacent elements in this subsequence is equal to difference.
4. A subsequence in arr can be formed by deleting some or no elements without changing the order of the remaining elements.
5. We are asked to return the length of the longest such subsequence.

Based on the ‘What’ components, this problem can be classified as a Dynamic Programming problem. The subproblems in this case are the longest subsequences ending at each index of the array, and these subproblems overlap because a subsequence ending at index i can be extended to a subsequence ending at index j > i by adding some elements from the array. The problem is to maximize the length of such a subsequence, which is a hallmark of optimization problems that Dynamic Programming is often used to solve.

Further, the problem has a clear optimal substructure, because the longest subsequence ending at index j includes the longest subsequence ending at some previous index i such that arr[j] - arr[i] = difference. This property of optimal substructure is another key characteristic of Dynamic Programming problems.

Thus, this problem is a combination of Sequence/Array manipulation, Combinatorics (specifically finding longest subsequence), and Dynamic Programming.

Distilling the Problem to Its Core Elements

1. Fundamental Concept: The problem is based upon the concept of “Dynamic Programming”. It involves maintaining an auxiliary data structure, in this case, a dictionary or hash map, to store and update the length of the longest arithmetic subsequence ending at each number in the array. The principle of “Optimal Substructure” underlies this problem, as the longest arithmetic subsequence ending at a number can be built upon the longest arithmetic subsequence ending at a previous number with the same difference.

2. Simple Description: Imagine you have a list of numbers and a target difference. You’re trying to find the longest chain of numbers in this list where each number is the previous number plus the target difference. You can skip numbers, but you can’t change their order.

3. Core Problem: The core problem is to find the longest subsequence in the given list of numbers where the difference between any two consecutive numbers in the subsequence is equal to the given difference.

4. Key Components: The key components of the problem are:

   • the input array of integers
   • the target difference
   • the concept of an arithmetic sequence or subsequence
   • the length of the longest such subsequence

5. Minimal Set of Operations:

   • Iterate over the array.
   • For each number, calculate the previous number in the sequence (current number - difference) and check if it exists in the data structure.
   • If it does, the longest sequence ending at the current number is one more than the longest sequence ending at the previous number.
   • If it doesn’t, the longest sequence ending at the current number is 1 (the number itself).
   • Keep track of the longest sequence found so far.

Visual Model of the Problem

You can visualize the problem using an array along with a hash map to demonstrate how the lengths of the longest arithmetic subsequences are tracked and updated.

Let’s use an example:

arr = [1,5,7,8,5,3,4,2,1], difference = -2

For simplicity, we represent the hash map as a list of tuples, each representing a key-value pair where the key is a number from arr and the value is the length of the longest subsequence ending at that number.

At the start, both arr and the hash map are empty:

arr: []
map: []

As we begin to iterate over the array, we update the hash map as follows:

arr: [1]
map: [(1, 1)]  # The longest subsequence ending at 1 is [1].

arr: [1, 5]
map: [(1, 1), (5, 1)]  # The longest subsequences ending at 1 and 5 are [1] and [5].

arr: [1, 5, 7]
map: [(1, 1), (5, 1), (7, 2)]  # The longest subsequence ending at 7 is [5, 7].

arr: [1, 5, 7, 8]
map: [(1, 1), (5, 1), (7, 2), (8, 1)]  # The longest subsequence ending at 8 is [8].

arr: [1, 5, 7, 8, 5]
map: [(1, 1), (5, 3), (7, 2), (8, 1)]  # The longest subsequence ending at 5 is [7, 5, 3].

arr: [1, 5, 7, 8, 5, 3]
map: [(1, 1), (5, 3), (7, 2), (8, 1), (3, 4)]  # The longest subsequence ending at 3 is [7, 5, 3, 1].

arr: [1, 5, 7, 8, 5, 3, 4]
map: [(1, 1), (5, 3), (7, 2), (8, 1), (3, 4), (4, 1)]  # The longest subsequence ending at 4 is [4].

arr: [1, 5, 7, 8, 5, 3, 4, 2]
map: [(1, 1), (5, 3), (7, 2), (8, 1), (3, 4), (4, 2), (2, 5)]  # The longest subsequence ending at 2 is [7, 5, 3, 1, -1].

arr: [1, 5, 7, 8, 5, 3, 4, 2, 1]
map: [(1, 1), (5, 3), (7, 2), (8, 1), (3, 4), (4, 2), (2, 5), (1, 6)]  # The longest subsequence ending at 1 is [7, 5, 3, 1, -1, -3].

The longest subsequence found is of length 6: [7, 5, 3, 1, -1, -3].

This visualization helps illustrate the process of dynamically tracking and updating the lengths of the longest subsequences, and how the lengths of subsequences ending at later numbers are built upon those ending at earlier numbers.

Problem Restatement

We have a list of numbers (we’ll call this “arr”) and a target difference. Our task is to find the longest sequence of numbers within this list where the difference between any two consecutive numbers in the sequence is equal to the target difference. The sequence doesn’t have to be continuous in the list but the order of the numbers in the sequence should be the same as their order in the list.

In other words, we are looking for the longest subsequence in “arr” where each pair of successive numbers subtract to give the target difference. The sequence can skip numbers in the list, but it can’t reorder them.

The problem gives us the length of the list (“arr.length”) which will be between 1 and 105, and also tells us that the numbers in the list and the target difference will be between -104 and 104. We need to take these constraints into account while solving the problem.

Abstract Representation of the Problem

We’re given an ordered list (sequence) of integers and a specific integer that we’ll refer to as “delta”. Our task is to identify the longest sub-sequence within this list, in which every pair of consecutive numbers has a difference equal to “delta”.

It’s important to note that a sub-sequence refers to an ordered subset of the original sequence that maintains the relative order of elements. This sub-sequence doesn’t have to be contiguous, meaning elements in the sub-sequence may not appear next to each other in the original sequence.

In formal terms, given a sequence S = {s1, s2, …, sn}, a delta value d, we need to find the longest subsequence S’ = {si1, si2, …, sim} where for every k in [1, m-1], si(k+1) - sik = d. The goal is to maximize m, the length of the sub-sequence.

The constraints tell us about the size of the sequence and the range of values in it, which are important for evaluating the performance and complexity of potential solutions.

Terminology

Here are a few technical terms and concepts crucial to understanding this problem:

1. Array (or List): An array (or in Python, a list) is a data structure that contains a group of elements. In this problem, the array represents the sequence of courses.

2. Subsequence: A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements. In this context, a subsequence represents a sequence of courses we are interested in which have the property of being an arithmetic sequence.

3. Arithmetic Sequence: An arithmetic sequence is a sequence of numbers such that the difference of any two successive members is a constant. In this problem, we’re looking for subsequences where the difference between any two consecutive elements is equal to a given “difference” value.

4. Longest: The term longest in this context refers to the subsequence with the most number of elements. Our goal is to find the longest arithmetic subsequence.

5. Difference: In the context of an arithmetic sequence, difference refers to the constant value between any two consecutive elements in the sequence. We’re given this value in the problem and our goal is to find the longest subsequence where this property holds true.

Understanding these concepts and how they fit together is key to solving this problem. They form the foundation of the problem’s requirements and constraints.

Problem Simplification and Explanation

Think of the given array as a sequence of stepping stones in a river and the task is to reach the stone that allows you to make the longest possible journey. Each stepping stone represents a number in the array, and you can only move to the next stone if the height difference between your current stone and the next one equals to a predefined value (this represents the difference parameter in the problem).

In more technical terms:

1. Array/List: The sequence of stepping stones in the river.
2. Subsequence: A possible path that you can take by stepping on the stones, without changing their order.
3. Arithmetic Sequence: A path where the height difference between each consecutive stone is constant.
4. Longest: The longest possible path you can take that follows the rules.
5. Difference: The constant height difference between each consecutive stone in a valid path.

In this problem, the main concept is the “subsequence” and its properties (being an arithmetic sequence of a given difference). The aim is to find the longest such subsequence in the given array, which translates to finding the longest possible valid path in the stepping stones analogy.

Constraints

Here are some key observations from the problem that can be exploited for an efficient solution:

1. Subsequence, not Subarray: The problem asks for the longest subsequence, not subarray. A subsequence does not require elements to be contiguous (next to each other) in the array, which allows more flexibility. If it were a subarray, elements would have to be in consecutive positions, which would be more restrictive.

2. Arithmetic Sequence: The fact that we are looking for an arithmetic sequence where the difference between adjacent elements in the subsequence equals a given difference allows us to make progress in a systematic way.

3. Difference Constraints: The difference is within the range of -10^4 to 10^4. We can leverage this in our solution, for example, by using the difference to index into an array or hashmap.

4. Possible Negative Difference: The difference between elements can be negative. This indicates that the sequence does not necessarily have to be increasing; it can be decreasing as well.

5. Repetition of Elements: The problem does not specify that the elements in the array will be distinct. This means that the same number could appear multiple times in the array, and this could be a part of the longest subsequence if it fits the difference criteria.

6. Length of Array: The length of the array could be up to 10^5, which means the solution needs to be at least O(n) or better. Any solution with a higher time complexity could result in a timeout error.

These observations suggest a dynamic programming approach could be viable, as we could potentially calculate the longest subsequence ending at each position and use these results to calculate subsequent values.

By analyzing the constraints, we can infer a few insights that might shape our approach to the problem:

1. Large Input Size: The length of the input array arr can go up to 10^5. This tells us that our algorithm needs to be efficient, ideally having a time complexity of O(n) or O(n log n) at worst. Anything slower will likely exceed the time limit on the test cases.

2. Range of Values: The elements in the array and the difference can range from -10^4 to 10^4. This wide range can be an indicator that the solution will need to handle negative numbers and large numbers effectively.

3. Arithmetic Sequence with Fixed Difference: The problem is asking for an arithmetic sequence with a fixed difference. This is a key insight because it narrows down the type of sequences we are looking for.

4. Negative Difference: The difference can be negative, which implies that we are looking for arithmetic sequences where each subsequent number can be smaller than the previous one.

5. Subsequence, not Subarray: The requirement for a subsequence (not necessarily contiguous) instead of a subarray (contiguous) allows for greater flexibility when determining the longest sequence.

6. Size of the Longest Subsequence: Since the length of the subsequence can’t be more than the length of the original array, we have an upper limit for the output. This information might be useful in setting initial values or boundaries in our algorithm.

These insights suggest that a dynamic programming or hashing-based approach could be a good starting point since these techniques often work well with problems requiring efficient solutions and manipulation of numerical values.

Case Analysis

I will create a few test cases, cover a variety of conditions, edge cases, and also identify what part of the problem they are highlighting:

1. Empty Sequence Case Input: arr = [], difference = 1 Output: 0 In this case, the array is empty so the longest arithmetic subsequence is 0.

2. Single Element Case Input: arr = [5], difference = 1 Output: 1 Even though there is only one element, it forms a subsequence of length 1.

3. All Elements Same Input: arr = [2,2,2,2,2], difference = 0 Output: 5 Since all the elements are the same and the difference is 0, the whole array forms an arithmetic subsequence.

4. Negative Difference Input: arr = [5,4,3,2,1], difference = -1 Output: 5 Here, the difference is negative which means we are looking for decreasing subsequences.

5. Multiple Possible Sequences Input: arr = [1,3,5,7,9,11,4,6,8,10], difference = 2 Output: 6 In this case, there are two possible subsequences - [1,3,5,7,9,11] and [4,6,8,10]. The function should return the length of the longer one.

6. Large Numbers Input: arr = [10000, 20000, 30000, 40000, 50000], difference = 10000 Output: 5 This is to ensure the function can handle large numbers correctly.

7. No Arithmetic Subsequence Case Input: arr = [1,3,5,7,9,2,4,6,8,10], difference = 3 Output: 1 In this case, no arithmetic subsequence of difference 3 exists, so the output is 1 as any single element forms a subsequence.

8. Subsequence Not Subarray Input: arr = [1, 100, 2, 200, 3, 300, 4, 400, 5, 500], difference = 1 Output: 5 This case tests if the solution correctly identifies subsequences that are not subarrays.

These cases cover a range of different scenarios and should give us confidence that the solution handles all edge cases if they all pass.

Visualizing these cases can be done by representing the elements of the array on a number line, and highlighting the elements that are part of the longest arithmetic subsequence. Here’s how you can visualize some of the test cases:

1. Empty Sequence Case The number line is empty.

2. Single Element Case The number line contains one point.

3. All Elements Same The number line contains multiple points at the same location.

4. Negative Difference The number line is descending.

5. Multiple Possible Sequences The number line has two separate ascending subsequences.

6. Large Numbers The number line has a wide range.

7. No Arithmetic Subsequence Case The number line has no subsequences with the desired difference.

8. Subsequence Not Subarray The number line has an arithmetic sequence with gaps in between.

For each case, you might also want to visualize the value of the difference. For example, in the “Negative Difference” case, you could visualize the difference as a negative distance on the number line. In the “Multiple Possible Sequences” case, you could visualize the difference as the distance between the elements in each subsequence.

This kind of visualization can help you understand what’s going on in each case and identify what you need to look for when finding the longest arithmetic subsequence.

Analyzing different cases gives us a few key insights:

1. Subsequence vs Subarray: This problem deals with subsequences, not subarrays. An arithmetic subsequence doesn’t necessarily have to be continuous in the array, so we need to consider all possible subsequences, not just contiguous ones.

2. Handling Negative Difference: The difference can be negative, which implies that the sequence can be decreasing. Our solution needs to handle this scenario.

3. All Elements Same: When all the elements are the same, and the difference is zero, the longest subsequence will be the length of the array itself.

4. No Arithmetic Subsequence Case: If no arithmetic subsequence with the given difference exists, our program should return 1 as per the problem statement, because every individual number can be considered a subsequence.

5. Handling Large Numbers: As the range of the input numbers and the difference can be quite large (-10^4 to 10^4), our solution needs to efficiently handle large numbers.

6. Length of Array: The length of the array could be as high as 10^5, meaning that our solution needs to be efficient in terms of both time and space complexity.

Understanding these key insights can help us come up with an optimized approach to solve the problem.

Identification of Applicable Theoretical Concepts

The problem can be simplified by applying the concept of Dynamic Programming (DP). Dynamic Programming is a method for solving problems by breaking them down into simpler subproblems and utilizing the fact that the optimal solution to the overall problem depends upon the optimal solution to its subproblems.

In the context of this problem, the Dynamic Programming approach can be used as follows:

1. State Definition: The state can be defined as a hashmap where the key is a number from the input array and the value is the maximum length of an arithmetic subsequence ending with that number.

2. State Transition: If we are at the number x in the array, we check if there exists a sequence ending with the number x-difference. If it does exist, we can append x to this sequence to form a new sequence. Therefore, the maximum length of the arithmetic sequence ending with x is one more than the maximum length of the sequence ending with x-difference.

This approach effectively reduces the problem to finding and updating the longest subsequence that ends with each number, which can be done in linear time by traversing the array once.

Remember, the key to applying DP is recognizing that the problem can be broken down into overlapping subproblems, and that the solution to the original problem can be found by combining the solutions of these smaller subproblems.

Also, when applying DP, it’s essential to define the state and state transition equation correctly, as the entire solution builds on these definitions.

Simple Explanation

Let’s think about this problem in terms of a game of dominoes.

In a game of dominoes, each piece has two numbers on it, and you can only place a piece next to another one if the numbers on their adjoining sides match. The objective of the game can be to create the longest possible chain of dominoes.

Now, let’s imagine a variant of this game, where each piece only has one number and the rule is that you can only place a piece next to another one if the difference between their numbers is a certain value. For instance, if the difference must be 1, you could place a ‘2’ next to a ‘1’, or a ‘3’ next to a ‘2’, and so on. The goal of this game is to create the longest chain possible.

Our problem is similar to this game. We’re given a set of “dominoes” (the numbers in the array) and a “difference” value. We need to figure out the longest “chain” we can make where the difference between each “domino” and the one next to it is the given difference.

And just like in the game, we can choose to use or not use any “domino”, and the “dominoes” don’t need to be used in the order they were given to us, but we can’t rearrange the “dominoes” within our “chain” once we’ve chosen them.

The challenge is figuring out the best way to choose and arrange our “dominoes” to make the longest “chain” possible, given these rules.

Problem Breakdown and Solution Methodology

Let’s break down the approach to solving this problem.

1. Understanding the Problem: The first thing we need to do is understand what the problem is asking us. We’re given an array of integers, and a difference value. We need to find the longest subsequence within the array where the difference between each consecutive pair of numbers in the subsequence is equal to the given difference.

2. Observations: This problem deals with sequences and their properties, specifically arithmetic sequences. An arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

3. Approach: A common approach to these types of problems is using dynamic programming. We could use a hashmap to store the longest sequence ending at each number.

4. Algorithm Steps:

   • Create an empty dictionary (let’s call it dp). This will serve as our dynamic programming table. Each key-value pair in the dictionary represents a number in the array (the key) and the length of the longest arithmetic subsequence that ends with that number (the value).

   • Iterate over the array. For each number num:

     • If num - difference is in dp, then we can extend the longest arithmetic subsequence ending at num - difference by num. So, we set dp[num] to be dp[num - difference] + 1.
     • If num - difference is not in dp, then the longest arithmetic subsequence ending at num is just num itself, with length 1. So, we set dp[num] to be 1.

   • Finally, we return the maximum value in dp. This represents the length of the longest arithmetic subsequence in the array.

5. Effect of Changes: If the difference value changes, it might affect the longest subsequence length. If the array values change or their order changes, that might also affect the result.

6. Example:

Let’s demonstrate this approach with an example. Consider arr = [1,5,7,8,5,3,4,2,1] and difference = -2.

• Initialize dp = {}.
• Iterating over arr:
  • For num = 1, 1 - (-2) = 3 is not in dp, so dp[1] = 1.
  • For num = 5, 5 - (-2) = 7 is not in dp, so dp[5] = 1.
  • For num = 7, 7 - (-2) = 9 is not in dp, so dp[7] = 1.
  • For num = 8, 8 - (-2) = 10 is not in dp, so dp[8] = 1.
  • For num = 5, 5 - (-2) = 7 is in dp, so dp[5] = dp[7] + 1 = 2.
  • For num = 3, 3 - (-2) = 5 is in dp, so dp[3] = dp[5] + 1 = 3.
  • For num = 4, 4 - (-2) = 6 is not in dp, so dp[4] = 1.
  • For num = 2, 2 - (-2) = 4 is in dp, so dp[2] = dp[4] + 1 = 2.
  • For num = 1, 1 - (-2) = 3 is in dp, so dp[1] = dp[3] + 1 = 4.
• The maximum value in dp is 4, so the length of the longest arithmetic subsequence in arr with a difference of -2 is 4.

Inference of Problem-Solving Approach from the Problem Statement

Let’s identify the key terms and concepts in this problem:

1. Array: This problem is about an array of integers. The order of the elements matters because a subsequence must maintain the original order of the array. We iterate over the array in order.

2. Subsequence: A subsequence of an array is a sequence that can be derived from the array by deleting some (possibly no) elements without changing the order of the remaining elements. We are looking for a specific kind of subsequence (an arithmetic sequence) and the longest such subsequence.

3. Arithmetic Sequence: This is a sequence of numbers such that the difference of any two successive members is a constant. We use this property in our dynamic programming approach. When we look at each number, we consider whether it can extend an existing arithmetic subsequence.

4. Difference: This is the difference between any two successive members in our target arithmetic sequence. It’s used to look back from the current number to see if an arithmetic sequence can be extended.

5. Dynamic Programming: Dynamic programming is a method for solving complex problems by breaking them down into simpler subproblems. It is used when the subproblems overlap. In this problem, the subproblems are finding the longest arithmetic subsequence ending at each number. These subproblems overlap because they all depend on the results of the previous numbers in the array.

To visualize this problem, we could draw a table where each row represents a number in the array and each column represents a possible difference. We would fill in each cell with the length of the longest arithmetic subsequence ending at that number with that difference. However, since the differences can be large, this may not be practical. Instead, we use a dictionary (dp) where the keys are the numbers and the values are the lengths of the longest arithmetic subsequences ending at those numbers. This provides a compact and efficient way to track our progress as we iterate through the array.

How did you infer from the problem statement that this problem can be solved using ?

Simple Explanation of the Proof

The proof for this algorithm is based on the principles of dynamic programming and induction.

The intuition behind this algorithm is that we’re trying to build the longest arithmetic sequence possible as we iterate over each number in the array. At each number, we ask ourselves: “Is there a previous number in the array such that the difference between it and the current number is equal to the target difference? If so, we can extend the longest arithmetic sequence ending at that previous number by including the current number.”

We store the longest arithmetic sequence ending at each number in a dictionary (dp). For each number arr[i], we look up arr[i] - difference in our dictionary. If it exists, then dp[arr[i]] should be dp[arr[i] - difference] + 1. Otherwise, the longest arithmetic sequence ending at arr[i] is just arr[i] itself, or 1 in length. Our final answer is the maximum length of all arithmetic sequences that we’ve found.

Here’s the induction step of the proof: Assume that for every index j less than i, we’ve correctly computed the length of the longest arithmetic subsequence ending at arr[j]. We’re trying to compute the length of the longest arithmetic subsequence ending at arr[i].

1. If arr[i] - difference is in dp, then there exists an arithmetic subsequence ending at arr[i] - difference with the same difference, and we can extend this subsequence by including arr[i]. The length of this extended subsequence would be dp[arr[i] - difference] + 1.

2. If arr[i] - difference is not in dp, then we cannot extend any existing subsequence to include arr[i], so the longest arithmetic subsequence ending at arr[i] is just arr[i] itself, which is 1 in length.

So at every step i, we’re making the optimal decision based on the results of previous steps. This is the essence of dynamic programming. Since we correctly compute dp[arr[i]] for all i, our final result, which is the maximum of all dp values, is also correct.

This algorithm is efficient because it only needs to look at each number once, and the lookups in dp take constant time on average. So the time complexity is linear in the size of the array. The space complexity is also linear because we need to store a dp entry for each number in the array.

Stepwise Refinement

1. Stepwise Refinement of the Approach:

Step 1: Initialize a dictionary (dp) to store the length of the longest arithmetic subsequence ending at each number in the array. This will serve as our dynamic programming table.

Step 2: Iterate over the array from left to right. For each number arr[i], do the following:

a. Check if `arr[i] - difference` exists in the `dp` dictionary. 
   
   - If it does, it means that we can extend the longest arithmetic subsequence ending at `arr[i] - difference` by including `arr[i]`. Update `dp[arr[i]]` to be `dp[arr[i] - difference] + 1`.

   - If it does not, it means that the longest arithmetic subsequence ending at `arr[i]` is `arr[i]` itself. Update `dp[arr[i]]` to be `1`.

Step 3: After the loop, the maximum of all the values in dp will be the length of the longest arithmetic subsequence in the array. Return this maximum value.

2. Distilling the High-Level Solution into Granular Steps:

The step-by-step approach detailed above already describes the solution in granular, actionable steps. Each step corresponds to a specific operation or a series of operations that can be translated directly into code.

3. Parts of the Problem That Can be Solved Independently:

The steps outlined above are sequential and each step depends on the result of the previous step. Hence, there are no parts of the problem that can be solved independently in parallel.

However, note that for each number arr[i], the computation of dp[arr[i]] is independent of the computation of dp[arr[j]] for j != i. This is because each dp value only depends on the dp value of arr[i] - difference, not on other dp values.

4. Repeatable Patterns within Our Solution:

The main repeatable pattern in our solution is the dynamic programming loop where for each number, we either extend the longest arithmetic subsequence ending at arr[i] - difference by including arr[i], or start a new subsequence ending at arr[i]. This pattern is repeated for each number in the array.

Solution Approach and Analysis

This problem can be compared to a game of connecting dots, where each dot is a number in the input array and you can only connect two dots if the difference between their values equals the given difference. The goal of the game is to form the longest chain of connected dots.

Here is the detailed process of solving this problem:

1. Set Up a Dictionary: Initialize a dictionary (dp) that will serve as a map between a number in the array and the length of the longest chain of dots (or arithmetic subsequence) that ends with this number. At the beginning, each number forms a chain by itself, so we could say that all numbers have a chain length of 1, but we will only add numbers to the dictionary as we encounter them in the array.

2. Iterate Over the Array: Go through the numbers in the array one by one. For each number, check if there is a dot you could have connected to. In other words, check if the number that would come just before it in an arithmetic sequence (current number - difference) is in the dp dictionary.

   • If it is, connect the current dot to that chain by setting dp[current number] to be dp[current number - difference] + 1. This essentially extends the chain ending at current number - difference by one dot.

   • If not, start a new chain with the current number as the only dot by setting dp[current number] to be 1.

3. Find the Longest Chain: After going through all numbers in the array, look in the dp dictionary for the longest chain, i.e., the maximum value in the dictionary, and return this as the length of the longest arithmetic subsequence in the array.

Now let’s apply this process to an example:

• Let’s say the input array is [1,5,7,8,5,3,4,2,1] and the difference is -2.

• We start by initializing an empty dictionary for dp.

• We then start iterating over the array. When we encounter 1, we don’t find 1 - (-2) = 3 in dp, so we start a new chain by setting dp[1] = 1.

• The next number is 5, we also don’t find 5 - (-2) = 7 in dp, so dp[5] = 1.

• The next number is 7, we don’t find 7 - (-2) = 9 in dp, so dp[7] = 1.

• The next number is 8, we don’t find 8 - (-2) = 10 in dp, so dp[8] = 1.

• The next number is 5 again, but this time we find 5 - (-2) = 7 in dp, so dp[5] = dp[7] + 1 = 2.

• We keep doing this until we reach the end of the array.

• Finally, we find that the maximum value in dp is 4, which corresponds to the chain [7,5,3,1]. Therefore, the length of the longest arithmetic subsequence in the array is 4.

Changes in the input array or the difference could affect the solution in various ways. For example, if all numbers in the array are the same, then the longest arithmetic subsequence would be the entire array if the difference is 0, or any single number if the difference is not 0. If the array is sorted in increasing order, then the longest arithmetic subsequence would be the entire array if the difference is arr[1] - arr[0], or any single number if the difference is not arr[1] - arr[0].

Identify Invariant

An invariant is a condition that can be relied upon to be true during the execution of a program, or during some portion of it.

In the context of this problem, the following can be considered as the invariant:

“The dictionary dp always stores the longest length of an arithmetic subsequence ending at each element encountered so far in the array.”

This means that at any point during the iteration through the array, for every key-value pair in the dictionary dp, the value is the length of the longest arithmetic subsequence with the given difference that ends with the key, considering only the elements that have been encountered in the array so far.

This invariant is crucial for the correct working of the algorithm. The algorithm maintains this invariant by properly updating the dictionary dp at each step. When the algorithm finishes iterating through the array, thanks to the maintained invariant, we can be sure that the maximum value in the dictionary dp is the length of the longest arithmetic subsequence in the entire array.

Identify Loop Invariant

In the context of the problem, a loop invariant can be the following:

“For each iteration i in the array, the dictionary dp keeps track of the longest arithmetic subsequence with the given difference, ending at element arr[i].”

Here, the loop invariant is maintained for every iteration in the array. Initially, when there are no elements processed (i=0), the dictionary dp is empty and the invariant holds as there are no subsequences yet.

During the loop, at each iteration, the dictionary dp is updated such that it keeps track of the longest arithmetic subsequence ending at arr[i] with the given difference. This is done by looking up the length of the longest subsequence ending at arr[i] - difference and incrementing it by one.

Finally, when the loop terminates after all elements in the array are processed, due to the maintained invariant, the maximum value in the dictionary dp would be the length of the longest arithmetic subsequence in the entire array.

Loop invariant and invariant in the context of this problem have the same concept. Both refer to a condition that initially holds true and remains unchanged as the loop or the process progresses.

In the case of the problem “Longest Arithmetic Subsequence of Given Difference”, the loop invariant could be phrased as:

“For each element in the array, the dictionary (or any data structure used) maintains the longest subsequence ending at that element, which is in arithmetic progression with a common difference given.”

This condition initially holds true when the dictionary is empty before we start scanning the array. It remains true as we traverse the array and update the dictionary based on the given rules (updating the length of the longest subsequence ending at current element). Finally, the condition still holds true after we’ve scanned the entire array. At this point, the dictionary’s maximum value gives us the length of the longest arithmetic subsequence in the array.

So, both invariants in general and loop invariants specifically refer to a condition or set of conditions that remain unchanged as the algorithm progresses, providing a means to prove the algorithm’s correctness.

Thought Process

This problem falls under the domain of Dynamic Programming (DP). Here are the step-by-step thought processes involved in solving this type of problem:

1. Understand the problem: The problem asks us to find the longest arithmetic subsequence in the given array with a common difference. A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.

2. Identify the sub-problems: As it involves finding subsequences and it has a longest/maximum component, Dynamic Programming might be a suitable approach. In this case, we are trying to find the longest subsequence for each element ending at that element.

3. Create an appropriate data structure: To keep track of the longest arithmetic subsequence for each number, we can use a dictionary in Python. The key would be the number, and the value would be the length of the longest arithmetic subsequence ending with that number.

4. Formulate a recurrence relation: While traversing the array from left to right, for each number we check if “current number - difference” exists in our dictionary. If it does, it means that the current number can extend an arithmetic subsequence which ends with “current number - difference”. So, we update the length of the longest arithmetic subsequence for the current number.

5. Initialize the data structure and solve the sub-problems: We initialize an empty dictionary and fill it according to the recurrence relation we’ve identified.

6. Extract the final result: After we’ve filled our dictionary, the answer would be the maximum value in the dictionary as it represents the length of the longest arithmetic subsequence in the array.

Here’s how you can implement this approach in Python:

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class Solution:
    def longestSubsequence(self, arr: List[int], difference: int) -> int:
        dp = {}
        for num in arr:
            if num - difference in dp:
                dp[num] = dp[num - difference] + 1
            else:
                dp[num] = 1
        return max(dp.values())

Cues in the problem statement that suggest this approach include keywords like “longest”, “subsequence”, and “difference”. These terms indicate that we’re looking for a maximum length sequence with a specific property (arithmetic sequence with given difference), which is a common scenario in Dynamic Programming problems.

Establishing Preconditions and Postconditions

1. Parameters:

   • The method takes two parameters: arr which is a list of integers and difference which is an integer.
   • arr represents the integer array given in the problem statement.
   • difference represents the difference between adjacent elements in the arithmetic subsequence.

2. Preconditions:

   • Before this method is called, it is assumed that arr is a non-empty list containing at least one integer.
   • The length of the list arr is between 1 and 105 (inclusive) and each element in arr as well as difference is between -10^4 and 10^4 (inclusive).

3. Method Functionality:

   • The method is expected to find the length of the longest subsequence in arr which is an arithmetic sequence such that the difference between adjacent elements in the subsequence equals difference.
   • The method does not interact with or change any program state outside its scope. It only works with the inputs provided and does not have side effects.

4. Postconditions:

   • After the method has been called and returned, the state of the program or the values of the parameters remain unchanged as the method does not modify any of its inputs or any global state.
   • The return value represents the length of the longest arithmetic subsequence in arr with a difference of difference.
   • The method does not have any side effects.

5. Error Handling:

   • If the preconditions are not met, the Python interpreter will raise an exception. For example, if arr is not a list, a TypeError will be raised. If arr is an empty list, the max() function will raise a ValueError because it cannot operate on an empty sequence.
   • The method itself does not handle errors or exceptions; it assumes that the inputs are valid and meet the preconditions.

Problem Decomposition

1. Problem Understanding:

   • We have an array of integers and a difference value. We need to find the longest subsequence of this array where the difference between consecutive elements equals the given difference. The subsequence doesn’t have to be contiguous in the array, but must maintain the same relative order.

2. Initial Breakdown:

   • The major parts of this problem are:
     1. Iterating through the array
     2. Keeping track of the longest arithmetic subsequence found so far
     3. Updating the longest arithmetic subsequence if a longer one is found

3. Subproblem Refinement:

   • Iterating through the array can be straightforward with a for-loop
   • Keeping track of the longest arithmetic subsequence can be done using a dictionary, where the keys are the array elements and the values are the lengths of the longest arithmetic subsequences ending at those elements.
   • Updating the longest arithmetic subsequence involves comparing the current longest length and the possible new length if the current element can extend an existing arithmetic subsequence.

4. Task Identification:

   • Two tasks that are repeated are looking up an element in the dictionary and updating the value in the dictionary. These could be abstracted into tasks like get_subsequence_length(element) and update_subsequence_length(element, length).

5. Task Abstraction:

   • The task get_subsequence_length(element) is abstracted enough as it is clear what it does - gets the length of the longest arithmetic subsequence ending at element.
   • The task update_subsequence_length(element, length) is also abstracted enough as it clearly shows that it updates the length of the longest arithmetic subsequence ending at element to length.

6. Method Naming:

   • Based on the tasks identified and abstracted, the methods can be named as get_subsequence_length(element) and update_subsequence_length(element, length).

7. Subproblem Interactions:

   • The iteration through the array should be done first.
   • Inside each iteration, get_subsequence_length(element) is called first to check if the current element can extend an existing arithmetic subsequence.
   • Then, update_subsequence_length(element, length) is called to update the length of the longest arithmetic subsequence ending at the current element.
   • These steps are repeated for every element in the array. There are dependencies between these steps as the output of get_subsequence_length(element) is used as input to update_subsequence_length(element, length).

From Brute Force to Optimal Solution

Let’s begin with the brute force solution.

The most straightforward way to solve this problem is to generate all possible subsequences of the array and check which one is the longest arithmetic sequence. Here’s how this could look:

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def longest_arithmetic_subsequence(arr, difference):
    all_subsequences = generate_all_subsequences(arr)
    max_length = 0
    for subsequence in all_subsequences:
        if is_arithmetic(subsequence, difference):
            max_length = max(max_length, len(subsequence))
    return max_length

Here, generate_all_subsequences(arr) generates all possible subsequences of arr, and is_arithmetic(subsequence, difference) checks whether subsequence is an arithmetic sequence with a difference of difference.

This solution works, but it’s highly inefficient. The time complexity is O(2^n), because we’re generating all subsequences, and then we also have to check each subsequence if it’s arithmetic or not which could also take up to O(n) time. Hence, the total time complexity could go up to O(n*2^n). The space complexity is also high, because we’re storing all subsequences.

Now, let’s think about how we can optimize this.

We can make a crucial observation that can help us - an arithmetic subsequence can be extended by adding an element to its end if the difference between this element and the last element in the subsequence is equal to the given difference. This means we can keep track of the longest arithmetic subsequence that ends at each element while iterating through the array, and then simply take the maximum of these lengths.

This leads to a dynamic programming solution where we use a dictionary to store the length of the longest arithmetic subsequence ending at each element. We can then update this dictionary as we iterate through the array:

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def longest_arithmetic_subsequence(arr, difference):
    dp = {}
    for num in arr:
        if num - difference in dp:
            dp[num] = dp[num - difference] + 1
        else:
            dp[num] = 1
    return max(dp.values())

The time complexity of this solution is O(n), because we’re doing a single pass through the array. The space complexity is also O(n), because we’re storing the longest arithmetic subsequence length for each element in a dictionary. This is a significant improvement over the brute force solution.

Code Explanation and Design Decisions

1. Initial parameters:

   The initial parameters are arr and difference. arr is an integer array representing the sequence of numbers where we need to find the longest arithmetic subsequence, and difference is the common difference of the arithmetic subsequence.

2. Primary loop:

   The primary loop in the optimized solution iterates over each element in the arr. Each iteration represents the process of trying to extend an existing arithmetic subsequence by the current number (or starting a new one if it cannot extend any). It advances the solution by possibly increasing the length of the longest arithmetic subsequence found so far.

3. Conditions or branches within the loop:

   The condition if num - difference in dp checks if there exists a previous number in the array such that the difference between the current number and the previous number is equal to the given difference. This is the condition for extending an arithmetic subsequence. The logical reasoning is that, to extend an arithmetic subsequence with a common difference, the difference between the current number and the last number in the subsequence must be equal to the common difference.

4. Updates or modifications to parameters within the loop:

   The modification dp[num] = dp[num - difference] + 1 extends the longest arithmetic subsequence ending with num - difference by adding num. This reflects the change in state of the solution because it updates the length of the longest arithmetic subsequence found so far. The else branch dp[num] = 1 starts a new arithmetic subsequence with num when it cannot extend any existing subsequences.

5. Invariant:

   The invariant in the code is that for each num in arr, dp[num] always represents the length of the longest arithmetic subsequence ending with num found so far. This invariant is maintained throughout the code and it ensures we correctly find the length of the longest arithmetic subsequence in the array.

6. Final output:

   The final output of the code is the maximum value in the dictionary dp. This represents the length of the longest arithmetic subsequence in arr with the common difference difference. It satisfies the problem’s requirement by providing the length of the longest arithmetic subsequence.

Coding Constructs

1. High-Level Problem-Solving Strategies:

   The code uses dynamic programming as its main problem-solving strategy. The dynamic programming approach builds up a solution by solving smaller subproblems and using these solved subproblems to solve larger ones. The solution to each subproblem is stored in a dictionary for later use, ensuring that each subproblem is solved only once.

2. Purpose of the Code:

   To explain to a non-programmer: this code is like a game where you are given a sequence of numbers and a difference, and your task is to find the longest sequence within the given sequence where the difference between any two consecutive numbers is the same and matches the given difference.

3. Logical Elements or Constructs:

   The constructs used in this code, independent of any programming language, include iteration (looping through the elements of the array), condition checking (if-else statements), and a data structure to store intermediate results (a dictionary or equivalent).

4. Algorithmic Approach:

   In plain English, the approach of the code is to loop through each number in the sequence, for each number it checks if there exists a previous number in the sequence such that their difference equals the given difference. If so, it extends the longest sequence ending with the previous number by the current number, otherwise, it starts a new sequence with the current number. It keeps track of the length of the longest sequence ending with each number, and finally returns the maximum length.

5. Key Steps or Operations:

   The key operations the code performs are: iterating over the input sequence, checking the existence of the required difference, updating the lengths of the sequences in the dictionary, and finding the maximum length among all sequences.

6. Algorithmic Patterns or Strategies:

   The code employs a dynamic programming strategy. This approach involves breaking down a problem into smaller subproblems, solving each subproblem only once, and storing their solutions in case they are needed later. In this case, the subproblems are the longest arithmetic subsequences ending with each number in the sequence.

Language Agnostic Coding Drills

1. Identifying Distinct Concepts:

   a. Iteration (Looping): This is a basic concept in programming, which involves traversing through data structures like arrays, lists, etc. It forms the base for many complex programming tasks.

   b. Condition Checking: An essential part of programming, where you decide what to do based on certain conditions.

   c. Data Structure Usage (Dictionaries/Maps): The usage of specific data structures to store and manipulate data. In this case, dictionaries or maps are used.

   d. Dynamic Programming: This is a method for solving complex problems by breaking them down into simpler subproblems. The solution to each subproblem is stored for later use.

2. Order of Increasing Difficulty:

   a. Iteration (Looping): This is typically one of the first things taught in any programming language. It’s a fundamental concept but not particularly difficult.

   b. Condition Checking: Slightly more complex than iteration, condition checking requires logical thinking to identify what conditions need to be checked.

   c. Data Structure Usage (Dictionaries/Maps): Understanding when and how to use specific data structures is a bit more complex. Dictionaries/maps require understanding of key-value pairs and hash functions.

   d. Dynamic Programming: This is a more advanced concept. It requires a good understanding of problem-solving, recursion, and often involves a deep understanding of the problem at hand.

3. Problem-Solving Approach:

   a. First, understand that we need to traverse the list of numbers, i.e., we need to iterate through each element of the array. This makes us think of the ‘Iteration’ concept.

   b. While iterating, for each number, we need to find if there exists a previous number with a difference that equals the given difference. This introduces the ‘Condition Checking’ concept.

   c. To efficiently keep track of the longest sequence ending with each number, we need some kind of data structure. Here comes the ‘Data Structure Usage’ concept, specifically using dictionaries or maps.

   d. Realizing that the problem can be broken down into subproblems, where each subproblem represents the longest arithmetic subsequence ending at each number, we recognize the need for ‘Dynamic Programming’. The solution to each subproblem is the length of the longest arithmetic subsequence ending with the corresponding number. We use these solutions to solve larger subproblems, i.e., subsequences ending with later numbers.

In summary, we start with iterating through the array. For each element, we check for a certain condition. If it is met, we update our data structure (dictionary or map), which is being used to store the solution of each subproblem. The application of dynamic programming allows us to build the solution iteratively, ultimately leading us to the overall solution.

Targeted Drills in Python

1. Iteration (Looping):

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numbers = [1, 2, 3, 4, 5]
for num in numbers:
    print(num)

This code iterates through a list of numbers and prints each one.

2. Condition Checking:

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num = 7
if num % 2 == 0:
    print("Even")
else:
    print("Odd")

This code checks whether a number is even or odd.

3. Data Structure Usage (Dictionaries/Maps):

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nums_dict = {1: 'one', 2: 'two', 3: 'three'}
print(nums_dict[2])

This code creates a dictionary mapping integers to their string representations and then accesses a value.

4. Dynamic Programming:

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def fibonacci(n):
    fib_values = [0, 1] + [0]*n
    for i in range(2, n+1):
        fib_values[i] = fib_values[i-1] + fib_values[i-2]
    return fib_values[n]

print(fibonacci(10))

This code calculates the nth Fibonacci number using dynamic programming.

For our specific problem, all these concepts are crucial.

Assembling the Concepts:

Now, let’s see how we can put these drills together to solve our problem.

1. We begin by initializing a dictionary that will store the maximum length of an arithmetic sequence ending with each difference and number. This represents the ‘Data Structure Usage’ drill.

2. Then, we iterate through the array, which represents the ‘Iteration’ drill.

3. For each number, we iterate through all numbers that come before it and calculate the difference. For each difference, we update our dictionary using a condition check to ensure that we always store the maximum length of the arithmetic sequence for any difference. This brings in the ‘Condition Checking’ drill.

4. As we’re updating our dictionary, we are essentially solving and storing the results of subproblems, which represent sequences ending at each number. Later, we use these subproblem solutions to solve larger subproblems. This embodies the ‘Dynamic Programming’ drill.

Putting these together, we can form the overall solution to our problem.

Remember, the main objective is to understand the problem, identify the key concepts involved, create separate drills for each concept, and then assemble these drills to form the solution.

Q&A

[3,0,-3,4,-4,7,6] 3

expect is 2. but 3, 0,-3 has length 3, no?

The given problem statement specifies that we are to return the length of the longest strictly increasing arithmetic subsequence. This means that each subsequent element in the subsequence should be greater than the previous one.

In your example of 3, 0, -3, the sequence is not strictly increasing. Thus, it is not considered an arithmetic subsequence according to the problem statement.

On the other hand, the sequence 0, 3, 6 is strictly increasing with a common difference of 3, which makes it an arithmetic subsequence. Therefore, the expected length is 2 (i.e., the common difference).

I saw nobody using recursion to solve this problem. Can anyone share their recursion solutions?

Sure, you could use recursion to solve this problem, however, it’s important to note that this approach might be less efficient and potentially cause a stack overflow for large inputs due to the recursion depth. Nevertheless, for the sake of understanding, here is a simple recursive approach in Python:

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def longestArithSeqLength(self, A):
    memo = {}
    def dp(i, diff):
        if (i, diff) in memo: return memo[(i, diff)]
        ans = 1
        for j in range(i+1, len(A)):
            if A[j]-A[i] == diff:
                ans = max(ans, 1 + dp(j, diff))
        memo[(i, diff)] = ans
        return ans
    return max(dp(i, A[j]-A[i]) for i in range(len(A)) for j in range(i+1, len(A)))

In this code, we define a recursive function dp(i, diff) which calculates the length of the longest arithmetic subsequence that ends at index i with a difference diff. For each index i and each subsequent index j, we compute the difference A[j] - A[i] and recursively calculate the maximum length of the arithmetic subsequence.

However, keep in mind that this approach results in overlapping subproblems and hence uses memoization to store the computed results for each (i, diff) pair. This somewhat transitions the approach to a top-down dynamic programming solution.

It’s worth mentioning again that for larger inputs, this approach might be less efficient and cause a stack overflow. A more efficient approach would be to use bottom-up dynamic programming which has a lower space complexity and doesn’t risk a stack overflow.

follow up : longest arithmetic subsequence

The longest arithmetic subsequence problem requires finding the longest subsequence within a given array such that all elements in the subsequence differ by the same difference. This problem can be considered a dynamic programming problem, where we iterate over the array and use a map to store the maximum length of an arithmetic subsequence that ends with each difference at each index.

The solution involves creating a 2D array dp of dictionaries at each index of the array A. dp[i][diff] represents the maximum length of an arithmetic subsequence that ends at index i with a difference diff.

Here is a Python solution using dynamic programming:

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def longestArithSeqLength(A):
    dp = [{} for _ in range(len(A))]
    max_len = 0
    for i in range(len(A)):
        for j in range(i+1, len(A)):
            diff = A[j] - A[i]
            if diff in dp[i]:
                dp[j][diff] = dp[i][diff] + 1
            else:
                dp[j][diff] = 2
            max_len = max(max_len, dp[j][diff])
    return max_len

In this solution, for each index i and each subsequent index j, we compute the difference A[j] - A[i] and set dp[j][diff] to be the maximum length of an arithmetic subsequence that ends at index j with a difference diff.

The maximum length of an arithmetic subsequence is then the maximum value in dp.

This solution works in O(n^2) time complexity where n is the length of the input array A, since we iterate over all pairs of indices (i, j). It also works in O(n^2) space complexity because in the worst case, we might end up storing information for all differences at all indices.

To get the actual subsequence, we could backtrack through dp starting from the maximum value, reconstructing the longest arithmetic subsequence by adding A[i] to the subsequence and moving to the previous index i with the same difference diff.

discription says : adjacent elements but examples jump around in the array.

Input: arr = [1,5,7,8,5,3,4,2,1], difference = -2 Output: 4 Explanation: The longest arithmetic subsequence is [7,5,3,1].

in this example, should 3 and 4 be swapped?

When we talk about a “subsequence” in this context, we don’t mean consecutive elements in the original array, rather we mean a sequence that can be derived from the original array by deleting some or no elements without changing the order of the remaining elements. So the elements in the subsequence don’t have to be adjacent in the original array.

In the example you provided: arr = [1,5,7,8,5,3,4,2,1], difference = -2, the longest arithmetic subsequence is [7,5,3,1] which has the length 4. Even though 7 and 5 are not adjacent in the original array, they are part of the arithmetic subsequence because they have the same difference (-2) as specified.

To your question: “should 3 and 4 be swapped?” - No, they don’t need to be swapped because we are looking for a subsequence, not a continuous subarray. The order of the original array should not be changed. We just need to find a sequence within this order where the difference between the elements remains constant and equals the given difference.

Man, the DP solutions always feel so elegant and clever. Yet I’m never fully sure how people come up with them on their own. Is there any trick to knowing when a problem can be solved with DP? Or is it just an intuition type thing?

Dynamic Programming (DP) can be a bit daunting when first encountered, but like most things in programming, practice and experience help build intuition over time.

Here are some signs that a problem might be a good candidate for a Dynamic Programming solution:

1. Overlapping Subproblems: If the problem can be broken down into smaller subproblems, and these subproblems are reused multiple times, then DP might be a good choice. If you find yourself computing the same values over and over again, that’s a clue.

2. Optimal Substructure: If an optimal solution of the problem can be obtained by using optimal solutions of its subproblems, then DP could be the way to go. For instance, in shortest path problems in graphs, if we know the shortest path from a point A to a point B, and from B to a point C, we can easily determine the shortest path from A to C.

3. Memoization or Tabulation: If you can see a way to store the results of each subproblem in a table (for bottom-up DP) or cache (for top-down DP), then you might be looking at a DP problem. This helps to avoid redundant calculation of subproblems.

4. Decision Making: If the problem involves making choices to reach the optimal answer and the choice at the current state depends on the future states, DP might be useful.

As you practice more DP problems, you will start to recognize these patterns more quickly. DP is all about understanding the structure of the problem and knowing how to break the problem down into smaller pieces that you can solve and combine to get the final solution. It’s a powerful technique and worth taking the time to learn well.

when the question is about optimality. there are two ways to solve , either you go by greedy or by DP. Now you have to figure it out on your own, could it be solved by Greedy?? if not go by dp.

Indeed, when the problem involves finding an optimal solution, both greedy algorithms and dynamic programming (DP) are common strategies. To decide which one to use, you’ll need to analyze the problem’s constraints and characteristics.

Here’s a simplified guide to help you decide:

1. Greedy Algorithms: Greedy algorithms work by making the “best” choice at each decision point in the hope that these local optimums will lead to a global optimum. They are generally easier to design and have lower computational requirements than DP solutions. However, they don’t always provide the optimal solution, especially when the problem requires considering future consequences of current choices. Greedy algorithms work well when the problem has a property called “greedy-choice property” (a global optimum can be arrived at by selecting a local optimum) and “optimal substructure” (an optimal solution to the problem contains an optimal solution to subproblems).

2. Dynamic Programming: DP, on the other hand, considers all possible choices at each decision point and uses the results of previous subproblems to make optimal decisions. DP is typically used when the problem has “overlapping subproblems” (the same subproblem is solved multiple times) and “optimal substructure” (an optimal solution to the problem contains an optimal solution to subproblems).

In summary, when dealing with optimization problems, first try to determine if a greedy approach can solve it. If not, or if the greedy algorithm fails to produce the correct answer on some test cases, then a DP approach may be necessary. Of course, this is a simplification and real-world problems might require deeper analysis or even a combination of various strategies. But it gives a general direction to start with. Practice and experience will refine this process further.

Why is this approach failing?

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class Solution {
public:
    int longestSubsequence(vector<int>& arr, int diff) {
        unordered_map<int, vector<int>> map;
        int n = arr.size(), ans = 1;
        for(int i = 0 ; i < n ; i++){
            map[arr[i]].push_back(i);
        }
        vector<bool> visited(n, false);
        for(int i = n-1 ; i >= 0 ; i--){
            int val = arr[i]-diff;
            int cnt = 1;
            int prev = i;
            while(!visited[prev] && map.count(val)){
                visited[prev] = true;
                int ind = upper_bound(map[val].begin(), map[val].end(), prev) - map[val].begin();
                if(ind != 0){
                    cnt++;
                    prev = map[val][ind-1];
                    val = val-diff;
                }else{
                    break;
                }
            }
            ans = max(ans, cnt);
        }
        return ans;
    }
};

This approach seems to be failing due to an incorrect assumption about the order of the sequence. The code is attempting to find the longest subsequence in a reverse order using the upper_bound function to search for the previous element in the sequence.

The problem with this approach is that it assumes that the longest subsequence will always continue in the backward direction. However, this isn’t necessarily the case. There could be a longer sequence that continues in the forward direction.

For example, let’s take a look at the following array: arr = [0,1,2,3,4,5] and diff = 1. Your algorithm will start from the last element and try to find a sequence going backward. However, the longest sequence actually goes forward from the first element.

So, your current solution misses out on such scenarios. A better approach would be to use dynamic programming (DP) to find the longest sequence, maintaining a hashmap to store the longest sequence at each element while iterating through the array in a forward direction.

Here’s a corrected version of your code using the DP approach:

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class Solution {
public:
    int longestSubsequence(vector<int>& arr, int difference) {
        unordered_map<int, int> dp;
        int res = 0;
        for (int a : arr) {
            dp[a] = dp[a - difference] + 1;
            res = max(res, dp[a]);
        }
        return res;
    }
};

This code works by maintaining a dp hashmap where for each element a in arr, it checks if a - difference is already in the map. If it is, it means there is a sequence with a - difference as the previous element and a can be the next element in that sequence. Therefore, dp[a] will be dp[a - difference] + 1. In the end, the longest subsequence will be the maximum value in the dp hashmap.

the problem only demands us to get increasing ap for d>0 and decreasing ap for d<0,which follows the order of array. However it is possible on some test cases to find ap of greater length if we don’t follow the order of array.

Yes, you’re correct that the order of the array matters for this problem. The problem is asking for the longest subsequence, not just any sequence. A subsequence maintains the order of the elements as they appear in the original sequence. So, we need to find an arithmetic progression (AP) within the given array, maintaining the original order of the array.

This is why the problem specifies an increasing AP for d>0 and a decreasing AP for d<0. It’s not about finding any AP in the array, but about finding an AP that is a subsequence of the original array.

Hence, if we don’t follow the order of the array, we might get a longer AP, but it won’t be a valid subsequence as per the original array, and it won’t satisfy the problem constraints.

When I created a solution, and then later when I read the similar editorial, it doesn’t feel like dynamic programming to me. Keeping track of the best of what we’ve seen so far as we iterate through an array seems different to me than exploring a divergent tree of possibilities. If we had to create our own difference variable, that would make it dynamic for me, but this is just keeping a variable number of maximums in storage.

Indeed, the term “Dynamic Programming” (DP) can sometimes be a bit confusing, as it’s often used to refer to a variety of slightly different problem-solving techniques. However, the core idea of DP involves solving complex problems by breaking them down into simpler subproblems, then combining the solutions of these subproblems to find the overall solution.

In the context of this problem, even though we’re not explicitly building a table of solutions to subproblems like we do in some classical DP problems (like the knapsack problem or the Fibonacci sequence), the technique still falls under the umbrella of DP.

Here’s why: the key idea in this problem is to keep track of the longest subsequence ending at each number, taking the difference into account. We’re effectively building a “table” (in this case, a hash table) where each entry keeps track of the maximum length of a subsequence ending with a specific difference. As we iterate through the array, we’re updating these lengths based on the lengths we’ve seen so far - that is, we’re using past information to make current decisions. This is essentially a DP approach, even if it doesn’t look exactly like the textbook examples.

So, even though we’re not exploring a divergent tree of possibilities, the key insight of using past information to solve the current subproblem is what makes this a DP problem. The use of the maximums in storage is a way to avoid redundant computation by reusing previously computed results, which is a hallmark of dynamic programming.

Can someone share how to do it recursively with memoization?

Although a recursive approach with memoization is not the most efficient for this problem, it’s still possible to solve it this way. Here’s a simple Python code that illustrates how you could implement it:

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def longestSubsequence(arr, difference):
    memo = {}
    def dp(i):
        if i in memo:
            return memo[i]
        max_len = 1
        for j in range(i + 1, len(arr)):
            if arr[j] - arr[i] == difference:
                max_len = max(max_len, dp(j) + 1)
        memo[i] = max_len
        return max_len
    return max(dp(i) for i in range(len(arr)))

In this code, dp(i) represents the length of the longest arithmetic subsequence that starts at index i. We initialize max_len to 1 (since a subsequence of one element is always an arithmetic sequence), then for every index j that is larger than i, if arr[j] - arr[i] equals difference, we update max_len to be the maximum of its current value and dp(j) + 1.

The use of the memo dictionary allows us to save results of subproblems so we don’t have to recompute them, which is the concept of memoization.

However, be aware that this approach has a time complexity of O(n^2) which may not pass the time limit for larger test cases. The most efficient way to solve this problem is still using dynamic programming with a hash table, as it provides a time complexity of O(n).

i die a little inside when my recursive memoized solution gets TLE or MLE

Don’t be discouraged! While recursive solutions with memoization are powerful techniques that can solve a wide range of problems, they’re not always the best or most efficient approach. Recursion tends to have higher time and space complexity compared to iterative approaches, which can lead to time limit exceeded (TLE) or memory limit exceeded (MLE) errors on larger inputs.

Here are a few tips that might help:

1. Identify the problem type: Not all problems are best solved with recursion and memoization. Dynamic programming, greedy algorithms, divide and conquer, or even simple iteration might be more suitable depending on the problem.

2. Analyze the problem’s time and space complexity: If the problem has a large input size and your recursive solution has high time or space complexity (e.g., O(2^n), O(n^2), etc.), it’s likely you’ll run into TLE or MLE errors.

3. Consider iterative solutions: If you find your recursive solution isn’t efficient enough, consider whether the problem can be solved iteratively. Dynamic programming is essentially an iterative version of memoized recursion and is often more efficient because it avoids the overhead of recursive function calls.

4. Optimize your memoization: Make sure your memoization is working effectively and you’re not storing unnecessary data. For example, in some cases, you may only need to keep track of the last two values instead of the whole array.

5. Try tail recursion: In some languages like Scala or Haskell, tail recursion can help optimize recursive calls by reducing the amount of stack space used.

Remember, programming and problem-solving is often about trade-offs. Recursion and memoization are powerful tools, but they’re not always the best tool for the job. Don’t let this discourage you! Keep practicing, keep learning, and keep improving.

Similar Problems

Here are ten problems from LeetCode that require similar problem-solving strategies or underlying concepts:

1. Longest Increasing Subsequence: This problem requires dynamic programming to track the longest increasing subsequence ending at each position. It’s similar in strategy to our original problem, which also tracked the longest arithmetic sequence ending at each position with a certain difference.

2. Longest Palindromic Subsequence: This problem also requires dynamic programming to track the longest palindromic subsequence within a string, similar to the original problem.

3. House Robber: This problem uses dynamic programming to determine the maximum amount of money you can rob from houses without robbing two adjacent houses. It requires tracking of optimal solutions to subproblems.

4. Climbing Stairs: This problem requires a dynamic programming approach to find the number of ways to reach the top of a stairwell when you can only climb one or two steps at a time.

5. Maximum Length of Pair Chain: This problem also asks for the longest chain you can make with given pairs, very similar to finding the longest sequence in our original problem.

6. Subarray Sum Equals K: This problem involves iterating through an array and keeping track of cumulative sums and their counts in a dictionary, similar to how we tracked differences and their counts in our problem.

7. Contiguous Array: This problem requires you to find a contiguous binary array with equal numbers of zeros and ones, similar to how we found arithmetic sequences in our problem.

8. Task Scheduler: This problem requires you to find the least number of units of time that the CPU will take to finish all the given tasks, and also involves handling frequency of tasks and idle time, similar to managing differences and sequences in our problem.

9. Best Time to Buy and Sell Stock: This problem requires tracking of the minimum price and maximum profit, similar to how we tracked the maximum length of arithmetic sequence.

10. Longest Harmonious Subsequence: This problem asks for the longest harmonious subsequence where the difference between the maximum and minimum value is exactly 1. It’s similar to our problem which also required finding a sequence with a specific difference.