Longest Arithmetic Progression (original) (raw)

Given an array **arr[] of **sorted integers and **distinct positive integers, find the **length of the Longest Arithmetic Progression in it.
**Note: A sequence **seq is an arithmetic progression if seq[i + 1] - seq[i] are all the same value ****(for 0 <= i < seq.length - 1)**.

**Examples:

**Input: arr[] = [1, 7, 10, 15, 27, 29]
**Output: 3
**Explanation: The longest arithmetic progression is [1, 15, 29] having common difference 14.

**Input: arr []= [5, 10, 15, 20, 25, 30]
**Output: 6
**Explanation: The whole set is in AP having common difference 5.

Table of Content

Using Recursion - O(n^3) Time and O(n) Space
Using Top-Down DP (Memoization) - O(n^2) Time and O(n^2) Space
Using Bottom-Up DP (Tabulation) - O(n^2) Time and O(n^2) Space

Using Recursion - O(n^3) Time and O(n) Space

For the **recursive approach to finding the length of the longest arithmetic progression (AP), there are two main cases:

**Include the current element in the AP

If the **difference between the **current element and a **previous element matches the desired difference, extend the **AP by considering the current element.

**Mathematically: solve(i, diff) = 1 + solve(j, diff) for all j < i such that arr[i] - arr[j] = diff.

**Exclude the current element from the AP:

Skip the current element and proceed with the remaining elements.

Mathematically: **solve(i, diff) = max(solve(i - 1, diff), solve(j, diff)) where j < i.

Base Case: **solve(i, diff) = 0 when i < 0. This means if there are no elements **left to form an AP, the length is zero.

For **every pair of **indices (i, j), compute the maximum AP length as:

**Length = max(Length, 2 + solve(i, arr[j] - arr[i]))

C++ ``

// C++ program to find the length of the longest // arithmetic progression (AP) using recursion #include <bits/stdc++.h> using namespace std;

// Recursive function to find the length of AP // ending at index index with difference diff int solve(int index, int diff, vector &arr) {

// Base case: if index goes out of bounds, 
// return 0
if (index < 0) 
    return 0;

int ans = 0;

for (int j = index - 1; j >= 0; j--) {
  
    // If the difference matches, extend the AP
    if (arr[index] - arr[j] == diff) {
        ans = max(ans, 1 + solve(j, diff, arr));
    }
}

return ans;

}

// Function to find the length of the longest // arithmetic progression (AP) in the array int lengthOfLongestAP(vector &arr) {

int n = arr.size();

if (n <= 2) 
    return n;

int ans = 0;

// Iterate through all pairs of elements as 
// possible first two terms
for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      
        // Calculate difference and find 
        // AP using recursion
        ans = max(ans, 2 + solve(i, 
                            arr[j] - arr[i], arr));
    }
}

return ans;

}

int main() {

vector<int> arr = {1, 7, 10, 15, 27, 29};

cout << lengthOfLongestAP(arr) << endl;

return 0;

}

Java

// Java program to find the length of the longest // arithmetic progression (AP) using recursion import java.util.ArrayList;

class GfG {

// Recursive function to find the length of AP 
// ending at index `index` with difference `diff`
static int solve(int index, int diff, 
                 ArrayList<Integer> arr) {

    // Base case: if index goes out of bounds, 
    // return 0
    if (index < 0) {
        return 0;
    }

    int ans = 0;

    // Iterate through previous elements 
    // to find matching differences
    for (int j = index - 1; j >= 0; j--) {

        // If the difference matches, extend the AP
        if (arr.get(index) - arr.get(j) == diff) {
            ans = Math.max(ans, 
                           1 + solve(j, diff, arr));
        }
    }

    return ans;
}

// Function to find the length of the longest 
// arithmetic progression (AP) in the array
static int lengthOfLongestAP(ArrayList<Integer> arr) {
    
    int n = arr.size();

    // If there are less than 3 elements, 
    // return the size
    if (n <= 2) {
        return n;
    }

    int ans = 0;

    // Iterate through all pairs of elements as 
    // possible first two terms
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {

            // Calculate difference and find AP 
            // using recursion
            int diff = arr.get(j) - arr.get(i);
            ans = Math.max(ans, 
                           2 + solve(i, diff, arr));
        }
    }

    return ans;
}

public static void main(String[] args) {

    ArrayList<Integer> arr = new ArrayList<>();
    arr.add(1);
    arr.add(7);
    arr.add(10);
    arr.add(15);
    arr.add(27);
    arr.add(29);

    System.out.println(lengthOfLongestAP(arr));
}

}

Python

Python program to find the length of the longest

arithmetic progression (AP) using recursion

Function to find the length of AP ending at

index `index` with difference `diff`

def solve(index, diff, arr):

# Base case: if index goes out of bounds,
# return 0
if index < 0:
    return 0

ans = 0

# Iterate through previous elements to find
# matching differences
for j in range(index - 1, -1, -1):

    # If the difference matches, extend the AP
    if arr[index] - arr[j] == diff:
        ans = max(ans, 1 + solve(j, diff, arr))

return ans

Function to find the length of the longest

arithmetic progression (AP) in the array

def lengthOfLongestAP(arr):

n = len(arr)

# If there are less than 3 elements,
# return the size
if n <= 2:
    return n

ans = 0

# Iterate through all pairs of elements as
# possible first two terms
for i in range(n):
    for j in range(i + 1, n):

        # Calculate difference and find AP
        # using recursion
        diff = arr[j] - arr[i]
        ans = max(ans, 2 + solve(i, diff, arr))

return ans

if name == "main":

arr = [1, 7, 10, 15, 27, 29]

print(lengthOfLongestAP(arr))

C#

// C# program to find the length of the longest // arithmetic progression (AP) using recursion using System; using System.Collections.Generic;

class GfG {

// Recursive function to find the length of AP 
// ending at index `index` with difference `diff`
static int Solve(int index, int diff, 
                        List<int> arr) {

    // Base case: if index goes out of bounds, 
    // return 0
    if (index < 0) {
        return 0;
    }

    int ans = 0;

    // Iterate through previous elements 
    // to find matching differences
    for (int j = index - 1; j >= 0; j--) {

        // If the difference matches, extend the AP
        if (arr[index] - arr[j] == diff) {
            ans = Math.Max(ans, 
                           1 + Solve(j, diff, arr));
        }
    }

    return ans;
}

// Function to find the length of the longest 
// arithmetic progression (AP) in the array
static int LengthOfLongestAP(List<int> arr) {

    int n = arr.Count;

    // If there are less than 3 elements, 
    // return the size
    if (n <= 2) {
        return n;
    }

    int ans = 0;

    // Iterate through all pairs of elements as 
    // possible first two terms
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {

            // Calculate difference and find AP 
            // using recursion
            int diff = arr[j] - arr[i];
            ans = Math.Max(ans, 
                           2 + Solve(i, diff, arr));
        }
    }

    return ans;
}

static void Main(string[] args) {

    List<int> arr = new List<int> { 1, 7, 10, 
                                    15, 27, 29 };

    Console.WriteLine(LengthOfLongestAP(arr));
}

}

JavaScript

// Javascript program to find the length of the longest // arithmetic progression (AP) using recursion

// Function to find the length of AP ending at // index index with difference diff function solve(index, diff, arr) {

// Base case: if index goes out of bounds, 
// return 0
if (index < 0) {
    return 0;
}

let ans = 0;

// Iterate through previous elements to find 
// matching differences
for (let j = index - 1; j >= 0; j--) {

    // If the difference matches, extend the AP
    if (arr[index] - arr[j] === diff) {
        ans = Math.max(ans, 
                       1 + solve(j, diff, arr));
    }
}

return ans;

}

// Function to find the length of the longest // arithmetic progression (AP) in the array function lengthOfLongestAP(arr) {

const n = arr.length;

// If there are less than 3 elements, 
// return the size
if (n <= 2) {
    return n;
}

let ans = 0;

// Iterate through all pairs of elements as 
// possible first two terms
for (let i = 0; i < n; i++) {
    for (let j = i + 1; j < n; j++) {

        // Calculate difference and find AP 
        // using recursion
        const diff = arr[j] - arr[i];
        ans = Math.max(ans, 
                       2 + solve(i, diff, arr));
    }
}

return ans;

}

const arr = [1, 7, 10, 15, 27, 29]; console.log(lengthOfLongestAP(arr));

Using Top-Down DP (Memoization) - O(n^2) Time and O(n^2) Space

**1. Optimal Substructure: The solution to the problem can be derived from optimal solutions of smaller subproblems. Specifically, If the **difference diff between two elements matches the desired difference for an **AP ending at index, we extend the progression:

**solve(index, diff) = 1 + solve(j, diff) where j < index and arr[index] - arr[j] = diff.

If no such j exists, the result for solve(index, diff) remains the same.

**2. Overlapping Subproblems: In the **recursive solution, many subproblems are recomputed **multiple times. For example, solve(i, diff) for a specific index and **difference can be computed repeatedly for different calls in the recursion tree. Memoization stores these results to avoid redundant calculations.
If the value for **solve(index, diff) is already computed and stored in **memo[index][diff], it is directly returned to avoid redundant computation:

**solve(index, diff) = memo[index][diff], if memo[index][diff] exists.

C++ ``

// C++ program to find the length of the longest // arithmetic progression (AP) using recursion // with memoization #include <bits/stdc++.h> using namespace std;

// Recursive function to find the length of AP // ending at index index with difference diff int solve(int index, int diff, vector &arr, unordered_map<int, unordered_map<int, int>> &memo) {

// Base case: if index goes out of bounds, 
// return 0
if (index < 0) 
    return 0;

// Check memo table for precomputed result
if (memo[index].count(diff)) 
    return memo[index][diff];

int ans = 0;

for (int j = index - 1; j >= 0; j--) {
    
    // If the difference matches, extend the AP
    if (arr[index] - arr[j] == diff) {
        ans = max(ans, 1 + solve(j, diff, arr, memo));
    }
}

// Store the result in the memo table
return memo[index][diff] = ans;

}

// Function to find the length of the longest // arithmetic progression (AP) in the array int lengthOfLongestAP(vector &arr) {

int n = arr.size();

if (n <= 2) 
    return n;

int ans = 0;

// Memoization table to store results for solve()
unordered_map<int, unordered_map<int, int>> memo;

// Iterate through all pairs of elements as 
// possible first two terms
for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {

        // Calculate difference and find 
        // AP using memoized recursion
        ans = max(ans, 2 + solve(i, 
                            arr[j] - arr[i], arr, memo));
    }
}

return ans;

}

int main() {

vector<int> arr = {1, 7, 10, 15, 27, 29};

cout << lengthOfLongestAP(arr) << endl;

return 0;

}

Java

// Java program to find the length of the longest // arithmetic progression (AP) using recursion // with memoization import java.util.ArrayList; import java.util.HashMap;

class GfG {

// Recursive function to find the length of AP 
// ending at index `index` with difference `diff`
static int solve(int index, int diff, 
                 ArrayList<Integer> arr, 
                 HashMap<Integer, HashMap<Integer, Integer>> memo) {

    // Base case: if index goes out of bounds, 
    // return 0
    if (index < 0) {
        return 0;
    }

    // Check memo table for precomputed result
    if (memo.containsKey(index) && 
        memo.get(index).containsKey(diff)) {
        return memo.get(index).get(diff);
    }

    int ans = 0;

    // Iterate through previous elements 
    // to find matching differences
    for (int j = index - 1; j >= 0; j--) {

        // If the difference matches, extend the AP
        if (arr.get(index) - arr.get(j) == diff) {
            ans = Math.max(ans, 
                           1 + solve(j, diff, arr, memo));
        }
    }

    // Store the result in the memo table
    memo.putIfAbsent(index, new HashMap<>());
    memo.get(index).put(diff, ans);

    return ans;
}

// Function to find the length of the longest 
// arithmetic progression (AP) in the array
static int lengthOfLongestAP(ArrayList<Integer> arr) {
    
    int n = arr.size();

    // If there are less than 3 elements, 
    // return the size
    if (n <= 2) {
        return n;
    }

    int ans = 0;

    // Memoization table to store results for solve()
    HashMap<Integer, HashMap<Integer, Integer>> memo = 
        new HashMap<>();

    // Iterate through all pairs of elements as 
    // possible first two terms
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {

            // Calculate difference and find AP 
            // using memoized recursion
            int diff = arr.get(j) - arr.get(i);
            ans = Math.max(ans, 
                           2 + solve(i, diff, arr, memo));
        }
    }

    return ans;
}

public static void main(String[] args) {

    ArrayList<Integer> arr = new ArrayList<>();
    arr.add(1);
    arr.add(7);
    arr.add(10);
    arr.add(15);
    arr.add(27);
    arr.add(29);

    System.out.println(lengthOfLongestAP(arr));
}

}

Python

Python program to find the length of the longest

arithmetic progression (AP) using recursion

with memoization

Function to find the length of AP ending at

index `index` with difference `diff`

def solve(index, diff, arr, memo):

# Base case: if index goes out of bounds,
# return 0
if index < 0:
    return 0

# Check if the result is already computed
if (index, diff) in memo:
    return memo[(index, diff)]

ans = 0

# Iterate through previous elements to find
# matching differences
for j in range(index - 1, -1, -1):

    # If the difference matches, extend the AP
    if arr[index] - arr[j] == diff:
        ans = max(ans, 1 + solve(j, diff, arr, memo))

# Store the result in the memo dictionary
memo[(index, diff)] = ans
return ans

Function to find the length of the longest

arithmetic progression (AP) in the array

def lengthOfLongestAP(arr):

n = len(arr)

# If there are less than 3 elements,
# return the size
if n <= 2:
    return n

ans = 0

# Memoization dictionary to store results for solve
memo = {}

# Iterate through all pairs of elements as
# possible first two terms
for i in range(n):
    for j in range(i + 1, n):

        # Calculate difference and find AP
        # using memoized recursion
        diff = arr[j] - arr[i]
        ans = max(ans, 2 + solve(i, diff, arr, memo))

return ans

if name == "main":

arr = [1, 7, 10, 15, 27, 29]

print(lengthOfLongestAP(arr))

C#

// C# program to find the length of the longest // arithmetic progression (AP) using recursion // with memoization using System; using System.Collections.Generic;

class GfG {

// Recursive function to find the length of AP 
// ending at index `index` with difference `diff`
static int Solve(int index, int diff, 
                        List<int> arr, 
                        Dictionary<(int, int), int> memo) {

    // Base case: if index goes out of bounds, 
    // return 0
    if (index < 0) {
        return 0;
    }

    // Check if the result is already computed
    if (memo.ContainsKey((index, diff))) {
        return memo[(index, diff)];
    }

    int ans = 0;

    // Iterate through previous elements 
    // to find matching differences
    for (int j = index - 1; j >= 0; j--) {

        // If the difference matches, extend the AP
        if (arr[index] - arr[j] == diff) {
            ans = Math.Max(ans, 
                           1 + Solve(j, diff, arr, memo));
        }
    }

    // Store the result in the memo dictionary
    memo[(index, diff)] = ans;
    return ans;
}

// Function to find the length of the longest 
// arithmetic progression (AP) in the array
static int LengthOfLongestAP(List<int> arr) {

    int n = arr.Count;

    // If there are less than 3 elements, 
    // return the size
    if (n <= 2) {
        return n;
    }

    int ans = 0;

    // Memoization dictionary to store results for Solve
    var memo = new Dictionary<(int, int), int>();

    // Iterate through all pairs of elements as 
    // possible first two terms
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {

            // Calculate difference and find AP 
            // using memoized recursion
            int diff = arr[j] - arr[i];
            ans = Math.Max(ans, 
                           2 + Solve(i, diff, arr, memo));
        }
    }

    return ans;
}

static void Main(string[] args) {

    List<int> arr = new List<int> { 1, 7, 10, 
                                    15, 27, 29 };

    Console.WriteLine(LengthOfLongestAP(arr));
}

}

JavaScript

// Javascript program to find the length of the longest // arithmetic progression (AP) using recursion with memoization

// Function to find the length of AP ending at // index index with difference diff function solve(index, diff, arr, memo) {

// Base case: if index goes out of bounds, 
// return 0
if (index < 0) {
    return 0;
}

// Check if the result is already computed
const key = `${index}-${diff}`;
if (memo[key] !== undefined) {
    return memo[key];
}

let ans = 0;

// Iterate through previous elements to find 
// matching differences
for (let j = index - 1; j >= 0; j--) {

    // If the difference matches, extend the AP
    if (arr[index] - arr[j] === diff) {
        ans = Math.max(ans, 
                       1 + solve(j, diff, arr, memo));
    }
}

// Store the result in the memo object
memo[key] = ans;
return ans;

}

// Function to find the length of the longest // arithmetic progression (AP) in the array function lengthOfLongestAP(arr) {

const n = arr.length;

// If there are less than 3 elements, 
// return the size
if (n <= 2) {
    return n;
}

let ans = 0;

// Memoization object to store results for solve
const memo = {};

// Iterate through all pairs of elements as 
// possible first two terms
for (let i = 0; i < n; i++) {
    for (let j = i + 1; j < n; j++) {

        // Calculate difference and find AP 
        // using memoized recursion
        const diff = arr[j] - arr[i];
        ans = Math.max(ans, 
                       2 + solve(i, diff, arr, memo));
    }
}

return ans;

}

const arr = [1, 7, 10, 15, 27, 29]; console.log(lengthOfLongestAP(arr));

Using Bottom-Up DP (Tabulation) - O(n^2) Time and O(n^2) Space

The **tabulation approach iteratively builds the solution in **a bottom-up manner, avoiding the use of recursion. Instead of solving smaller subproblems recursively, we use a table to store and update solutions for progressively larger subproblems.

We will create a **2D table of size **n x diffRange, where **dp[i][diff] represents the **length of the arithmetic progression (AP) **ending at index i with a common difference of **diff.

The dynamic programming relation is as follows:

For each pair of indices (j, i) where j < i: Calculate the common difference: diff = arr[i] - arr[j]

If there is an **AP ending at index j with the same diff: dp[i][diff] = dp[j][diff] + 1

Otherwise, start a **new AP of length 2: dp[i][diff] = 2

C++ `

// C++ program to find the length of the longest // arithmetic progression (AP) using tabulation #include <bits/stdc++.h> using namespace std;

// Function to find the length of the longest // arithmetic progression (AP) in the array int lengthOfLongestAP(vector &arr) { int n = arr.size();

// If there are less than 3 elements, 
// return the size
if (n <= 2) 
    return n;

int ans = 2;

// Create a 2D dp table where dp[i][j] stores the
// length of the AP ending at indices i and j
vector<unordered_map<int, int>> dp(n);

// Iterate through all pairs of elements as 
// possible first two terms
for (int i = 1; i < n; i++) {
    for (int j = 0; j < i; j++) {

        // Calculate the common difference
        int diff = arr[i] - arr[j];

        // If there's an AP ending at j with this diff,
        // extend it; otherwise, start a new AP of length 2
        dp[i][diff] = dp[j].count(diff) ? dp[j][diff] + 1 : 2;

        // Update the overall maximum length
        ans = max(ans, dp[i][diff]);
    }
}

return ans;

}

int main() { vector arr = {1, 7, 10, 15, 27, 29};

cout << lengthOfLongestAP(arr) << endl;

return 0;

}

` Java ``

// Java program to find the length of the longest // arithmetic progression (AP) using tabulation import java.util.ArrayList; import java.util.HashMap;

class GfG {

// Function to find the length of the longest 
// arithmetic progression (AP) in the array
static int lengthOfLongestAP(ArrayList<Integer> arr) {
    
    int n = arr.size();

    // If there are less than 3 elements, 
    // return the size
    if (n <= 2) {
        return n;
    }

    int ans = 0;

    // Create a 2D HashMap to store the lengths of 
    // APs ending at each index with a given difference
    HashMap<Integer, Integer>[] dp = new HashMap[n];

    for (int i = 0; i < n; i++) {
        dp[i] = new HashMap<>();
    }

    // Iterate through all pairs of elements as 
    // possible first two terms
    for (int i = 1; i < n; i++) {
        for (int j = 0; j < i; j++) {

            // Calculate the common difference
            int diff = arr.get(i) - arr.get(j);

            // Check if this difference was seen before
            int length = dp[j].getOrDefault(diff, 1);

            // Update the length of the AP ending at index `i`
            dp[i].put(diff, length + 1);

            // Update the global maximum length
            ans = Math.max(ans, dp[i].get(diff));
        }
    }

    return ans;
}

public static void main(String[] args) {

    ArrayList<Integer> arr = new ArrayList<>();
    arr.add(1);
    arr.add(7);
    arr.add(10);
    arr.add(15);
    arr.add(27);
    arr.add(29);

    System.out.println(lengthOfLongestAP(arr));
}

}

Python

Python program to find the length of the longest

arithmetic progression (AP) using tabulation

Function to find the length of the longest

AP in the array

def lengthOfLongestAP(arr):

n = len(arr)

# If there are less than 3 elements,
# return the size
if n <= 2:
    return n

# Create a list of dictionaries to store
# lengths of APs
dp = [{} for _ in range(n)]

ans = 0

# Iterate through all pairs of elements as
# possible first two terms
for i in range(1, n):
    for j in range(i):
        
        # Calculate the common difference
        diff = arr[i] - arr[j]

        # Get the previous length or start with 1
        length = dp[j].get(diff, 1)

        # Update the length of the AP ending at index `i`
        dp[i][diff] = length + 1

        # Update the global maximum length
        ans = max(ans, dp[i][diff])

return ans

if name == "main":

arr = [1, 7, 10, 15, 27, 29]

print(lengthOfLongestAP(arr))

C#

// C# program to find the length of the longest // arithmetic progression (AP) using tabulation using System; using System.Collections.Generic;

class GfG {

// Function to find the length of the longest 
// arithmetic progression (AP) in the array
static int LengthOfLongestAP(List<int> arr) {

    int n = arr.Count;

    // If there are less than 3 elements,
      // return the size
    if (n <= 2) {
        return n;
    }

    // Create a list of dictionaries to store 
      // lengths of APs
    var dp = new List<Dictionary<int, int>>();
    for (int i = 0; i < n; i++) {
        dp.Add(new Dictionary<int, int>());
    }

    int ans = 0;

    // Iterate through all pairs of elements as
    // possible first two terms
    for (int i = 1; i < n; i++) {
        for (int j = 0; j < i; j++) {

            // Calculate the common difference
            int diff = arr[i] - arr[j];

            // Get the previous length or start with 1
            int length = dp[j].ContainsKey(diff) 
                         ? dp[j][diff] : 1;

            // Update the length of the AP ending 
              // at index `i`
            if (!dp[i].ContainsKey(diff)) {
                dp[i][diff] = length + 1;
            } else {
                dp[i][diff] = Math.Max(dp[i][diff], 
                                       length + 1);
            }

            // Update the global maximum length
            ans = Math.Max(ans, dp[i][diff]);
        }
    }

    return ans;
}

static void Main(string[] args) {

    List<int> arr = new List<int> { 1, 7, 10, 
                                    15, 27, 29 };

    Console.WriteLine(LengthOfLongestAP(arr));
}

}

JavaScript

// Javascript program to find the length of the longest // arithmetic progression (AP) using tabulation

function lengthOfLongestAP(arr) { const n = arr.length;

// If there are less than 3 elements, 
// return the size
if (n <= 2) {
    return n;
}

// Create an array of maps to store lengths of APs
const dp = Array.from({ length: n }, () => new Map());

let ans = 0;

// Iterate through all pairs of elements as
// possible first two terms
for (let i = 1; i < n; i++) {
    for (let j = 0; j < i; j++) {

        // Calculate the common difference
        const diff = arr[i] - arr[j];

        // Get the previous length or start with 1
        const length = dp[j].get(diff) || 1;

        // Update the length of the AP ending at index `i`
        dp[i].set(diff, Math.max(dp[i].get(diff) || 0, length + 1));

        // Update the global maximum length
        ans = Math.max(ans, dp[i].get(diff));
    }
}

return ans;

}

const arr = [1, 7, 10, 15, 27, 29]; console.log(lengthOfLongestAP(arr));

Longest Arithmetic Progression (original) (raw)

Using Recursion - O(n^3) Time and O(n) Space

Python program to find the length of the longest

arithmetic progression (AP) using recursion

Function to find the length of AP ending at

index index with difference diff

Function to find the length of the longest

arithmetic progression (AP) in the array

Using Top-Down DP (Memoization) - O(n^2) Time and O(n^2) Space

Python program to find the length of the longest

arithmetic progression (AP) using recursion

with memoization

Function to find the length of AP ending at

index index with difference diff

Function to find the length of the longest

arithmetic progression (AP) in the array

Using Bottom-Up DP (Tabulation) - O(n^2) Time and O(n^2) Space

Python program to find the length of the longest

arithmetic progression (AP) using tabulation

Function to find the length of the longest

AP in the array

index `index` with difference `diff`

index `index` with difference `diff`