# Longest subsequence forming an Arithmetic Progression (AP)

• Difficulty Level : Hard
• Last Updated : 14 Jun, 2021

Given an array arr[] consisting of N integers, the task is to find the length of the longest subsequence that forms an Arithmetic Progression.

Examples:

Input: arr[] = {5, 10, 15, 20, 25, 30}
Output: 6
Explanation:
The whole set is in AP having common difference = 5.
Therefore, the length is 4.

Input: arr[] = { 20, 1, 15, 3, 10, 5, 8 }
Output: 4
Explanation:
The longest subsequence having the same difference is { 20, 15, 10, 5 }.
The above subsequence has same difference for every consecutive pairs i.e., (15 – 20) = (10 – 15) = (5 – 10) = -5.
Therefore, the length is 4.

Naive Approach: The simplest approach to solve the problem is to generate all the possible subsequences of the given array and print the length of the longest subsequence having the same difference between adjacent pairs of elements.Time

Complexity: O(N*2N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized using Dynamic Programming. Below are the steps:

1. Initialize an auxiliary matrix dp[][] where dp[i][j] denotes the length of subsequence starting at i and having a common difference as j.
2. Iterate over the array using nested loops i from N – 1 till 0 and j from i+1 to N.
• Assume that arr[i] and arr[j] are the first two elements of the sequence and let their difference be d.
• Now, two cases arise:
1. dp[j][d] = 0: No such sequence exists that starts with arr[j] and has its difference as d.In this case dp[i][d] = 2, as we can have only arr[i] and arr[j] in the sequence.
2. dp[j][d] > 0: A sequence exists that starts with arr[j] and has difference between adjacent elements as d.In this case, the relation is dp[i][d] = 1 + dp[j][d].
3. Finally, print the maximum length of all subsequences formed.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach   #include using namespace std;   // Function that finds the longest // arithmetic subsequence having the // same absolute difference int lenghtOfLongestAP(int A[], int n) {       // Stores the length of sequences     // having same difference     unordered_map >         dp;       // Stores the resultant length     int res = 2;       // Iterate over the array     for (int i = 0; i < n; ++i) {           for (int j = i + 1; j < n; ++j) {               int d = A[j] - A[i];               // Update length of subsequence             dp[d][j] = dp[d].count(i)                            ? dp[d][i] + 1                            : 2;               res = max(res, dp[d][j]);         }     }       // Return res     return res; }   // Driver Code int main() {     // Given array arr[]     int arr[] = { 20, 1, 15, 3, 10, 5, 8 };       int N = sizeof(arr) / sizeof(arr[0]);       // Function Call     cout << lenghtOfLongestAP(arr, N);     return 0; }

## Java

 // Java program for the above approach import java.io.*; import java.util.*;   class GFG{   // Function that finds the longest // arithmetic subsequence having the // same absolute difference static int lenghtOfLongestAP(int A[], int n) {           // Stores the length of sequences     // having same difference     Map> dp = new HashMap>();         // Stores the resultant length     int res = 2;       // Iterate over the array     for(int i = 0; i < n; ++i)     {         for(int j = i + 1; j < n; ++j)         {             int d = A[j] - A[i];             Map temp;               // Update length of subsequence             if (dp.containsKey(d))             {                 temp = dp.get(d);                   if (temp.containsKey(i))                     temp.put(j, temp.get(i) + 1);                 else                     temp.put(j, 2);             }             else             {                 temp = new HashMap();                 temp.put(j, 2);             }             dp.put(d, temp);             res = Math.max(res, temp.get(j));         }     }           // Return res     return res; }   // Driver Code public static void main(String[] args) {           // Given array arr[]     int arr[] = { 20, 1, 15, 3, 10, 5, 8 };       int N = arr.length;       // Function Call     System.out.println(lenghtOfLongestAP(arr, N)); } }   // This code is contributed by jithin

## Python3

 # Python3 program for the above approach   # Function that finds the longest # arithmetic subsequence having the # same absolute difference def lenghtOfLongestAP(A, n) :       # Stores the length of sequences     # having same difference     dp = {}       # Stores the resultant length     res = 2       # Iterate over the array     for i in range(n) :           for j in range(i + 1, n) :               d = A[j] - A[i]               # Update length of subsequence             if d in dp :                 if i in dp[d] :                     dp[d][j] = dp[d][i] + 1                 else :                     dp[d][j] = 2             else :                 dp[d] = {}                 dp[d][j] = 2               if d in dp :                 if j in dp[d] :                     res = max(res, dp[d][j])       # Return res     return res   # Given array arr[] arr = [ 20, 1, 15, 3, 10, 5, 8 ]   N = len(arr)   # Function Call print(lenghtOfLongestAP(arr, N))   # This code is contributed by divyeshrabadiya07

## C#

 // C# program for the above approach using System; using System.Collections.Generic;    class GFG{     // Function that finds the longest // arithmetic subsequence having the // same absolute difference static int lenghtOfLongestAP(int []A, int n) {         // Stores the length of sequences     // having same difference     Dictionary> dp = new Dictionary>();         // Stores the resultant length     int res = 2;        // Iterate over the array     for(int i = 0; i < n; ++i)     {         for(int j = i + 1; j < n; ++j)         {             int d = A[j] - A[i];                // Update length of subsequence             if (dp.ContainsKey(d))             {                  if (dp[d].ContainsKey(i))                 {                     dp[d][j] = dp[d][i] + 1;                    }                 else                 {                     dp[d][j] = 2;                 }             }             else             {                 dp[d] = new Dictionary();                 dp[d][j] = 2;             }             res = Math.Max(res, dp[d][j]);         }     }        // Return res     return res; }     // Driver Code public static void Main(string[] args) {         // Given array arr[]     int []arr = { 20, 1, 15, 3, 10, 5, 8 };        int N = arr.Length;        // Function call     Console.Write(lenghtOfLongestAP(arr, N)); } }   // This code is contributed by rutvik_56

## Javascript



Output:

4

Time Complexity: O(N2)
Auxiliary Space: O(N2)

My Personal Notes arrow_drop_up
Related Articles