Longest subsequence forming an Arithmetic Progression (AP)

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*2^N)
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 <bits/stdc++.h>
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<int,
                  unordered_map<int, int> >
        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<Integer,
    Map<Integer,
        Integer>> dp = new HashMap<Integer,
                               Map<Integer,
                                   Integer>>();

    // 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<Integer, Integer> 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<Integer, Integer>();
                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<int,
    Dictionary<int, int>> dp = new Dictionary<int,
                                   Dictionary<int, int>>();

    // 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<int, int>();
                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

<script>

// JavaScript program for the above approach

// Function that finds the longest
// arithmetic subsequence having the
// same absolute difference
function lenghtOfLongestAP(A, n)
{

    // Stores the length of sequences
    // having same difference
    var dp = new Map();

    // Stores the resultant length
    var res = 2;

    // Iterate over the array
    for (var i = 0; i < n; ++i) {

        for (var j = i + 1; j < n; ++j) {

            var d = A[j] - A[i];

             // Update length of subsequence
             if (dp.has(d))
            {
                 if (dp.get(d).has(i))
                {
                    var tmp = dp.get(d);
                    tmp.set(j, dp.get(d).get(i)+1);
                }
                else
                {
                    var tmp = new Map();
                    tmp.set(j, 2);
                    dp.set(d, tmp);
                }
            }
            else
            {
                var tmp = new Map();
                tmp.set(j, 2);
                dp.set(d, tmp);
            }
            res = Math.max(res, dp.get(d).get(j));
        }
    }

    // Return res
    return res;
}

// Driver Code

// Given array arr[]
var arr = [20, 1, 15, 3, 10, 5, 8];
var N = arr.length;

// Function Call
document.write( lenghtOfLongestAP(arr, N));

</script>

Output:

4

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