# Longest Arithmetic Progression

Given an array Set[ ] of sorted integers having no duplicates, find the length of the Longest Arithmetic Progression (LLAP) subsequence in it.

Examples:

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

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

Approach 1: Brute forcing the solution in O(N^3)

A simple solution is to one by one consider every pair as first two elements of AP and check for the remaining elements in sorted array. To consider all pairs as first two elements, we need to run a O(n^2) nested loop. Inside the nested loops, we need a third loop which linearly looks for the more elements in Arithmetic Progression (AP). This process takes O(n3) time.
We can solve this problem in O(n2) time using Dynamic Programming. To get idea of the DP solution, let us first discuss solution of following simpler problem.

Given a sorted Array, find if there exist three elements in Arithmetic Progression or not:

Please note that, the answer is true if there are 3 or more elements in AP, otherwise false. To find the three elements, we first fix an element as middle element and search for other two (one smaller and one greater). We start from the second element and fix every element as middle element.

For an element set[j] to be middle of AP, there must exist elements ‘set[i]’ and ‘set[k]’ such that set[i] + set[k] = 2*set[j] where 0 <= i < j and j < k <=n-1.

Algorithm to efficiently find i and k for a given j:

We can find i and k in linear time using following simple algorithm.

• Initialize i as j-1 and k as j+1
• Do following while i >= 0 and k <= n-1
• If set[i] + set[k] is equal to 2*set[j], then we are done.
• If set[i] + set[k] > 2*set[j], then decrement i (do i–).
• Else if set[i] + set[k] < 2*set[j], then increment k (do k++).

Following is the implementation of the above algorithm for the simpler problem.

## C++

 `// The function returns true if there exist three ``// elements in AP Assumption: set[0..n-1] is sorted. ``// The code strictly implements the algorithm provided ``// in the reference.``bool` `arithmeticThree(vector<``int``> set, ``int` `n)``{``    ` `    ``// One by fix every element as middle element``    ``for``(``int` `j = 1; j < n - 1; j++)``    ``{``        ` `        ``// Initialize i and k for the current j``        ``int` `i = j - 1, k = j + 1;` `        ``// Find if there exist i and k that form AP``        ``// with j as middle element``        ``while` `(i >= 0 && k <= n-1)``        ``{``            ``if` `(set[i] + set[k] == 2 * set[j])``                ``return` `true``;``                ` `            ``(set[i] + set[k] < 2 * set[j]) ? k++ : i--;``        ``}``    ``}``    ``return` `false``;``}` `// This code is contributed by Samim Hossain Mondal.`

## C

 `// The function returns true if there exist three elements in AP``// Assumption: set[0..n-1] is sorted. ``// The code strictly implements the algorithm provided in the reference.``bool` `arithmeticThree(``int` `set[], ``int` `n)``{``    ``// One by fix every element as middle element``    ``for` `(``int` `j=1; j= 0 && k <= n-1)``        ``{``            ``if` `(set[i] + set[k] == 2*set[j])``                ``return` `true``;``            ``(set[i] + set[k] < 2*set[j])? k++ : i--;``        ``}``    ``}` `    ``return` `false``;``}`

## Java

 `// The function returns true if there exist three elements in AP``// Assumption: set[0..n-1] is sorted. ``// The code strictly implements the algorithm provided in the reference.``static` `boolean` `arithmeticThree(``int` `set[], ``int` `n)``{``    ``// One by fix every element as middle element``    ``for` `(``int` `j = ``1``; j < n - ``1``; j++)``    ``{``        ``// Initialize i and k for the current j``        ``int` `i = j - ``1``, k = j + ``1``;` `        ``// Find if there exist i and k that form AP``        ``// with j as middle element``        ``while` `(i >= ``0` `&& k <= n-``1``)``        ``{``            ``if` `(set[i] + set[k] == ``2``*set[j])``                ``return` `true``;``            ``(set[i] + set[k] < ``2``*set[j])? k++ : i--;``        ``}``    ``}` `    ``return` `false``;``}` `// This code is contributed by gauravrajput1 `

## Python3

 `# The function returns true if there exist three elements in AP``# Assumption: set[0..n-1] is sorted. ``# The code strictly implements the algorithm provided in the reference.``def` `arithematicThree(set_,n):` `    ``# One by fix every element as middle element``    ``for` `j ``in` `range``(n):``        ` `        ``# Initialize i and k for the current j``        ``i,k``=``j``-``1``,j``+``1` `        ``# Find if there exist i and k that form AP``        ``# with j as middle element``        ``while` `i>``-``1` `and` `k

## C#

 `// The function returns true if there exist three elements in AP``// Assumption: set[0..n-1] is sorted. ``// The code strictly implements the algorithm provided in the reference.``static` `bool` `arithmeticThree(``int` `[]``set``, ``int` `n)``{``  ` `    ``// One by fix every element as middle element``    ``for` `(``int` `j = 1; j < n - 1; j++)``    ``{``      ` `        ``// Initialize i and k for the current j``        ``int` `i = j - 1, k = j + 1;` `        ``// Find if there exist i and k that form AP``        ``// with j as middle element``        ``while` `(i >= 0 && k <= n-1)``        ``{``            ``if` `(``set``[i] + ``set``[k] == 2*``set``[j])``                ``return` `true``;``            ``if``(``set``[i] + ``set``[k] < 2*``set``[j]) ``                ``k++;``            ``else``                ``i--;``        ``}``    ``}` `    ``return` `false``;``}` `// This code is contributed by gauravrajput1`

## Javascript

 ``

## Longest Arithmetic Progression using Dynamic Programming

Iteratively examining all possible pairs of elements as the first two elements of an AP and then extending the AP to find the longest one. The algorithm uses a dynamic programming approach to store the length of the LLAP ending at each pair of elements. It starts with a minimum LLAP length of 2 (two elements), and as it iterates through the elements, it updates and keeps track of the maximum LLAP length found so far. Finally, it returns the length of the longest arithmetic progression found in the set.

Step-by-step approach:

• Initialize a 2D array L[][] to store the length of LLAPs ending at pairs of elements.
• For each element in the set, fill the last column of L[][] with 2 since any pair forms an AP of length 2.
• Starting from the second-to-last column, iterate through pairs of elements (set[i], set[j]).
• Within the loop, compare set[i] + set[k] to 2 * set[j]:
• If less, increment k to consider a larger element.
• If greater, decrement i to consider a smaller element.
• If equal, extend the LLAP by setting L[i][j] to L[j][k] + 1 and update the maximum LLAP length (llap).
• Handle cases where k exceeds the last index by filling in remaining cells in the column with 2.
• Return the maximum LLAP length (llap) found in the set.

Following is the implementation of the Dynamic Programming algorithm.

## C++

 `// C++ program to find Length of the Longest AP (llap) in a given sorted set.``// The code strictly implements the algorithm provided in the reference.``#include ``using` `namespace` `std;` `// Returns length of the longest AP subset in a given set``int` `lenghtOfLongestAP(``int` `set[], ``int` `n)``{``    ``if` `(n <= 2)  ``return` `n;` `    ``// Create a table and initialize all values as 2. The value of``    ``// L[i][j] stores LLAP with set[i] and set[j] as first two``    ``// elements of AP. Only valid entries are the entries where j>i``    ``int` `L[n][n];``    ``int` `llap = 2;  ``// Initialize the result` `    ``// Fill entries in last column as 2. There will always be``    ``// two elements in AP with last number of set as second``    ``// element in AP``    ``for` `(``int` `i = 0; i < n; i++)``        ``L[i][n-1] = 2;` `    ``// Consider every element as second element of AP``    ``for` `(``int` `j=n-2; j>=1; j--)``    ``{``        ``// Search for i and k for j``        ``int` `i = j-1, k = j+1;``        ``while` `(i >= 0 && k <= n-1)``        ``{``           ``if` `(set[i] + set[k] < 2*set[j])``               ``k++;` `           ``// Before changing i, set L[i][j] as 2``           ``else` `if` `(set[i] + set[k] > 2*set[j])``           ``{   L[i][j] = 2, i--;   }` `           ``else``           ``{``               ``// Found i and k for j, LLAP with i and j as first two``               ``// elements is equal to LLAP with j and k as first two``               ``// elements plus 1. L[j][k] must have been filled``               ``// before as we run the loop from right side``               ``L[i][j] = L[j][k] + 1;` `               ``// Update overall LLAP, if needed``               ``llap = max(llap, L[i][j]);` `               ``// Change i and k to fill more L[i][j] values for``               ``// current j``               ``i--; k++;``           ``}``        ``}` `        ``// If the loop was stopped due to k becoming more than``        ``// n-1, set the remaining entities in column j as 2``        ``while` `(i >= 0)``        ``{``            ``L[i][j] = 2;``            ``i--;``        ``}``    ``}``    ``return` `llap;``}` `/* Driver program to test above function*/``int` `main()``{``    ``int` `set1[] = {1, 7, 10, 13, 14, 19};``    ``int` `n1 = ``sizeof``(set1)/``sizeof``(set1[0]);``    ``cout <<   lenghtOfLongestAP(set1, n1) << endl;` `    ``int` `set2[] = {1, 7, 10, 15, 27, 29};``    ``int` `n2 = ``sizeof``(set2)/``sizeof``(set2[0]);``    ``cout <<   lenghtOfLongestAP(set2, n2) << endl;` `    ``int` `set3[] = {2, 4, 6, 8, 10};``    ``int` `n3 = ``sizeof``(set3)/``sizeof``(set3[0]);``    ``cout <<   lenghtOfLongestAP(set3, n3) << endl;` `    ``return` `0;``}`

## Java

 `// Java program to find Length of the``// Longest AP (llap) in a given sorted set.``// The code strictly implements the ``// algorithm provided in the reference.``import` `java.io.*;` `class` `GFG ``{``    ``// Returns length of the longest ``    ``// AP subset in a given set``    ``static` `int` `lenghtOfLongestAP(``int` `set[], ``int` `n)``    ``{``        ``if` `(n <= ``2``) ``return` `n;``    ` `        ``// Create a table and initialize all ``        ``// values as 2. The value ofL[i][j] stores ``        ``// LLAP with set[i] and set[j] as first two``        ``// elements of AP. Only valid entries are ``        ``// the entries where j>i``        ``int` `L[][] = ``new` `int``[n][n];``        ` `         ``// Initialize the result``        ``int` `llap = ``2``;``    ` `        ``// Fill entries in last column as 2. ``        ``// There will always be two elements in ``        ``// AP with last number of set as second``        ``// element in AP``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``L[i][n - ``1``] = ``2``;``    ` `        ``// Consider every element as second element of AP``        ``for` `(``int` `j = n - ``2``; j >= ``1``; j--)``        ``{``            ``// Search for i and k for j``            ``int` `i = j -``1` `, k = j + ``1``;``            ``while` `(i >= ``0` `&& k <= n - ``1``)``            ``{``            ``if` `(set[i] + set[k] < ``2` `* set[j])``                ``k++;``    ` `            ``// Before changing i, set L[i][j] as 2``            ``else` `if` `(set[i] + set[k] > ``2` `* set[j])``            ``{ ``                ``L[i][j] = ``2``; i--; ``                ` `            ``}``    ` `            ``else``            ``{``                ``// Found i and k for j, LLAP with i and j as first two``                ``// elements is equal to LLAP with j and k as first two``                ``// elements plus 1. L[j][k] must have been filled``                ``// before as we run the loop from right side``                ``L[i][j] = L[j][k] + ``1``;``    ` `                ``// Update overall LLAP, if needed``                ``llap = Math.max(llap, L[i][j]);``    ` `                ``// Change i and k to fill ``                ``// more L[i][j] values for current j``                ``i--; k++;``            ``}``            ``}``    ` `            ``// If the loop was stopped due``            ``// to k becoming more than``            ``// n-1, set the remaining ``            ``// entities in column j as 2``            ``while` `(i >= ``0``)``            ``{``                ``L[i][j] = ``2``;``                ``i--;``            ``}``        ``}``        ``return` `llap;``    ``}``    ` `    ``// Driver program ``    ``public` `static` `void` `main (String[] args) ``    ``{``        ``int` `set1[] = {``1``, ``7``, ``10``, ``13``, ``14``, ``19``};``        ``int` `n1 = set1.length;``        ``System.out.println ( lenghtOfLongestAP(set1, n1));``    ` `        ``int` `set2[] = {``1``, ``7``, ``10``, ``15``, ``27``, ``29``};``        ``int` `n2 = set2.length;``        ``System.out.println(lenghtOfLongestAP(set2, n2));``    ` `        ``int` `set3[] = {``2``, ``4``, ``6``, ``8``, ``10``};``        ``int` `n3 = set3.length;``        ``System.out.println(lenghtOfLongestAP(set3, n3)) ;``    ` `    ` `    ``}``}` `// This code is contributed by vt_m`

## Python3

 `# Python 3 program to find Length of the``# Longest AP (llap) in a given sorted set.``# The code strictly implements the algorithm``# provided in the reference` `# Returns length of the longest AP``# subset in a given set``def` `lenghtOfLongestAP(``set``, n):` `    ``if` `(n <``=` `2``):``        ``return` `n` `    ``# Create a table and initialize all ``    ``# values as 2. The value of L[i][j]``    ``# stores LLAP with set[i] and set[j] ``    ``# as first two elements of AP. Only ``    ``# valid entries are the entries where j>i``    ``L ``=` `[[``0` `for` `x ``in` `range``(n)]``            ``for` `y ``in` `range``(n)]``    ``llap ``=` `2` `# Initialize the result` `    ``# Fill entries in last column as 2. ``    ``# There will always be two elements ``    ``# in AP with last number of set as ``    ``# second element in AP``    ``for` `i ``in` `range``(n):``        ``L[i][n ``-` `1``] ``=` `2` `    ``# Consider every element as second ``    ``# element of AP``    ``for` `j ``in` `range``(n ``-` `2``, ``0``, ``-``1``):` `        ``# Search for i and k for j``        ``i ``=` `j ``-` `1``        ``k ``=` `j ``+` `1``        ``while``(i >``=` `0` `and` `k <``=` `n ``-` `1``):` `            ``if` `(``set``[i] ``+` `set``[k] < ``2` `*` `set``[j]):``                ``k ``+``=` `1` `            ``# Before changing i, set L[i][j] as 2``            ``elif` `(``set``[i] ``+` `set``[k] > ``2` `*` `set``[j]):``                ``L[i][j] ``=` `2``                ``i ``-``=` `1` `            ``else``:` `                ``# Found i and k for j, LLAP with i and j ``                ``# as first two elements are equal to LLAP ``                ``# with j and k as first two elements plus 1. ``                ``# L[j][k] must have been filled before as ``                ``# we run the loop from right side``                ``L[i][j] ``=` `L[j][k] ``+` `1` `                ``# Update overall LLAP, if needed``                ``llap ``=` `max``(llap, L[i][j])` `                ``# Change i and k to fill more L[i][j] ``                ``# values for current j``                ``i ``-``=` `1``                ``k ``+``=` `1` `                ``# If the loop was stopped due to k ``                ``# becoming more than n-1, set the``                ``# remaining entities in column j as 2``                ``while` `(i >``=` `0``):``                    ``L[i][j] ``=` `2``                    ``i ``-``=` `1``    ``return` `llap` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``set1 ``=` `[``1``, ``7``, ``10``, ``13``, ``14``, ``19``]``    ``n1 ``=` `len``(set1)``    ``print``(lenghtOfLongestAP(set1, n1))` `    ``set2 ``=` `[``1``, ``7``, ``10``, ``15``, ``27``, ``29``]``    ``n2 ``=` `len``(set2)``    ``print``(lenghtOfLongestAP(set2, n2))` `    ``set3 ``=` `[``2``, ``4``, ``6``, ``8``, ``10``]``    ``n3 ``=` `len``(set3)``    ``print``(lenghtOfLongestAP(set3, n3))` `# This code is contributed by ita_c`

## C#

 `// C# program to find Length of the ``// Longest AP (llap) in a given sorted set. ``// The code strictly implements the ``// algorithm provided in the reference.``using` `System;` `class` `GFG``{``// Returns length of the longest ``// AP subset in a given set ``static` `int` `lenghtOfLongestAP(``int` `[]``set``, ``                             ``int` `n) ``{ ``    ``if` `(n <= 2) ``return` `n; ` `    ``// Create a table and initialize ``    ``// all values as 2. The value of ``    ``// L[i][j] stores LLAP with set[i]  ``    ``// and set[j] as first two elements ``    ``// of AP. Only valid entries are ``    ``// the entries where j>i ``    ``int` `[,]L = ``new` `int``[n, n]; ``    ` `    ``// Initialize the result ``    ``int` `llap = 2; ` `    ``// Fill entries in last column as 2. ``    ``// There will always be two elements``    ``// in AP with last number of set as ``    ``// second element in AP ``    ``for` `(``int` `i = 0; i < n; i++) ``        ``L[i, n - 1] = 2; ` `    ``// Consider every element as ``    ``// second element of AP ``    ``for` `(``int` `j = n - 2; j >= 1; j--) ``    ``{ ``        ``// Search for i and k for j ``        ``int` `i = j - 1 , k = j + 1; ``        ``while` `(i >= 0 && k <= n - 1) ``        ``{ ``        ``if` `(``set``[i] + ``set``[k] < 2 * ``set``[j]) ``            ``k++; ` `        ``// Before changing i, set L[i][j] as 2 ``        ``else` `if` `(``set``[i] + ``set``[k] > 2 * ``set``[j]) ``        ``{ ``            ``L[i, j] = 2; i--; ``            ` `        ``} ` `        ``else``        ``{ ``            ``// Found i and k for j, LLAP with ``            ``// i and j as first two elements ``            ``// is equal to LLAP with j and k ``            ``// as first two elements plus 1. ``            ``// L[j][k] must have been filled ``            ``// before as we run the loop from ``            ``// right side ``            ``L[i, j] = L[j, k] + 1; ` `            ``// Update overall LLAP, if needed ``            ``llap = Math.Max(llap, L[i, j]); ` `            ``// Change i and k to fill ``            ``// more L[i][j] values for current j ``            ``i--; k++; ``        ``} ``        ``} ` `        ``// If the loop was stopped due ``        ``// to k becoming more than ``        ``// n-1, set the remaining ``        ``// entities in column j as 2 ``        ``while` `(i >= 0) ``        ``{ ``            ``L[i, j] = 2; ``            ``i--; ``        ``} ``    ``} ``    ``return` `llap; ``} ` `// Driver Code ``static` `public` `void` `Main ()``{``    ``int` `[]set1 = {1, 7, 10, 13, 14, 19}; ``    ``int` `n1 = set1.Length; ``    ``Console.WriteLine(lenghtOfLongestAP(set1, n1)); ` `    ``int` `[]set2 = {1, 7, 10, 15, 27, 29}; ``    ``int` `n2 = set2.Length; ``    ``Console.WriteLine(lenghtOfLongestAP(set2, n2)); ` `    ``int` `[]set3 = {2, 4, 6, 8, 10}; ``    ``int` `n3 = set3.Length; ``    ``Console.WriteLine(lenghtOfLongestAP(set3, n3)) ;``}``}` `// This code is contributed by Sach_Code`

## Javascript

 ``

## PHP

 `i ``    ``\$L``[``\$n``][``\$n``] = ``array``(``array``()); ``    ``\$llap` `= 2; ``// Initialize the result ` `    ``// Fill entries in last column as 2.``    ``// There will always be two elements ``    ``// in AP with last number of set as ``    ``// second element in AP ``    ``for` `(``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++) ``        ``\$L``[``\$i``][``\$n` `- 1] = 2; ` `    ``// Consider every element as ``    ``// second element of AP ``    ``for` `(``\$j` `= ``\$n` `- 2; ``\$j` `>= 1; ``\$j``--) ``    ``{ ``        ``// Search for i and k for j ``        ``\$i` `= ``\$j` `- 1;``        ``\$k` `= ``\$j` `+ 1; ``        ``while` `(``\$i` `>= 0 && ``\$k` `<= ``\$n` `- 1) ``        ``{ ``        ``if` `(``\$set``[``\$i``] + ``\$set``[``\$k``] < 2 * ``\$set``[``\$j``]) ``            ``\$k``++; ` `        ``// Before changing i, set L[i][j] as 2 ``        ``else` `if` `(``\$set``[``\$i``] + ``\$set``[``\$k``] > 2 * ``\$set``[``\$j``]) ``        ``{ ``            ``\$L``[``\$i``][``\$j``] = 2;``            ``\$i``--; } ` `        ``else``        ``{ ``            ``// Found i and k for j, LLAP with ``            ``// i and j as first two elements ``            ``// is equal to LLAP with j and k ``            ``// as first two elements plus 1.``            ``// L[j][k] must have been filled ``            ``// before as we run the loop from``            ``// right side ``            ``\$L``[``\$i``][``\$j``] = ``\$L``[``\$j``][``\$k``] + 1; ` `            ``// Update overall LLAP, if needed ``            ``\$llap` `= max(``\$llap``, ``\$L``[``\$i``][``\$j``]); ` `            ``// Change i and k to fill more ``            ``// L[i][j] values for current j ``            ``\$i``--; ``            ``\$k``++; ``        ``} ``        ``} ` `        ``// If the loop was stopped due to k ``        ``// becoming more than n-1, set the``        ``// remaining entities in column j as 2 ``        ``while` `(``\$i` `>= 0) ``        ``{ ``            ``\$L``[``\$i``][``\$j``] = 2; ``            ``\$i``--; ``        ``} ``    ``} ``    ``return` `\$llap``; ``} ` `// Driver Code``\$set1` `= ``array``(1, 7, 10, 13, 14, 19); ``\$n1` `= sizeof(``\$set1``); ``echo` `lenghtOfLongestAP(``\$set1``, ``\$n1``),``"\n"``; ` `\$set2` `= ``array``(1, 7, 10, 15, 27, 29); ``\$n2` `= sizeof(``\$set2``); ``echo` `lenghtOfLongestAP(``\$set2``, ``\$n2``),``"\n"``; ` `\$set3` `= ``array``(2, 4, 6, 8, 10); ``\$n3` `= sizeof(``\$set3``); ``echo` `lenghtOfLongestAP(``\$set3``, ``\$n3``),``"\n"``; ` `// This code is contributed by Sach_Code``?>`

Output
```4
3
5
```

Time Complexity: O(n2
Auxiliary Space: O(n2)

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