# Longest Geometric Progression

Given a set of numbers, find the Length of the Longest Geometrix Progression (LLGP) in it. The common ratio of GP must be an integer.

Examples:

```set[] = {5, 7, 10, 15, 20, 29}
output = 3
The longest geometric progression is {5, 10, 20}

set[] = {3, 9, 27, 81}
output = 4
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

This problem is similar to Longest Arithmetic Progression Problem. We can solve this problem using Dynamic Programming.

We first sort the given set. We use an auxiliary table L[n][n] to store results of subproblems. An entry L[i][j] in this table stores LLGP with set[i] and set[j] as first two elements of GP and j > i. The table is filled from bottom right to top left. To fill the table, j (second element in GP) is first fixed. i and k are searched for a fixed j. If i and k are found such that i, j, k form an GP, then the value of L[i][j] is set as L[j][k] + 1. Note that the value of L[j][k] must have been filled before as the loop traverses from right to left columns.

Following is implementation of the Dynamic Programming algorithm.

## C++

 `// C++ program to find length of the longest geometric ` `// progression in a given set ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `// Returns length of the longest GP subset of set[] ` `int` `lenOfLongestGP(``int` `set[], ``int` `n) ` `{ ` `    ``// Base cases ` `    ``if` `(n < 2) ` `        ``return` `n; ` `    ``if` `(n == 2) ` `        ``return` `(set % set == 0); ` ` `  `    ``// Let us sort the set first ` `    ``sort(set, set+n); ` ` `  `    ``// An entry L[i][j] in this table stores LLGP with ` `    ``// set[i] and set[j] as first two elements of GP ` `    ``// and j > i. ` `    ``int` `L[n][n]; ` ` `  `    ``// Initialize result (A single element is always a GP) ` `    ``int` `llgp = 1; ` ` `  `    ``// Initialize values of last column ` `    ``for` `(``int` `i = 0; i < n; ++i) ` `        ``if` `(set[n-1] % set[i] == 0) ` `            ``L[i][n-1] = 2; ` `        ``else` `            ``L[i][n-1] = 1; ` ` `  ` `  `    ``// Consider every element as second element of GP ` `    ``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) ` `        ``{ ` `            ``// Two cases when i, j and k don't form ` `            ``// a GP. ` `            ``if` `(set[i] * set[k] < set[j]*set[j]) ` `                ``++k; ` ` `  `            ``else` `if` `(set[i] * set[k] > set[j]*set[j]) ` `            ``{ ` `                ``if` `(set[j] % set[i] == 0) ` `                    ``L[i][j] = 2; ` `                ``else` `                    ``L[i][j] = 1; ` `                ``--i; ` `            ``} ` ` `  ` `  `            ``// i, j and k form GP, LLGP with i and j as ` `            ``// first two elements is equal to LLGP 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 ` `            ``else` `            ``{ ` `                ``L[i][j] = L[j][k] + 1; ` ` `  `                ``// Update overall LLGP ` `                ``if` `(L[i][j] > llgp) ` `                    ``llgp = 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 entties ` `        ``// in column j as 1 or 2 based on divisibility ` `        ``// of set[j] by set[i] ` `        ``while` `(i >= 0) ` `        ``{ ` `            ``if` `(set[j] % set[i] == 0) ` `                ``L[i][j] = 2; ` `            ``else` `                ``L[i][j] = 1; ` `            ``--i; ` `        ``} ` `    ``} ` ` `  `    ``// Return result ` `    ``return` `llgp; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `set1[] = {1, 3, 9, 27, 81, 243}; ` `    ``int` `n1 = ``sizeof``(set1)/``sizeof``(set1); ` `    ``cout << lenOfLongestGP(set1, n1) << ``"\n"``; ` ` `  `    ``int` `set2[] = {1, 3, 4, 9, 7, 27}; ` `    ``int` `n2 = ``sizeof``(set2)/``sizeof``(set2); ` `    ``cout << lenOfLongestGP(set2, n2) << ``"\n"``; ` ` `  `    ``int` `set3[] = {2, 3, 5, 7, 11, 13}; ` `    ``int` `n3 = ``sizeof``(set3)/``sizeof``(set3); ` `    ``cout << lenOfLongestGP(set3, n3) << ``"\n"``; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find length  ` `// of the longest geometric  ` `// progression in a given set  ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Returns length of the longest GP subset of set[]  ` `    ``static` `int` `lenOfLongestGP(``int` `set[], ``int` `n)  ` `    ``{ ` `        ``// Base cases  ` `        ``if` `(n < ``2``)  ` `        ``{ ` `            ``return` `n; ` `        ``} ` `        ``if` `(n == ``2``) ` `        ``{ ` `            ``return` `(set[``1``] % set[``0``] == ``0` `? ``1` `: ``0``); ` `        ``} ` ` `  `        ``// Let us sort the set first  ` `        ``Arrays.sort(set); ` ` `  `        ``// An entry L[i][j] in this table  ` `        ``// stores LLGP with set[i] and set[j] ` `        ``// as first two elements of GP  ` `        ``// and j > i.  ` `        ``int` `L[][] = ``new` `int``[n][n]; ` ` `  `        ``// Initialize result (A single  ` `        ``// element is always a GP)  ` `        ``int` `llgp = ``1``; ` ` `  `        ``// Initialize values of last column  ` `        ``for` `(``int` `i = ``0``; i < n; ++i)  ` `        ``{ ` `            ``if` `(set[n - ``1``] % set[i] == ``0``) ` `            ``{ ` `                ``L[i][n - ``1``] = ``2``; ` `            ``}  ` `            ``else`  `            ``{ ` `                ``L[i][n - ``1``] = ``1``; ` `            ``} ` `        ``} ` ` `  `        ``// Consider every element as second element of GP  ` `        ``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``)  ` `            ``{ ` `                ``// Two cases when i, j and k  ` `                ``// don't form a GP.  ` `                ``if` `(set[i] * set[k] < set[j] * set[j]) ` `                ``{ ` `                    ``++k; ` `                ``}  ` `                ``else` `if` `(set[i] * set[k] > set[j] * set[j])  ` `                ``{ ` `                    ``if` `(set[j] % set[i] == ``0``)  ` `                    ``{ ` `                        ``L[i][j] = ``2``; ` `                    ``}  ` `                    ``else`  `                    ``{ ` `                        ``L[i][j] = ``1``; ` `                    ``} ` `                    ``--i; ` `                ``}  ` `                 `  `                ``// i, j and k form GP, LLGP with i and j as  ` `                ``// first two elements is equal to LLGP 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  ` `                ``else`  `                ``{ ` `                    ``L[i][j] = L[j][k] + ``1``; ` ` `  `                    ``// Update overall LLGP  ` `                    ``if` `(L[i][j] > llgp)  ` `                    ``{ ` `                        ``llgp = 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 entties  ` `            ``// in column j as 1 or 2 based on divisibility  ` `            ``// of set[j] by set[i]  ` `            ``while` `(i >= ``0``)  ` `            ``{ ` `                ``if` `(set[j] % set[i] == ``0``)  ` `                ``{ ` `                    ``L[i][j] = ``2``; ` `                ``}  ` `                ``else`  `                ``{ ` `                    ``L[i][j] = ``1``; ` `                ``} ` `                ``--i; ` `            ``} ` `        ``} ` ` `  `        ``// Return result  ` `        ``return` `llgp; ` `    ``} ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `set1[] = {``1``, ``3``, ``9``, ``27``, ``81``, ``243``}; ` `        ``int` `n1 = set1.length; ` `        ``System.out.print(lenOfLongestGP(set1, n1) + ``"\n"``); ` ` `  `        ``int` `set2[] = {``1``, ``3``, ``4``, ``9``, ``7``, ``27``}; ` `        ``int` `n2 = set2.length; ` `        ``System.out.print(lenOfLongestGP(set2, n2) + ``"\n"``); ` ` `  `        ``int` `set3[] = {``2``, ``3``, ``5``, ``7``, ``11``, ``13``}; ` `        ``int` `n3 = set3.length; ` `        ``System.out.print(lenOfLongestGP(set3, n3) + ``"\n"``); ` `    ``} ` `} ` ` `  `/* This code contributed by PrinciRaj1992 */`

## Python3

 `# Python3 program to find length of the longest geometric ` `# progression in a given sett  ` ` `  `# Returns length of the longest GP subset of sett[] ` `def` `lenOfLongestGP(sett,n): ` `    ``# Base cases  ` `    ``if` `n<``2``: ` `        ``return` `n ` `    ``if` `n``=``=``2``: ` `        ``return` `(sett[``1``] ``%` `sett[``0``] ``=``=` `0``) ` `    ``# let us sort the sett first  ` `    ``sett.sort() ` `     `  `    ``# An entry L[i][j] in this table stores LLGP with  ` `    ``# sett[i] and sett[j] as first two elements of GP ` `    ``# and j > i. ` `    ``L``=``[[``0` `for` `i ``in` `range``(n)] ``for` `i ``in` `range``(n)] ` ` `  `    ``# Initialize result (A single element is always a GP)  ` `    ``llgp``=``1` ` `  `    ``# Initialize values of last column  ` `    ``for` `i ``in` `range``(``0``,n): ` `        ``if` `sett[n``-``1``] ``%` `sett[i] ``=``=` `0``: ` `            ``L[i][n``-``1``] ``=` `2` `        ``else``: ` `            ``L[i][n``-``1``] ``=` `1` `             `  `    ``# Consider every element as second element of GP  ` `    ``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``: ` `             `  `            ``# Two cases when i, j and k don't form  ` `            ``# a GP. ` `            ``if` `sett[i] ``*` `sett[k] < sett[j]``*``sett[j]: ` `                ``k``+``=``1` `            ``elif` `sett[i] ``*` `sett[k] > sett[j]``*``sett[j]: ` `                ``if` `sett[j] ``%` `sett[i] ``=``=` `0``: ` `                    ``L[i][j]``=``2` `                ``else``: ` `                    ``L[i][j]``=``1` `                ``i``-``=``1` `                 `  `            ``# i, j and k form GP, LLGP with i and j as  ` `            ``# first two elements is equal to LLGP 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 ` `            ``else``: ` `                ``L[i][j] ``=` `L[j][k] ``+` `1` `                 `  `                ``# Update overall LLGP ` `                ``if` `L[i][j] > llgp: ` `                    ``llgp ``=` `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, sett the remaining entties  ` `        ``# in column j as 1 or 2 based on divisibility  ` `        ``# of sett[j] by sett[i]  ` `        ``while` `i>``=``0``: ` `            ``if` `sett[j] ``%` `sett[i] ``=``=` `0``: ` `                ``L[i][j]``=``2` `            ``else``: ` `                ``L[i][j]``=``1` `            ``i``-``=``1` `    ``return` `llgp ` `# Driver code  ` `if` `__name__``=``=``'__main__'``: ` `    ``set1``=``[``1``, ``3``, ``9``, ``27``, ``81``, ``243``] ` `    ``n1``=``len``(set1) ` `    ``print``(lenOfLongestGP(set1,n1)) ` ` `  `    ``set2 ``=` `[``1``, ``3``, ``4``, ``9``, ``7``, ``27``] ` `    ``n2``=``len``(set2) ` `    ``print``(lenOfLongestGP(set2,n2)) ` ` `  `    ``set3 ``=` `[``2``, ``3``, ``5``, ``7``, ``11``, ``13``] ` `    ``n3``=``len``(set3) ` `    ``print``(lenOfLongestGP(set3,n3)) ` ` `  `# this code is contrubuted by sahilshelangia `

## C#

 `// C# program to find length  ` `// of the longest geometric  ` `// progression in a given Set  ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Returns length of the  ` `    ``// longest GP subset of Set[]  ` `    ``static` `int` `lenOfLongestGP(``int` `[]Set, ``int` `n)  ` `    ``{ ` `        ``// Base cases  ` `        ``if` `(n < 2)  ` `        ``{ ` `            ``return` `n; ` `        ``} ` `        ``if` `(n == 2) ` `        ``{ ` `            ``return` `(Set % Set == 0 ? 1 : 0); ` `        ``} ` ` `  `        ``// Let us sort the Set first  ` `        ``Array.Sort(Set); ` ` `  `        ``// An entry L[i,j] in this table  ` `        ``// stores LLGP with Set[i] and Set[j] ` `        ``// as first two elements of GP  ` `        ``// and j > i.  ` `        ``int` `[,]L = ``new` `int``[n, n]; ` ` `  `        ``// Initialize result (A single  ` `        ``// element is always a GP)  ` `        ``int` `llgp = 1; ` ` `  `        ``// Initialize values of last column  ` `        ``for` `(``int` `i = 0; i < n; ++i)  ` `        ``{ ` `            ``if` `(Set[n - 1] % Set[i] == 0) ` `            ``{ ` `                ``L[i, n - 1] = 2; ` `            ``}  ` `            ``else` `            ``{ ` `                ``L[i, n - 1] = 1; ` `            ``} ` `        ``} ` ` `  `        ``// Consider every element  ` `        ``// as second element of GP  ` `        ``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)  ` `            ``{ ` `                ``// Two cases when i, j and k  ` `                ``// don't form a GP.  ` `                ``if` `(Set[i] * Set[k] < Set[j] * Set[j]) ` `                ``{ ` `                    ``++k; ` `                ``}  ` `                ``else` `if` `(Set[i] * Set[k] > Set[j] * Set[j])  ` `                ``{ ` `                    ``if` `(Set[j] % Set[i] == 0)  ` `                    ``{ ` `                        ``L[i,j] = 2; ` `                    ``}  ` `                    ``else` `                    ``{ ` `                        ``L[i,j] = 1; ` `                    ``} ` `                    ``--i; ` `                ``}  ` `                 `  `                ``// i, j and k form GP, LLGP with i and j as  ` `                ``// first two elements is equal to LLGP 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  ` `                ``else` `                ``{ ` `                    ``L[i, j] = L[j, k] + 1; ` ` `  `                    ``// Update overall LLGP  ` `                    ``if` `(L[i, j] > llgp)  ` `                    ``{ ` `                        ``llgp = 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 entties  ` `            ``// in column j as 1 or 2 based on divisibility  ` `            ``// of Set[j] by Set[i]  ` `            ``while` `(i >= 0)  ` `            ``{ ` `                ``if` `(Set[j] % Set[i] == 0)  ` `                ``{ ` `                    ``L[i, j] = 2; ` `                ``}  ` `                ``else` `                ``{ ` `                    ``L[i, j] = 1; ` `                ``} ` `                ``--i; ` `            ``} ` `        ``} ` ` `  `        ``// Return result  ` `        ``return` `llgp; ` `    ``} ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``int` `[]set1 = {1, 3, 9, 27, 81, 243}; ` `        ``int` `n1 = set1.Length; ` `        ``Console.Write(lenOfLongestGP(set1, n1) + ``"\n"``); ` ` `  `        ``int` `[]set2 = {1, 3, 4, 9, 7, 27}; ` `        ``int` `n2 = set2.Length; ` `        ``Console.Write(lenOfLongestGP(set2, n2) + ``"\n"``); ` ` `  `        ``int` `[]set3 = {2, 3, 5, 7, 11, 13}; ` `        ``int` `n3 = set3.Length; ` `        ``Console.Write(lenOfLongestGP(set3, n3) + ``"\n"``); ` `    ``} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

Output:

```6
4
1```

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

This article is contributed by Vivek Pandya. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.