Given a set of numbers, find the **L**ength of the **L**ongest **G**eometrix **P**rogression (**LLGP**) in it. The common ratio of GP must be an integer.

Examples:

set[] = {5, 7, 10, 15, 20, 29} output = 3 The longest arithmetic progression is {5, 10, 20} set[] = {3, 9, 27, 81} output = 4

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 <iostream> ` `#include <algorithm> ` `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[1] % set[0] == 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[0]); ` ` ` `cout << lenOfLongestGP(set1, n1) << ` `"\n"` `; ` ` ` ` ` `int` `set2[] = {1, 3, 4, 9, 7, 27}; ` ` ` `int` `n2 = ` `sizeof` `(set2)/` `sizeof` `(set2[0]); ` ` ` `cout << lenOfLongestGP(set2, n2) << ` `"\n"` `; ` ` ` ` ` `int` `set3[] = {2, 3, 5, 7, 11, 13}; ` ` ` `int` `n3 = ` `sizeof` `(set3)/` `sizeof` `(set3[0]); ` ` ` `cout << lenOfLongestGP(set3, n3) << ` `"\n"` `; ` ` ` ` ` `return` `0; ` `} ` |

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## 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 ` |

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**Output:**

6 4 1

**Time Complexity:** O(n^{2})

**Auxiliary Space:** O(n^{2})

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.

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