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
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 the 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) ? 2 : 1; // 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 - 1; ++i) { if (set[n-1] % set[i] == 0) { L[i][n-1] = 2; if (2 > llgp) llgp = 2; } else { L[i][n-1] = 1; } } L[n-1][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 { if (set[j] % set[i] == 0) { L[i][j] = L[j][k] + 1; // Update overall LLGP if (L[i][j] > llgp) llgp = L[i][j]; } else { L[i][j] = 1; } // 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 entries // 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; if (2 > llgp) llgp = 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; } |
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 ? 2 : 1); } // 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 - 1; ++i) { if (set[n - 1] % set[i] == 0) { L[i][n - 1] = 2; if (2 > llgp) llgp = 2; } else { L[i][n - 1] = 1; } } L[n - 1][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; if (2 > llgp) llgp = 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 { if (set[j] % set[i] == 0) { L[i][j] = L[j][k] + 1; // Update overall LLGP if (L[i][j] > llgp) { llgp = L[i][j]; } } else { L[i][j] = 1; } // 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 entries // 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; if (2 > llgp) llgp = 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 set # Returns length of the longest GP # subset of sett[] def lenOfLongestGP(sett, n): # Base cases if n < 2: return n if n == 2: return 2 if (sett[1] % sett[0] == 0) else 1 # 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-1): if sett[n-1] % sett[i] == 0: L[i][n-1] = 2 if 2 > llgp: llgp = 2 else: L[i][n-1] = 1 L[n-1][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 else if 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: if sett[j] % sett[i] == 0: L[i][j] = L[j][k] + 1 # Update overall LLGP if L[i][j] > llgp: llgp = L[i][j] else: L[i][j] = 1 # 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 entries # 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 contributed 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[1] % Set[0] == 0 ? 2 : 1); } // 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 - 1; ++i) { if (Set[n - 1] % Set[i] == 0) { L[i, n - 1] = 2; if (2 > llgp) llgp = 2; } else { L[i, n - 1] = 1; } } L[n - 1, 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; if (2 > llgp) llgp = 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 { if (Set[j] % Set[i] == 0) { L[i, j] = L[j, k] + 1; // Update overall LLGP if (L[i, j] > llgp) { llgp = L[i, j]; } } else { L[i, j] = 1; } // 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 entries // 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; if (2 > llgp) llgp = 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 |
Javascript
<script> // Javascript program to find length // of the longest geometric // progression in a given set // Returns length of the longest GP subset of set[] function lenOfLongestGP(set, n) { // Base cases if (n < 2) { return n; } if (n == 2) { return (set[1] % set[0] == 0 ? 2 : 1); } // Let us sort the set first set.sort(function(a, b){return a - b}); // 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. let L = new Array(n); for (let i = 0; i < n; ++i) { L[i] = new Array(n); for (let j = 0; j < n; ++j) { L[i][j] = 0; } } // Initialize result (A single // element is always a GP) let llgp = 1; // Initialize values of last column for (let i = 0; i < n - 1; ++i) { if (set[n - 1] % set[i] == 0) { L[i][n - 1] = 2; if (2 > llgp) llgp = 2; } else { L[i][n - 1] = 1; } } L[n - 1][n - 1] = 1; // Consider every element as second element of GP for (let j = n - 2; j >= 1; --j) { // Search for i and k for j let 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; if (2 > llgp) llgp = 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 { if (set[j] % set[i] == 0) { L[i][j] = L[j][k] + 1; // Update overall LLGP if (L[i][j] > llgp) { llgp = L[i][j]; } } else { L[i][j] = 1; } // 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 entries // 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; if (2 > llgp) llgp = 2; } else { L[i][j] = 1; } --i; } } // Return result return llgp; } let set1 = [ 1, 3, 9, 27, 81, 243 ]; let n1 = set1.length; document.write(lenOfLongestGP(set1, n1) + "</br>"); let set2 = [ 1, 3, 4, 9, 7, 27 ]; let n2 = set2.length; document.write(lenOfLongestGP(set2, n2) + "</br>"); let set3 = [ 2, 3, 5, 7, 11, 13 ]; let n3 = set3.length; document.write(lenOfLongestGP(set3, n3) + "</br>"); // This code is contributed by rameshtravel07. </script> |
Output:
6 4 1
Time Complexity: O(n2)
Auxiliary Space: O(n2)
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