Find all triplets in a sorted array that forms Geometric Progression
Given a sorted array of distinct positive integers, print all triplets that forms Geometric Progression with integral common ratio.
A geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. For example, the sequence 2, 6, 18, 54,… is a geometric progression with common ratio 3.
Examples:
Input: arr = [1, 2, 6, 10, 18, 54] Output: 2 6 18 6 18 54 Input: arr = [2, 8, 10, 15, 16, 30, 32, 64] Output: 2 8 32 8 16 32 16 32 64 Input: arr = [ 1, 2, 6, 18, 36, 54] Output: 2 6 18 1 6 36 6 18 54
The idea is to start from the second element and fix every element as middle element and search for the other two elements in a triplet (one smaller and one greater). For an element arr[j] to be middle of geometric progression, there must exist elements arr[i] and arr[k] such that –
arr[j] / arr[i] = r and arr[k] / arr[j] = r where r is an positive integer and 0 <= i < j and j < k <= n - 1
Below is the implementation of above idea –
C++
// C++ program to find if there exist three elements in // Geometric Progression or not #include <iostream> using namespace std; // The function prints three elements in GP if exists // Assumption: arr[0..n-1] is sorted. void findGeometricTriplets(int arr[], 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 all i and k such that (i, j, k) // forms a triplet of GP while (i >= 0 && k <= n - 1) { // if arr[j]/arr[i] = r and arr[k]/arr[j] = r // and r is an integer (i, j, k) forms Geometric // Progression while (arr[j] % arr[i] == 0 && arr[k] % arr[j] == 0 && arr[j] / arr[i] == arr[k] / arr[j]) { // print the triplet cout << arr[i] << " " << arr[j] << " " << arr[k] << endl; // Since the array is sorted and elements // are distinct. k++ , i--; } // if arr[j] is multiple of arr[i] and arr[k] is // multiple of arr[j], then arr[j] / arr[i] != // arr[k] / arr[j]. We compare their values to // move to next k or previous i. if(arr[j] % arr[i] == 0 && arr[k] % arr[j] == 0) { if(arr[j] / arr[i] < arr[k] / arr[j]) i--; else k++; } // else if arr[j] is multiple of arr[i], then // try next k. Else, try previous i. else if (arr[j] % arr[i] == 0) k++; else i--; } } } // Driver code int main() { // int arr[] = {1, 2, 6, 10, 18, 54}; // int arr[] = {2, 8, 10, 15, 16, 30, 32, 64}; // int arr[] = {1, 2, 6, 18, 36, 54}; int arr[] = {1, 2, 4, 16}; // int arr[] = {1, 2, 3, 6, 18, 22}; int n = sizeof(arr) / sizeof(arr[0]); findGeometricTriplets(arr, n); return 0; } |
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Java
// Java program to find if there exist three elements in // Geometric Progression or not import java.util.*; class GFG { // The function prints three elements in GP if exists // Assumption: arr[0..n-1] is sorted. static void findGeometricTriplets(int arr[], 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 all i and k such that (i, j, k) // forms a triplet of GP while (i >= 0 && k <= n - 1) { // if arr[j]/arr[i] = r and arr[k]/arr[j] = r // and r is an integer (i, j, k) forms Geometric // Progression while (i >= 0 && arr[j] % arr[i] == 0 && arr[k] % arr[j] == 0 && arr[j] / arr[i] == arr[k] / arr[j]) { // print the triplet System.out.println(arr[i] +" " + arr[j] + " " + arr[k]); // Since the array is sorted and elements // are distinct. k++ ; i--; } // if arr[j] is multiple of arr[i] and arr[k] is // multiple of arr[j], then arr[j] / arr[i] != // arr[k] / arr[j]. We compare their values to // move to next k or previous i. if(i >= 0 && arr[j] % arr[i] == 0 && arr[k] % arr[j] == 0) { if(i >= 0 && arr[j] / arr[i] < arr[k] / arr[j]) i--; else k++; } // else if arr[j] is multiple of arr[i], then // try next k. Else, try previous i. else if (i >= 0 && arr[j] % arr[i] == 0) k++; else i--; } } } // Driver code public static void main(String[] args) { // int arr[] = {1, 2, 6, 10, 18, 54}; // int arr[] = {2, 8, 10, 15, 16, 30, 32, 64}; // int arr[] = {1, 2, 6, 18, 36, 54}; int arr[] = {1, 2, 4, 16}; // int arr[] = {1, 2, 3, 6, 18, 22}; int n = arr.length; findGeometricTriplets(arr, n); } } // This code is contributed by Rajput-Ji |
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Python 3
# Python 3 program to find if # there exist three elements in # Geometric Progression or not # The function prints three elements # in GP if exists. # Assumption: arr[0..n-1] is sorted. def findGeometricTriplets(arr, n): # One by fix every element # as middle element for j in range(1, n - 1): # Initialize i and k for # the current j i = j - 1 k = j + 1 # Find all i and k such that # (i, j, k) forms a triplet of GP while (i >= 0 and k <= n - 1): # if arr[j]/arr[i] = r and # arr[k]/arr[j] = r and r # is an integer (i, j, k) forms # Geometric Progression while (arr[j] % arr[i] == 0 and arr[k] % arr[j] == 0 and arr[j] // arr[i] == arr[k] // arr[j]): # print the triplet print( arr[i] , " " , arr[j], " " , arr[k]) # Since the array is sorted and # elements are distinct. k += 1 i -= 1 # if arr[j] is multiple of arr[i] # and arr[k] is multiple of arr[j], # then arr[j] / arr[i] != arr[k] / arr[j]. # We compare their values to # move to next k or previous i. if(arr[j] % arr[i] == 0 and arr[k] % arr[j] == 0): if(arr[j] // arr[i] < arr[k] // arr[j]): i -= 1 else: k += 1 # else if arr[j] is multiple of # arr[i], then try next k. Else, # try previous i. elif (arr[j] % arr[i] == 0): k += 1 else: i -= 1 # Driver code if __name__ =="__main__": arr = [1, 2, 4, 16] n = len(arr) findGeometricTriplets(arr, n) # This code is contributed # by ChitraNayal |
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C#
// C# program to find if there exist three elements // in Geometric Progression or not using System; class GFG { // The function prints three elements in GP if exists // Assumption: arr[0..n-1] is sorted. static void findGeometricTriplets(int []arr, 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 all i and k such that (i, j, k) // forms a triplet of GP while (i >= 0 && k <= n - 1) { // if arr[j]/arr[i] = r and arr[k]/arr[j] = r // and r is an integer (i, j, k) forms Geometric // Progression while (i >= 0 && arr[j] % arr[i] == 0 && arr[k] % arr[j] == 0 && arr[j] / arr[i] == arr[k] / arr[j]) { // print the triplet Console.WriteLine(arr[i] +" " + arr[j] + " " + arr[k]); // Since the array is sorted and elements // are distinct. k++ ; i--; } // if arr[j] is multiple of arr[i] and arr[k] is // multiple of arr[j], then arr[j] / arr[i] != // arr[k] / arr[j]. We compare their values to // move to next k or previous i. if(i >= 0 && arr[j] % arr[i] == 0 && arr[k] % arr[j] == 0) { if(i >= 0 && arr[j] / arr[i] < arr[k] / arr[j]) i--; else k++; } // else if arr[j] is multiple of arr[i], then // try next k. Else, try previous i. else if (i >= 0 && arr[j] % arr[i] == 0) k++; else i--; } } } // Driver code static public void Main () { // int arr[] = {1, 2, 6, 10, 18, 54}; // int arr[] = {2, 8, 10, 15, 16, 30, 32, 64}; // int arr[] = {1, 2, 6, 18, 36, 54}; int []arr = {1, 2, 4, 16}; // int arr[] = {1, 2, 3, 6, 18, 22}; int n = arr.Length; findGeometricTriplets(arr, n); } } // This code is contributed by ajit. |
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Output:
1 2 4 1 4 16
Time complexity of above solution is O(n2) as for every j, we are finding i and k in linear time.
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