Find all triplets in a sorted array that forms Geometric Progression - GeeksforGeeks

08 Apr, 2021

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++
Java
Python 3
C#
Javascript
 

 
 










// 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;
}






 

 
 










// 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






 

 
 










# 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






 

 
 










// 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.






 

 
 










<script>
// Javascript 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.
    function findGeometricTriplets(arr,n)
    {
     
        // One by fix every element as middle element
        for (let j = 1; j < n - 1; j++)
        {
         
            // Initialize i and k for the current j
            let 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
                    document.write(arr[i] +" " + arr[j]
                        + " " + arr[k]+"<br>");
       
                    // 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
    // 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};
     
    let arr = [1, 2, 4, 16];
     
    // int arr[] = {1, 2, 3, 6, 18, 22};
    let n = arr.length;
    findGeometricTriplets(arr, n);
     
    // This code is contributed by avanitrachhadiya2155
</script>






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.This article is contributed by Aditya Goel. 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.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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