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Longest subarray forming a Geometic Progression (GP)
  • Last Updated : 20 Apr, 2021

Given a sorted array arr[] consisting of distinct numbers, the task is to find the length of the longest subarray that forms a Geometric Progression.

Examples:

Input: arr[]={1, 2, 4, 7, 14, 28, 56, 89}
Output: 4
Explanation:
The subarrays {1, 2, 4} and {7, 14, 28, 56} forms a GP.
Since {7, 14, 28, 56} is he longest, the required output is 4.

Input: arr[]={3, 6, 7, 12, 24, 28, 56}
Output: 2

Naive Approach: The simplest approach to solve the problem is to generate all possible subarrays and for each subarray, check if it forms a GP or not. Keep updating the maximum length of such subarrays found. Finally, print the maximum length obtained.
Time Complexity: O(N3)
Auxiliary Space: O(N)

Efficient Approach: The above approach can be optimized by the following steps:



  • Traverse the array and select a pair of adjacent elements, i.e., arr[i] and arr[i+1], as the first two terms of the Geometric Progression.
  • If arr[i+1] is not divisible by arr[i], then it cannot be considered for the common ratio. Otherwise, take arr[i+1] / arr[i] as the common ratio for the current Geometric Progression.
  • Increase and store the length of the Geometric Progression if the subsequent elements have the same common ratio. Otherwise, update the common ratio equal to the ratio of the new pair of adjacent elements.
  • Finally, return the length of the longest subarray that forms a Geometric Progression as the output.

Below is the implementation of the above approach:

C++




// C++ Program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the length of
// the longest subarray forming a
// GP in a sorted array
int longestGP(int A[], int N)
{
    // Base Case
    if (N < 2)
        return N;
 
    // Stores the length of GP
    // and the common ratio
    int length = 1, common_ratio = 1;
 
    // Stores the maximum
    // length of the GP
    int maxlength = 1;
 
    // Traverse the array
    for (int i = 0; i < N - 1; i++) {
 
        // Check if the common ratio
        // is valid for GP
        if (A[i + 1] % A[i] == 0) {
 
            // If the current common ratio
            // is equal to previous common ratio
            if (A[i + 1] / A[i] == common_ratio) {
 
                // Increment the length of the GP
                length = length + 1;
 
                // Store the max length of GP
                maxlength
                    = max(maxlength, length);
            }
 
            // Otherwise
            else {
 
                // Update the common ratio
                common_ratio = A[i + 1] / A[i];
 
                // Update the length of GP
                length = 2;
            }
        }
        else {
 
            // Store the max length of GP
            maxlength
                = max(maxlength, length);
 
            // Update the length of GP
            length = 1;
        }
    }
 
    // Store the max length of GP
    maxlength = max(maxlength, length);
 
    // Return the max length of GP
    return maxlength;
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 1, 2, 4, 7, 14, 28, 56, 89 };
 
    // Length of the array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << longestGP(arr, N);
 
    return 0;
}

Java




// Java program to implement
// the above approach
import java.io.*;
 
class GFG{
 
// Function to return the length of
// the longest subarray forming a
// GP in a sorted array
static int longestGP(int A[], int N)
{
     
    // Base Case
    if (N < 2)
        return N;
 
    // Stores the length of GP
    // and the common ratio
    int length = 1, common_ratio = 1;
 
    // Stores the maximum
    // length of the GP
    int maxlength = 1;
 
    // Traverse the array
    for(int i = 0; i < N - 1; i++)
    {
 
        // Check if the common ratio
        // is valid for GP
        if (A[i + 1] % A[i] == 0)
        {
 
            // If the current common ratio
            // is equal to previous common ratio
            if (A[i + 1] / A[i] == common_ratio)
            {
 
                // Increment the length of the GP
                length = length + 1;
 
                // Store the max length of GP
                maxlength = Math.max(maxlength, length);
            }
 
            // Otherwise
            else
            {
                 
                // Update the common ratio
                common_ratio = A[i + 1] / A[i];
 
                // Update the length of GP
                length = 2;
            }
        }
        else
        {
 
            // Store the max length of GP
            maxlength = Math.max(maxlength, length);
 
            // Update the length of GP
            length = 1;
        }
    }
 
    // Store the max length of GP
    maxlength = Math.max(maxlength, length);
 
    // Return the max length of GP
    return maxlength;
}
 
// Driver code
public static void main (String[] args)
{
     
    // Given array    
    int arr[] = { 1, 2, 4, 7, 14, 28, 56, 89 };
     
    // Length of the array
    int N = arr.length;
     
    // Function call
    System.out.println(longestGP(arr, N));
}
}
 
// This code is contributed by jana_sayantan   

Python3




# Python3 program to implement
# the above approach
 
# Function to return the length of
# the longest subarray forming a
# GP in a sorted array
def longestGP(A, N):
     
    # Base Case
    if (N < 2):
        return N
 
    # Stores the length of GP
    # and the common ratio
    length = 1
    common_ratio = 1
 
    # Stores the maximum
    # length of the GP
    maxlength = 1
 
    # Traverse the array
    for i in range(N - 1):
 
        # Check if the common ratio
        # is valid for GP
        if (A[i + 1] % A[i] == 0):
 
            # If the current common ratio
            # is equal to previous common ratio
            if (A[i + 1] // A[i] == common_ratio):
 
                # Increment the length of the GP
                length = length + 1
 
                # Store the max length of GP
                maxlength = max(maxlength, length)
             
            # Otherwise
            else:
 
                # Update the common ratio
                common_ratio = A[i + 1] // A[i]
 
                # Update the length of GP
                length = 2
             
        else:
 
            # Store the max length of GP
            maxlength = max(maxlength, length)
 
            # Update the length of GP
            length = 1
         
    # Store the max length of GP
    maxlength = max(maxlength, length)
 
    # Return the max length of GP
    return maxlength
 
# Driver Code
 
# Given array
arr = [ 1, 2, 4, 7, 14, 28, 56, 89 ]
 
# Length of the array
N = len(arr)
 
# Function call
print(longestGP(arr, N))
 
# This code is contributed by sanjoy_62

C#




// C# program to implement
// the above approach
using System;
class GFG{
 
// Function to return the length of
// the longest subarray forming a
// GP in a sorted array
static int longestGP(int []A, int N)
{    
    // Base Case
    if (N < 2)
        return N;
 
    // Stores the length of GP
    // and the common ratio
    int length = 1, common_ratio = 1;
 
    // Stores the maximum
    // length of the GP
    int maxlength = 1;
 
    // Traverse the array
    for(int i = 0; i < N - 1; i++)
    {
        // Check if the common ratio
        // is valid for GP
        if (A[i + 1] % A[i] == 0)
        {
            // If the current common ratio
            // is equal to previous common ratio
            if (A[i + 1] / A[i] == common_ratio)
            {
                // Increment the length of the GP
                length = length + 1;
 
                // Store the max length of GP
                maxlength = Math.Max(maxlength,
                                     length);
            }
 
            // Otherwise
            else
            {               
                // Update the common ratio
                common_ratio = A[i + 1] /
                               A[i];
 
                // Update the length of GP
                length = 2;
            }
        }
        else
        {
            // Store the max length of GP
            maxlength = Math.Max(maxlength,
                                 length);
 
            // Update the length of GP
            length = 1;
        }
    }
 
    // Store the max length of GP
    maxlength = Math.Max(maxlength,
                         length);
 
    // Return the max length of GP
    return maxlength;
}
 
// Driver code
public static void Main(String[] args)
{    
    // Given array    
    int []arr = {1, 2, 4, 7,
                 14, 28, 56, 89};
     
    // Length of the array
    int N = arr.Length;
     
    // Function call
    Console.WriteLine(longestGP(arr, N));
}
}
 
// This code is contributed by shikhasingrajput

Javascript




<script>
 
// JavaScript implementation of the above approach
 
// Function to return the length of
// the longest subarray forming a
// GP in a sorted array
function longestGP(A, N)
{
       
    // Base Case
    if (N < 2)
        return N;
   
    // Stores the length of GP
    // and the common ratio
    let length = 1, common_ratio = 1;
   
    // Stores the maximum
    // length of the GP
    let maxlength = 1;
   
    // Traverse the array
    for(let i = 0; i < N - 1; i++)
    {
   
        // Check if the common ratio
        // is valid for GP
        if (A[i + 1] % A[i] == 0)
        {
   
            // If the current common ratio
            // is equal to previous common ratio
            if (A[i + 1] / A[i] == common_ratio)
            {
   
                // Increment the length of the GP
                length = length + 1;
   
                // Store the max length of GP
                maxlength = Math.max(maxlength, length);
            }
   
            // Otherwise
            else
            {
                   
                // Update the common ratio
                common_ratio = A[i + 1] / A[i];
   
                // Update the length of GP
                length = 2;
            }
        }
        else
        {
   
            // Store the max length of GP
            maxlength = Math.max(maxlength, length);
   
            // Update the length of GP
            length = 1;
        }
    }
   
    // Store the max length of GP
    maxlength = Math.max(maxlength, length);
   
    // Return the max length of GP
    return maxlength;
}
   
// Driver code
         
    // Given array    
    let arr = [ 1, 2, 4, 7, 14, 28, 56, 89 ];
       
    // Length of the array
    let N = arr.length;
       
    // Function call
    document.write(longestGP(arr, N));
   
  // This code is contributed by code_hunt.
</script>
Output: 
4

Time Complexity: O(N)
Space Complexity: O(1)

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