Longest subarray forming a Geometic Progression (GP)

• Last Updated : 04 Jan, 2022

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 the 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 ` `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

 ``

Output:

`4`

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

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