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# GCD of all subarrays of size K

• Last Updated : 06 May, 2021

Given an array, arr[] of size N, the task is to print the GCD of all subarrays of size K.

Examples:

Input: arr[] = {2, 4, 3, 9, 14, 20, 25, 17}, K = 2
Output: 2 1 3 1 2 5 1
Explanation:
gcd(2, 4}) = 2
gcd(4, 3) = 1
gcd(3, 9) = 3
gcd(9, 14) = 1
gcd(14, 20) = 2
gcd(20, 25) = 5
gcd(25, 17) = 1
Therefore, the required output is {2, 1, 3, 1, 2, 5, 1}

Input: arr[] = {2, 4, 8, 24, 14, 20, 25, 35, 7, 49, 7}, K = 3
Output: 2 4 2 2 1 5 1 7 7

Approach: The idea is to generate all subarrays of size K and print the GCD of each subarray. To efficiently compute the GCD of each subarray, the idea is to use the following property of GCD.

GCD(A1, A2, A3, …, AK) = GCD(A1, GCD(A2, A3, A4, …., AK))

Follow the steps below to solve the problem:

1. Initialize a variable, say gcd, to store the GCD of the current subarray.
2. Generate K-length subarrays from the given array.
3. Applying the above property of GCD, compute the GCD of each subarray, and print the obtained result.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach` `#include ``using` `namespace` `std;` `// Function to print the gcd``// of each subarray of length K``void` `printSub(``int` `arr[], ``int` `N,``              ``int` `K)``{``    ``for` `(``int` `i = 0; i <= N - K; i++) {` `        ``// Store GCD of subarray``        ``int` `gcd = arr[i];` `        ``for` `(``int` `j = i + 1; j < i + K;``             ``j++) {` `            ``// Update GCD of subarray``            ``gcd = __gcd(gcd, arr[j]);``        ``}` `        ``// Print GCD of subarray``        ``cout << gcd << ``" "``;``    ``}``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 2, 4, 3, 9, 14,``                  ``20, 25, 17 };``    ``int` `K = 2;``    ``int` `N = ``sizeof``(arr)``            ``/ ``sizeof``(arr);` `    ``printSub(arr, N, K);``}`

## Java

 `// Java program to implement``// the above approach``class` `GFG{` `static` `int` `__gcd(``int` `a, ``int` `b)``{``  ``if` `(b == ``0``)``    ``return` `a;``  ``return` `__gcd(b, a % b);``}``  ` `// Function to print the gcd``// of each subarray of length K``static` `void` `printSub(``int` `arr[],``                     ``int` `N, ``int` `K)``{``  ``for` `(``int` `i = ``0``; i <= N - K; i++)``  ``{``    ``// Store GCD of subarray``    ``int` `gcd = arr[i];` `    ``for` `(``int` `j = i + ``1``; j < i + K; j++)``    ``{``      ``// Update GCD of subarray``      ``gcd = __gcd(gcd, arr[j]);``    ``}` `    ``// Print GCD of subarray``    ``System.out.print(gcd + ``" "``);``  ``}``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``  ``int` `arr[] = {``2``, ``4``, ``3``, ``9``,``               ``14``, ``20``, ``25``, ``17``};``  ``int` `K = ``2``;``  ``int` `N = arr.length;``  ``printSub(arr, N, K);``}``}` `// This code is contributed by Chitranayal`

## Python3

 `# Python3 program to implement``# the above approach``from` `math ``import` `gcd` `# Function to print the gcd``# of each subarray of length K``def` `printSub(arr, N, K):``    ` `    ``for` `i ``in` `range``(N ``-` `K ``+` `1``):` `        ``# Store GCD of subarray``        ``g ``=` `arr[i]` `        ``for` `j ``in` `range``(i ``+` `1``, i ``+` `K):``            ` `            ``# Update GCD of subarray``            ``g ``=` `gcd(g, arr[j])` `        ``# Print GCD of subarray``        ``print``(g, end ``=` `" "``)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``arr ``=` `[ ``2``, ``4``, ``3``, ``9``, ``14``,``            ``20``, ``25``, ``17` `]``    ``K ``=` `2``    ``N ``=` `len``(arr)` `    ``printSub(arr, N, K)` `# This code is contributed by mohit kumar 29`

## C#

 `// C# program to implement``// the above approach``using` `System;``class` `GFG{` `static` `int` `__gcd(``int` `a, ``int` `b)``{``  ``if` `(b == 0)``    ``return` `a;``  ``return` `__gcd(b, a % b);``}``  ` `// Function to print the gcd``// of each subarray of length K``static` `void` `printSub(``int` `[]arr,``                     ``int` `N, ``int` `K)``{``  ``for` `(``int` `i = 0; i <= N - K; i++)``  ``{``    ``// Store GCD of subarray``    ``int` `gcd = arr[i];` `    ``for` `(``int` `j = i + 1; j < i + K; j++)``    ``{``      ``// Update GCD of subarray``      ``gcd = __gcd(gcd, arr[j]);``    ``}` `    ``// Print GCD of subarray``    ``Console.Write(gcd + ``" "``);``  ``}``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``  ``int` `[]arr = {2, 4, 3, 9,``               ``14, 20, 25, 17};``  ``int` `K = 2;``  ``int` `N = arr.Length;``  ``printSub(arr, N, K);``}``}`  `// This code is contributed by Princi Singh`

## Javascript

 ``
Output:
`2 1 3 1 2 5 1`

Time Complexity: O((N – K + 1) * K)
Auxiliary Space: O(1)

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