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:
- Initialize a variable, say gcd, to store the GCD of the current subarray.
- Generate K-length subarrays from the given array.
- 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 <bits/stdc++.h>using namespace std;// Function to print the gcd// of each subarray of length Kvoid 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 Codeint main(){ int arr[] = { 2, 4, 3, 9, 14, 20, 25, 17 }; int K = 2; int N = sizeof(arr) / sizeof(arr[0]); printSub(arr, N, K);} |
Java
// Java program to implement// the above approachclass 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 Kstatic 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 Codepublic 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 approachfrom math import gcd# Function to print the gcd# of each subarray of length Kdef 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 Codeif __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 approachusing 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 Kstatic 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 Codepublic 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
<script>// Javascript program to implement// the above approach function __gcd(a, b){ if (b == 0) return a; return __gcd(b, a % b);} // Function to print the gcd// of each subarray of length Kfunction prletSub(arr, N, K){ for (let i = 0; i <= N - K; i++) { // Store GCD of subarray let gcd = arr[i]; for (let j = i + 1; j < i + K; j++) { // Update GCD of subarray gcd = __gcd(gcd, arr[j]); } // Print GCD of subarray document.write(gcd + " "); }}// Driver Code let arr = [2, 4, 3, 9, 14, 20, 25, 17]; let K = 2; let N = arr.length; prletSub(arr, N, K);// This code is contributed by avijitmondal1998.</script> |
Output:
2 1 3 1 2 5 1
Time Complexity: O((N – K + 1) * K)
Auxiliary Space: O(1)
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