GCD of all subarrays of size K

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++

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// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
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[0]);
 
    printSub(arr, N, K);
}

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Java

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// 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

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Python3

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# Python3 program to implement
# the above approach
from math import gcd
 
# Function to prthe 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

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C#

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// 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

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Output: 

2 1 3 1 2 5 1







 

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

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