GCD of all subarrays of size K
Last Updated :
23 Nov, 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:
- 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++
#include <bits/stdc++.h>
using namespace std;
void printSub( int arr[], int N,
int K)
{
for ( int i = 0; i <= N - K; i++) {
int gcd = arr[i];
for ( int j = i + 1; j < i + K;
j++) {
gcd = __gcd(gcd, arr[j]);
}
cout << gcd << " " ;
}
}
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);
}
|
Java
class GFG{
static int __gcd( int a, int b)
{
if (b == 0 )
return a;
return __gcd(b, a % b);
}
static void printSub( int arr[],
int N, int K)
{
for ( int i = 0 ; i <= N - K; i++)
{
int gcd = arr[i];
for ( int j = i + 1 ; j < i + K; j++)
{
gcd = __gcd(gcd, arr[j]);
}
System.out.print(gcd + " " );
}
}
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);
}
}
|
Python3
from math import gcd
def printSub(arr, N, K):
for i in range (N - K + 1 ):
g = arr[i]
for j in range (i + 1 , i + K):
g = gcd(g, arr[j])
print (g, end = " " )
if __name__ = = '__main__' :
arr = [ 2 , 4 , 3 , 9 , 14 ,
20 , 25 , 17 ]
K = 2
N = len (arr)
printSub(arr, N, K)
|
C#
using System;
class GFG{
static int __gcd( int a, int b)
{
if (b == 0)
return a;
return __gcd(b, a % b);
}
static void printSub( int []arr,
int N, int K)
{
for ( int i = 0; i <= N - K; i++)
{
int gcd = arr[i];
for ( int j = i + 1; j < i + K; j++)
{
gcd = __gcd(gcd, arr[j]);
}
Console.Write(gcd + " " );
}
}
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);
}
}
|
Javascript
<script>
function __gcd(a, b)
{
if (b == 0)
return a;
return __gcd(b, a % b);
}
function prletSub(arr, N, K)
{
for (let i = 0; i <= N - K; i++)
{
let gcd = arr[i];
for (let j = i + 1; j < i + K; j++)
{
gcd = __gcd(gcd, arr[j]);
}
document.write(gcd + " " );
}
}
let arr = [2, 4, 3, 9,
14, 20, 25, 17];
let K = 2;
let N = arr.length;
prletSub(arr, N, K);
</script>
|
Time Complexity: O((N – K + 1) * K)
Auxiliary Space: O(1)
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