Find element in array that divides all array elements
Given an array of n non-negative integers. Find such element in the array, that all array elements are divisible by it.
Examples :
Input : arr[] = {2, 2, 4}
Output : 2
Input : arr[] = {2, 1, 3, 1, 6}
Output : 1
Input: arr[] = {2, 3, 5}
Output : -1
Brute Force Approach:
The brute force approach to solve this problem would be to iterate through all the elements of the array and check if any of them can divide all other elements of the array. If such a number is found, return it as the answer. Otherwise, return -1 as no such number exists.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int findNumber( int arr[], int n)
{
for ( int i = 0; i < n; i++) {
int j;
for (j = 0; j < n; j++)
if (arr[j] % arr[i] != 0)
break ;
if (j == n)
return arr[i];
}
return -1;
}
int main()
{
int arr[] = { 2, 2, 4 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << findNumber(arr, n) << endl;
return 0;
}
|
Java
import java.util.*;
public class Main {
static int findNumber( int arr[], int n) {
for ( int i = 0 ; i < n; i++) {
int j;
for (j = 0 ; j < n; j++) {
if (arr[j] % arr[i] != 0 )
break ;
}
if (j == n)
return arr[i];
}
return - 1 ;
}
public static void main(String[] args) {
int arr[] = { 2 , 2 , 4 };
int n = arr.length;
System.out.println(findNumber(arr, n));
}
}
|
Python3
def findNumber(arr):
n = len (arr)
for i in range (n):
j = 0
while j < n:
if arr[j] % arr[i] ! = 0 :
break
j + = 1
if j = = n:
return arr[i]
return - 1
arr = [ 2 , 2 , 4 ]
print (findNumber(arr))
|
C#
using System;
class MainClass {
public static int FindNumber( int [] arr)
{
for ( int i = 0; i < arr.Length; i++) {
int j;
for (j = 0; j < arr.Length; j++)
if (arr[j] % arr[i] != 0)
break ;
if (j == arr.Length)
return arr[i];
}
return -1;
}
public static void Main( string [] args)
{
int [] arr = { 2, 2, 4 };
Console.WriteLine(FindNumber(arr));
}
}
|
Javascript
function findNumber(arr, n) {
for (let i = 0; i < n; i++) {
let j;
for (j = 0; j < n; j++)
if (arr[j] % arr[i] != 0)
break ;
if (j == n)
return arr[i];
}
return -1;
}
let arr = [2, 2, 4];
let n = arr.length;
console.log(findNumber(arr, n));
|
Output: 2
Time Complexity: O(N^2)
Space Complexity: O(1)
The approach is to calculate GCD of the entire array and then check if there exist an element equal to the GCD of the array. For calculating the gcd of the entire array we will use Euclidean algorithm.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
int gcd( int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
int findNumber( int arr[], int n)
{
int ans = arr[0];
for ( int i = 0; i < n; i++)
ans = gcd(ans, arr[i]);
for ( int i = 0; i < n; i++)
if (arr[i] == ans)
return ans;
return -1;
}
int main()
{
int arr[] = { 2, 2, 4 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << findNumber(arr, n) << endl;
return 0;
}
|
Java
import java.io.*;
class GFG {
static int gcd( int a, int b)
{
if (a == 0 )
return b;
return gcd(b % a, a);
}
static int findNumber( int arr[], int n)
{
int ans = arr[ 0 ];
for ( int i = 0 ; i < n; i++)
ans = gcd(ans, arr[i]);
for ( int i = 0 ; i < n; i++)
if (arr[i] == ans)
return ans;
return - 1 ;
}
public static void main(String args[])
{
int arr[] = { 2 , 2 , 4 };
int n = arr.length;
System.out.println(findNumber(arr, n));
}
}
|
Python3
def gcd (a, b) :
if (a = = 0 ) :
return b
return gcd (b % a, a)
def findNumber (arr, n) :
ans = arr[ 0 ]
for i in range ( 0 , n) :
ans = gcd (ans, arr[i])
for i in range ( 0 , n) :
if (arr[i] = = ans) :
return ans
return - 1
arr = [ 2 , 2 , 4 ];
n = len (arr)
print (findNumber(arr, n))
|
Javascript
<script>
function gcd(a, b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
function findNumber(arr, n)
{
let ans = arr[0];
for (let i = 0; i < n; i++)
ans = gcd(ans, arr[i]);
for (let i = 0; i < n; i++)
if (arr[i] == ans)
return ans;
return -1;
}
let arr = [ 2, 2, 4 ];
let n = arr.length;
document.write(findNumber(arr, n));
</script>
|
C#
using System;
class GFG {
static int gcd( int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static int findNumber( int [] arr, int n)
{
int ans = arr[0];
for ( int i = 0; i < n; i++)
ans = gcd(ans, arr[i]);
for ( int i = 0; i < n; i++)
if (arr[i] == ans)
return ans;
return -1;
}
public static void Main()
{
int [] arr = { 2, 2, 4 };
int n = arr.Length;
Console.WriteLine(findNumber(arr, n));
}
}
|
PHP
<?php
function gcd ( $a , $b )
{
if ( $a == 0)
return $b ;
return gcd ( $b % $a , $a );
}
function findNumber ( $arr , $n )
{
$ans = $arr [0];
for ( $i = 0; $i < $n ; $i ++)
$ans = gcd ( $ans , $arr [ $i ]);
for ( $i = 0; $i < $n ; $i ++)
if ( $arr [ $i ] == $ans )
return $ans ;
return -1;
}
$arr = array (2, 2, 4);
$n = sizeof( $arr );
echo findNumber( $arr , $n ), "\n" ;
?>
|
Time complexity: O(n*logn)
Auxiliary space: O(Amax), where Amax is the maximum element in the given array.
Last Updated :
19 Apr, 2023
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