Find an integer X which is divisor of all except exactly one element in an array

Given an array of integers. Find an integer X which is the divisor of all except for exactly one element in the given array.

Note: The GCD of all the elements is not 1.

Examples:

Input : arr[] = {6, 18, 3, 12}
Output : 6
6 is the divisor of all except 3.

Input : arr[] = {40, 15, 30, 42}
Output : 3
3 is the divisor of all except 40.


Approach: Make a prefix array P such that index i contains the GCD of all the elements from 1 to i. Similarly, make a suffix array S such that index i contains the GCD of all the elements from i to n-1 (last index). If the GCD of P[i-1] and S[i+1] is not the divisor of the element at i, then it is the required answer.

Below is the implementation of the above approach:

C++

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// C++ program to find the  divisor  of all
// except for exactly one element in an array.
  
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns the  divisor  of all
// except for exactly one element in an array.
int getDivisor(int a[], int n)
{
    // There's only one element in the array
    if (n == 1)
        return (a[0] + 1);
  
    int P[n], S[n];
  
    // Creating prefix array of GCD
    P[0] = a[0];
    for (int i = 1; i < n; i++) 
         P[i] = __gcd(a[i], P[i - 1]);    
  
    // Creating suffix array of GCD 
    S[n-1] = a[n-1];
    for (int i = n - 2; i >= 0; i--) 
         S[i] = __gcd(S[i + 1], a[i]);    
  
    // Iterate through the array
    for (int i = 0; i <= n; i++) {
  
        // Variable to store the divisor
        int cur;
  
        // Getting the divisor
        if (i == 0)
            cur = S[i + 1];
        else if (i == n - 1)
            cur = P[i - 1];
        else
            cur = __gcd(P[i - 1], S[i + 1]);
  
        // Check if it is not a divisor of a[i]
        if (a[i] % cur != 0)
            return cur;
    }
  
    return 0;
}
  
// Driver code
int main()
{
    int a[] = { 12, 6, 18, 12, 16 };
  
    int n = sizeof(a) / sizeof(a[0]);
  
    cout << getDivisor(a, n);
  
    return 0;
}

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Java

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// Java  program to find the divisor of all
// except for exactly one element in an array.
import java.io.*;
  
class GFG {
      
// Recursive function to return gcd of a and b 
    static int __gcd(int a, int b) 
    
        // Everything divides 0  
        if (a == 0
          return b; 
        if (b == 0
          return a; 
         
        // base case 
        if (a == b) 
            return a; 
         
        // a is greater 
        if (a > b) 
            return __gcd(a-b, b); 
        return __gcd(a, b-a); 
    
// Function that returns the divisor of all
// except for exactly one element in an array.
static int getDivisor(int a[], int n)
{
    // There's only one element in the array
    if (n == 1)
        return (a[0] + 1);
  
    int P[] = new int[n];
    int    S[] = new int[n];
  
    // Creating prefix array of GCD
    P[0] = a[0];
    for (int i = 1; i < n; i++) 
        P[i] = __gcd(a[i], P[i - 1]); 
  
    // Creating suffix array of GCD 
    S[n-1] = a[n-1];
    for (int i = n - 2; i >= 0; i--) 
        S[i] = __gcd(S[i + 1], a[i]); 
  
    // Iterate through the array
    for (int i = 0; i <= n; i++) {
  
        // Variable to store the divisor
        int cur;
  
        // Getting the divisor
        if (i == 0)
            cur = S[i + 1];
        else if (i == n - 1)
            cur = P[i - 1];
        else
            cur = __gcd(P[i - 1], S[i + 1]);
  
        // Check if it is not a divisor of a[i]
        if (a[i] % cur != 0)
            return cur;
    }
  
    return 0;
}
  
// Driver code
  
  
    public static void main (String[] args) {
            int a[] = { 12, 6, 18, 12, 16 };
  
    int n = a.length;
  
    System.out.println(getDivisor(a, n));
    }
}
// This code is contributed by anuj_67..

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Python 3

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# Python 3 program to find the divisor of all 
# except for exactly one element in an array. 
from math import gcd
  
# Function to find the divisor of all 
# except for exactly one element in an array. 
def getDivisor(a, n):
      
    # There's only one element in the array 
    if (n == 1): 
        return (a[0] + 1
  
    P = [0] *
    S = [0] * n
  
    # Creating prefix array of GCD 
    P[0] = a[0]
    for i in range(1, n): 
        P[i] = gcd(a[i], P[i - 1]) 
  
    # Creating suffix array of GCD 
    S[n - 1] = a[n - 1]
    for i in range(n - 2, -1, -1): 
        S[i] = gcd(S[i + 1], a[i]) 
  
    # Iterate through the array 
    for i in range(0, n + 1): 
  
        # Variable to store the divisor 
        cur = 0
  
        # Getting the divisor 
        if (i == 0):
            cur = S[i + 1
        elif (i == n - 1):
            cur = P[i - 1]
        else:
            cur = gcd(P[i - 1], S[i + 1])
  
        # Check if it is not a divisor of a[i] 
        if (a[i] % cur != 0): 
            return cur
  
    return 0
  
# Driver Code
if __name__=='__main__'
    a = [12, 6, 18, 12, 16]
    n = len(a) 
      
    print(getDivisor(a, n))
      
# This code is contributed by Rupesh Rao

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C#

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// C# program to find the divisor of all
// except for exactly one element in an array.
using System;
  
class GFG 
{
      
// Recursive function to return gcd of a and b 
static int __gcd(int a, int b) 
    // Everything divides 0 
    if (a == 0) 
        return b; 
    if (b == 0) 
        return a; 
      
    // base case 
    if (a == b) 
        return a; 
      
    // a is greater 
    if (a > b) 
        return __gcd(a - b, b); 
    return __gcd(a, b - a); 
  
// Function that returns the divisor of all
// except for exactly one element in an array.
static int getDivisor(int[] a, int n)
{
      
    // There's only one element in the array
    if (n == 1)
        return (a[0] + 1);
      
    int[] P = new int[n];
    int[] S = new int[n];
      
    // Creating prefix array of GCD
    P[0] = a[0];
    for (int i = 1; i < n; i++) 
        P[i] = __gcd(a[i], P[i - 1]); 
      
    // Creating suffix array of GCD 
    S[n - 1] = a[n - 1];
    for (int i = n - 2; i >= 0; i--) 
        S[i] = __gcd(S[i + 1], a[i]); 
      
    // Iterate through the array
    for (int i = 0; i <= n; i++) 
    {
      
        // Variable to store the divisor
        int cur;
      
        // Getting the divisor
        if (i == 0)
            cur = S[i + 1];
        else if (i == n - 1)
            cur = P[i - 1];
        else
            cur = __gcd(P[i - 1], S[i + 1]);
      
        // Check if it is not a divisor of a[i]
        if (a[i] % cur != 0)
            return cur;
    }
      
    return 0;
}
  
// Driver code
public static void Main ()
{
    int[] a = { 12, 6, 18, 12, 16 };
  
    int n = a.Length;
      
    Console.WriteLine(getDivisor(a, n));
}
}
  
// This code is contributed 
// by Akanksha Rai

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PHP

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<?php
// PHP program to find the divisor of all 
// except for exactly one element in an array. 
  
// Recursive function to return 
// gcd of a and b 
function __gcd($a, $b
    // Everything divides 0 
    if ($a == 0) 
        return b; 
    if ($b == 0) 
        return $a
      
    // base case 
    if ($a == $b
        return $a
      
    // a is greater 
    if ($a > $b
        return __gcd($a - $b, $b); 
          
    return __gcd($a, $b - $a); 
  
// Function that returns the divisor of all 
// except for exactly one element in an array. 
function getDivisor($a, $n
    // There's only one element in the array 
    if ($n == 1) 
        return ($a[0] + 1); 
  
    $P = array() ;
    $S = array() ;
  
    // Creating prefix array of GCD 
    $P[0] = $a[0]; 
      
    for ($i = 1; $i < $n; $i++) 
        $P[$i] = __gcd($a[$i], $P[$i - 1]); 
  
    // Creating suffix array of GCD 
    $S[$n - 1] = $a[$n - 1]; 
    for ($i = $n - 2; $i >= 0; $i--) 
        $S[$i] = __gcd($S[$i + 1], $a[$i]); 
  
    // Iterate through the array 
    for ($i = 0; $i <= $n; $i++) 
    {
  
        // Getting the divisor 
        if ($i == 0) 
            $cur = $S[$i + 1]; 
        else if ($i == $n - 1) 
            $cur = $P[$i - 1]; 
        else
            $cur = __gcd($P[$i - 1], $S[$i + 1]); 
  
        // Check if it is not a divisor of a[i] 
        if ($a[$i] % $cur != 0) 
            return $cur
    
  
    return 0; 
  
// Driver code 
$a = array( 12, 6, 18, 12, 16 );
  
$n = sizeof($a);
  
echo getDivisor($a, $n);
  
// This code is contributed by Ryuga 
?>

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Output:

6


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