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

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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++




// 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;
}


Java




// 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..


Python 3




# 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] * n
    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


C#




// 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


PHP




<?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
?>


Javascript




<script>
 
// Javascript 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);
 
    var P = new Array(n);
    var S = new Array(n);
 
    // Creating prefix array of GCD
    P[0] = a[0];
    for(var 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 (var i = n - 2; i >= 0; i--)
        S[i] = __gcd(S[i + 1], a[i]);
 
    // Iterate through the array
    for (var i = 0; i < n; i++) {
 
        // Variable to store the divisor
        var 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
var a = [ 12, 6, 18, 12, 16 ];
var n = a.length;
     
document.write(getDivisor(a, n));
 
// This code is contributed by kirti
 
</script>


Output

6

Complexity Analysis:

  • Time Complexity: O(nlogn)
  • Auxiliary Space: O(n)


Last Updated : 12 Sep, 2022
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