Find an element which is coprime with all elements of a given array

Given an array arr[] consisting of N positive integers, the task is to find an integer greater than 1 which is coprime with all the given array elements.

Examples:

Input: arr[ ] = {10,13,17,19}
Output: 23
Explanation:
The GCD of 23 with every array element is 1. Therefore, 23 is coprime with all the given array elements.

Input: arr[] = {13, 17, 23, 24, 50}
Output: 53
Explanation:
The GCD of 53 with every array element is 1. Therefore, 53 is coprime with all the given array elements.

Approach: The idea is to use the fact that a prime number greater than the maximum array element will be coprime with all the given array elements. Therefore, simply find the prime number greater than the largest element present in the array.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find an element which
// is coprime with all array elements
int find_X(int arr[], int N)
{

    // Stores maximum array element
    int R = INT_MIN;
    for (int i = 0; i < N; i++)
        R = max(R, arr[i]);

    // Stores if index of an array is prime or not
    bool prime[1000001];
    for (int i = 0; i < 1000001; i++)
        prime[i] = true;

    int p = 2;
    while (p * p <= 1000002)
    {

        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == true)
        {

            // Update all multiples of p
            for (int i = p * 2; i < 1000001; i += p)
            {
                prime[i] = false;
            }
        }

        // Increment p by 1
        p = p + 1;
    }

    prime[0] = false;
    prime[1] = false;

    // Traverse the range [R, 10000000 + 1]
    for (int i = R; i < 1000001; i++) {

        // If i is greater than R and prime
        if (i > R and prime[i] == true) {

            // Return i
            return i;
        }
    }

    // Dummy value to omit return error
    return -1;
}

// Driven Program
int main()
{
    // Given array
    int arr[] = { 10, 13, 17, 19 };

    // stores the length of array
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function call
    cout << find_X(arr, N);

    return 0;
}

// This code is contributed by Kingash.

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG {

  // Function to find an element which
  // is coprime with all array elements
  static int find_X(int arr[])
  {

    // Stores maximum array element
    int R = Integer.MIN_VALUE;
    for (int i = 0; i < arr.length; i++)
      R = Math.max(R, arr[i]);

    // Stores if index of an array is prime or not
    boolean prime[] = new boolean[1000001];
    Arrays.fill(prime, true);

    int p = 2;
    while (p * p <= 1000002) {

      // If prime[p] is not changed,
      // then it is a prime
      if (prime[p] == true) {

        // Update all multiples of p
        for (int i = p * 2; i < 1000001; i += p) {
          prime[i] = false;
        }
      }
      // Increment p by 1
      p = p + 1;
    }

    prime[0] = false;
    prime[1] = false;

    // Traverse the range [R, 10000000 + 1]
    for (int i = R; i < 1000001; i++) {

      // If i is greater than R and prime
      if (i > R && prime[i] == true) {

        // Return i
        return i;
      }
    }

    // Dummy value to omit return error
    return -1;
  }

  // Driver Code
  public static void main(String[] args)
  {
    // Given array
    int arr[] = { 10, 13, 17, 19 };

    // Function call
    System.out.println(find_X(arr));
  }
}

// This code is contributed by Kingash.

Python3

# Python3 program for the above approach

import math

# Function to find an element which
# is coprime with all array elements
def find_X(arr):

    # Stores maximum array element
    R = max(arr)

    # Stores if index of an array is prime or not
    prime = [True for i in range(0, 1000001)]

    p = 2
    while (p * p <= 1000002):

        # If prime[p] is not changed,
        # then it is a prime
        if (prime[p] == True):

            # Update all multiples of p
            for i in range(p * 2, 1000001, p):
                prime[i] = False

        # Increment p by 1
        p = p + 1

    prime[0] = False
    prime[1] = False

    # Traverse the range [R, 10000000 + 1]
    for i in range(R, 1000001):

        # If i is greater than R and prime
        if i > R and prime[i] == True:

           # Return i
            return i


# Driver Code
arr = [10, 13, 17, 19]
print(find_X(arr))

C#

// C# program for the above approach
using System;

class GFG{

// Function to find an element which
// is coprime with all array elements
static int find_X(int[] arr)
{

    // Stores maximum array element
    int R = Int32.MinValue;

    for(int i = 0; i < arr.Length; i++)
        R = Math.Max(R, arr[i]);

    // Stores if index of an array is prime or not
    bool[] prime = new bool[1000001];

    for(int i = 0; i < 1000001; i++)
    {
        prime[i] = true;
    }

    int p = 2;

    while (p * p <= 1000002)
    {

        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == true)
        {

            // Update all multiples of p
            for (int i = p * 2; i < 1000001; i += p)
            {
                prime[i] = false;
            }
        }

        // Increment p by 1
        p = p + 1;
    }

    prime[0] = false;
    prime[1] = false;

    // Traverse the range [R, 10000000 + 1]
    for(int i = R; i < 1000001; i++)
    {

        // If i is greater than R and prime
        if (i > R && prime[i] == true)
        {

            // Return i
            return i;
        }
    }

    // Dummy value to omit return error
    return -1;
}

// Driver Code
public static void Main(String []args)
{

    // Given array
    int[] arr = { 10, 13, 17, 19 };

    // Function call
    Console.WriteLine(find_X(arr));
}
}

// This code is contributed by souravghosh0416

Javascript

<script>

// Javascript program for the above approach

// Function to find an element which
// is coprime with all array elements
function find_X(arr)
{

    // Stores maximum array element
    let R = Number.MIN_VALUE;
    for(let i = 0; i < arr.length; i++)
        R = Math.max(R, arr[i]);

    // Stores if index of an array is prime or not
    let prime = Array(1000001).fill(true);

    let p = 2;
    while (p * p <= 1000002)
    {

        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == true)
        {

            // Update all multiples of p
            for(let i = p * 2; i < 1000001; i += p)
            {
                prime[i] = false;
            }
        }

        // Increment p by 1
        p = p + 1;
    }
    prime[0] = false;
    prime[1] = false;

    // Traverse the range [R, 10000000 + 1]
    for(let i = R; i < 1000001; i++)
    {

        // If i is greater than R and prime
        if (i > R && prime[i] == true)
        {

            // Return i
            return i;
        }
    }

    // Dummy value to omit return error
    return -1;
}

// Driver code

// Given array
let arr = [ 10, 13, 17, 19 ];

// Function call
document.write(find_X(arr));

// This code is contributed by target_2

</script>

Output:

23

Time Complexity: O(N*log(N))
Auxiliary Space: O(N)