Smallest prime giving remainder K when divided by any Array element

• Difficulty Level : Basic
• Last Updated : 12 Jun, 2022

Given an array of integers arr[] of size N and an integer K, the task is to find the smallest prime such that it gives remainder K when it is divided by any of the elements of the array.

Note: The prime number should be in the range [1, 106]

Examples:

Input: arr[]= {3, 4, 5}, K = 1
Output: 61
Explanation: The smallest prime that satisfies the condition is 61.
61%3 = 1, 61%4 = 1 and 61%5 =1.

Input: arr[] = {2, 4, 5}, K = 3
Output: -1
Explanation: The output should not be possible because
no number can have remainder 3 when divided by 2.

Input: arr[] = {10, 4, 5}, K = 3
Output: 23
Explanation: The smallest prime that satisfies the condition is 23.
23%10 = 3, 23%4 = 3 and 23%5 = 3

Approach: The idea to solve the problem is mentioned below:

If any element of the array is smaller than K then solution is not possible. Because no number can have a remainder higher than itself when dividing a number.

The LCM of numbers is the smallest number divisible by all the other numbers. So find the LCM of the array and check for each multiple of the LCM whether LCM + K is prime or not to get the smallest prime which gives the remainder as K.

Follow the below steps for the above:

• First check,If minimum element of the array is less than k {ie; min(arr)>=K}, then solution is not possible, then return -1
• Find the lcm of all the elements of the array(say lcm).
• Iterate for each multiple of lcm which are less than or equal to 106:
• Check whether (lcm + K) is prime or not for each multiple of lcm
• If the condition holds true, return the sum of K  and the current multiple of lcm as the smallest_prime.
• If no prime number is found, return -1.

Below is the implementation of the above approach:

C++

 // C++ code for the above approach   #include using namespace std;   // Function to find the // gcd of two integers long long gcd(long long int a,               long long int b) {     if (b == 0)         return a;     return gcd(b, a % b); } // Function to check if an integer // is prime or not bool checkPrime(long long n) {     long long ub = sqrt(n);     for (long long i = 2; i <= ub; i++) {         if (n % i == 0) {             return false;         }     }     return true; }   // Function to find the smallest prime // that gives the same remainder when // divided by any of the element in arr. long long smallestPrime(vector arr,                         int q) {       // Finding the lcm of the array     long long lcm = arr[0];     for (int i : arr) {         lcm = (lcm * i) / (gcd(i, lcm));     }       // Edge case, if the value of any     // in the array is less than remainder     for (auto i : arr) {         if (i < q)             return -1;     }     // Check whether (lcm + remainder) is     // prime or not, for all multiples of lcm     for (long long i = lcm; i <= 1e9;          i += lcm) {         if (checkPrime(i + q)) {             return i + q;         }     }       // If any prime satisfying the     // condition is not found     // simply return -1;     return -1; }   // Driver code int main() {       vector arr = { 2, 5, 4 };     int q = 3;     cout << smallestPrime(arr, q);     return 0; }

Java

 // Java code for the above approach import java.io.*;   class GFG {     // Function to find the   // gcd of two integers   public static long gcd(long a, long b)   {     if (b == 0)       return a;     return gcd(b, a % b);   }     // Function to check if an integer   // is prime or not   public static boolean checkPrime(long n)   {     long ub = (long)Math.sqrt(n);     for (int i = 2; i <= ub; i++) {       if (n % i == 0) {         return false;       }     }     return true;   }     // Function to find the smallest prime   // that gives the same remainder when   // divided by any of the element in arr.   public static long smallestPrime(int arr[], int q)   {       // Finding the lcm of the array     long lcm = (long)arr[0];     for (int i : arr) {       lcm =(long)(lcm * (long)i) / (long)(gcd((long)i,(long)lcm));     }       // Edge case, if the value of any     // in the array is less than remainder     for (int i : arr) {       if (i < q)         return -1;     }     // Check whether (lcm + remainder) is     // prime or not, for all multiples of lcm     for (long i = lcm; i <= 1e9; i += lcm) {       if (checkPrime(i + q)) {         return i + q;       }     }       // If any prime satisfying the     // condition is not found     // simply return -1;     return -1;   }     public static void main(String[] args)   {     int arr[] = { 2, 5, 4 };     int q = 3;     System.out.print(smallestPrime(arr, q));   } }   // This code is contributed by Rohit Pradhan

Python3

 # Python code for the above approach import math   # Function to find the # gcd of two integers def gcd( a, b):       if b == 0:         return a;     return gcd(b, a % b);   # Function to check if an integer # is prime or not def checkPrime(n):       ub = math.sqrt(n);     for i in range(2,ub):         if (n % i == 0):             return 0;     return 1;   # Function to find the smallest prime # that gives the same remainder when # divided by any of the element in arr. def smallestPrime(arr, q):       # Finding the lcm of the array     lcm = arr[0];     for i in range(len(arr)):         lcm = (lcm * i) / (gcd(arr[i], lcm));           # Edge case, if the value of any     # in the array is less than remainder     for i in range(len(arr)):         if (arr[i] < q):             return -1;           # Check whether (lcm + remainder) is     # prime or not, for all multiples of lcm     for i in range(lcm, 1e9, lcm):         if (checkPrime(i + q)):             return i + q;               # If any prime satisfying the     # condition is not found     # simply return -1;     return -1;   # Driver code arr = [2, 5, 4]; q = 3; print(smallestPrime(arr, q));     # This code is contributed by Potta Lokesh

C#

 // C# code for the above approach using System;   class GFG {     // Function to find the   // gcd of two integers   static long gcd(long a, long b)   {     if (b == 0)       return a;     return gcd(b, a % b);   }     // Function to check if an integer   // is prime or not   static bool checkPrime(long n)   {     long ub = (long)Math.Sqrt(n);     for (int i = 2; i <= ub; i++) {       if (n % i == 0) {         return false;       }     }     return true;   }     // Function to find the smallest prime   // that gives the same remainder when   // divided by any of the element in arr.   static long smallestPrime(int []arr, int q)   {       // Finding the lcm of the array     long lcm = (long)arr[0];     for (int i = 0; i < arr.Length; i++) {       lcm =(long)(lcm * (long)arr[i]) / (long)(gcd((long)arr[i],(long)lcm));     }       // Edge case, if the value of any     // in the array is less than remainder     for (int i = 0; i < arr.Length; i++) {       if (arr[i] < q)         return -1;     }     // Check whether (lcm + remainder) is     // prime or not, for all multiples of lcm     for (long i = lcm; i <= 1e9; i += lcm) {       if (checkPrime(i + q)) {         return i + q;       }     }       // If any prime satisfying the     // condition is not found     // simply return -1;     return -1;   }     public static void Main()   {     int []arr = { 2, 5, 4 };     int q = 3;     Console.Write(smallestPrime(arr, q));   } }   // This code is contributed by Samim Hossain Mondal

Javascript



Output

-1

Time Complexity: O(N * logD + (M/lcm)*sqrt(M)) where D is max of array, M is upper possible limit of prime, lcm is LCM of all array elements
Auxiliary Space: O(1)

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