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Smallest prime number missing in an array

  • Difficulty Level : Basic
  • Last Updated : 21 May, 2021

Given an array containing n distinct numbers. The task is to find the smallest prime which is not present in the array. 
Note: If there is no prime number missing up to the maximum element of the array then print “No prime number missing”.
Examples: 
 

Input: arr[] = {9, 11, 4, 2, 3, 7, 0, 1}
Output: 5
5 is the smallest prime, which is not present in array.

Input: arr[] = {3, 0, 2, 5}
Output: No prime number missing
As 5 is the maximum element and all prime numbers upto 5 
are present in the array.

 

Approach: First of all, find all prime numbers using Sieve of Eratosthenes then sequentially check which one is not present there. Just iterate over the array and check whether the number is there or not using hashing.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
 
// this store all prime number
// upto 10^5
// this function find all prime
vector<ll> findPrime(int MAX)
{
    bool pm[MAX + 1];
    memset(pm, true, sizeof(pm));
 
    // use sieve to find prime
    pm[0] = pm[1] = false;
    for (int i = 2; i <= MAX; i++)
        if (pm[i])
            for (int j = 2 * i; j <= MAX; j += i)
                pm[j] = false;
 
    // if number is prime then
    // store it in prime vector
    vector<ll> prime;
    for (int i = 0; i <= MAX; i++)
        if (pm[i])
            prime.push_back(i);
 
    return prime;
}
 
// Function to find the smallest prime missing
int findSmallest(int arr[], int n)
{
    int MAX = *max_element(arr, arr + n);
 
    // first of all find all prime
    vector<ll> prime = findPrime(MAX);
 
    // now store all number as index of freq arr
    // so that we can improve searching
    unordered_set<int> s;
    for (int i = 0; i < n; i++)
        s.insert(arr[i]);
 
    // now check for each prime
    int ans = -1;
    for (int i = 0; i < prime.size(); i++)
        if (s.find(prime[i]) == s.end()) {
            ans = prime[i];
            break;
        }
    return ans;
}
 
// Driver code
int main()
{
    int arr[] = { 3, 0, 1, 2, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // find smallest prime
    // which is not present
    if (findSmallest(arr, n) == -1)
        cout << "No prime number missing";
    else
        cout << findSmallest(arr, n);
 
    return 0;
}

Java




// Java implementation of the above approach
import java.util.*;
 
class GFG {
 
    // this store all prime number
    // upto 10^5
    // this function find all prime
    static Vector<Integer> findPrime(int MAX)
    {
        boolean pm[] = new boolean[MAX + 1];
        for (int i = 0; i < pm.length; i++)
            pm[i] = true;
 
        // use sieve to find prime
        pm[0] = pm[1] = false;
        for (int i = 2; i <= MAX; i++)
            if (pm[i])
                for (int j = 2 * i; j <= MAX; j += i)
                    pm[j] = false;
 
        // if number is prime then
        // store it in prime vector
        Vector<Integer> prime = new Vector<Integer>();
        for (int i = 0; i <= MAX; i++)
            if (pm[i])
                prime.add(i);
 
        return prime;
    }
 
    static int max_element(int arr[])
    {
        int max = arr[0];
 
        for (int i = 0; i < arr.length; i++)
            max = Math.max(max, arr[i]);
 
        return max;
    }
 
    // Function to find the smallest prime missing
    static int findSmallest(int arr[], int n)
    {
        int MAX = max_element(arr);
 
        // first of all find all prime
        Vector<Integer> prime = findPrime(MAX);
 
        // now store all number as index of freq arr
        // so that we can improve searching
        Set<Integer> s = new HashSet<Integer>();
        for (int i = 0; i < arr.length; i++)
            s.add(arr[i]);
 
        // now check for each prime
        long ans = -1;
        for (int i = 0; i < prime.size(); i++) {
            if (!s.contains(prime.get(i))) {
 
                ans = (prime.get(i));
                break;
            }
        }
        return (int)ans;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int arr[] = { 3, 0, 1, 2, 7 };
        int n = arr.length;
 
        // find smallest prime
        // which is not present
        if (findSmallest(arr, n) == -1)
            System.out.print("No prime number missing");
        else
            System.out.print(findSmallest(arr, n));
    }
}
 
// This code is contributed by Arnab Kundu

Python3




# Python3 implementation of the above approach
 
# This function finds all
# prime numbers upto 10 ^ 5
def findPrime(MAX):
 
    pm = [True] * (MAX + 1)
 
    # use sieve to find prime
    pm[0], pm[1] = False, False
     
    for i in range(2, MAX + 1):
        if pm[i] == True:
             
            for j in range(2 * i, MAX + 1, i):
                pm[j] = False
 
    # If number is prime, then
    # store it in prime list
    prime = []
    for i in range(0, MAX + 1):
        if pm[i] == True:
            prime.append(i)
 
    return prime
 
# Function to find the smallest prime missing
def findSmallest(arr, n):
 
    MAX = max(arr)
     
    # first of all find all prime
    prime = findPrime(MAX)
 
    # now store all number as index of freq
    # arr so that we can improve searching
    s = set()
    for i in range(0, n):
        s.add(arr[i])
 
    # now check for each prime
    ans = -1
    for i in range(0, len(prime)):
        if prime[i] not in s:
            ans = prime[i]
            break
         
    return ans
 
# Driver Code
if __name__ == "__main__":
 
    arr = [3, 0, 1, 27]
    n = len(arr)
 
    # find smallest prime
    # which is not present
    if(findSmallest(arr, n) == -1):
        print("No prime number missing")
    else:
        print(findSmallest(arr, n))
 
# This code is contributed by Rituraj Jain

C#




// C# implementation of the above approach
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG
{
 
    // this store all prime number
    // upto 10^5
    // this function find all prime
    static ArrayList findPrime(int MAX)
    {
        bool[] pm = new bool[MAX + 1];
        for (int i = 0; i < MAX + 1; i++)
            pm[i] = true;
 
        // use sieve to find prime
        pm[0] = pm[1] = false;
        for (int i = 2; i <= MAX; i++)
            if (pm[i])
                for (int j = 2 * i; j <= MAX; j += i)
                    pm[j] = false;
 
        // if number is prime then
        // store it in prime vector
        ArrayList prime = new ArrayList();
        for (int i = 0; i <= MAX; i++)
            if (pm[i])
                prime.Add(i);
 
        return prime;
    }
 
    static int max_element(int []arr)
    {
        int max = arr[0];
 
        for (int i = 0; i < arr.Length; i++)
            max = Math.Max(max, arr[i]);
 
        return max;
    }
 
    // Function to find the smallest prime missing
    static int findSmallest(int []arr, int n)
    {
        int MAX = max_element(arr);
 
        // first of all find all prime
        ArrayList prime = findPrime(MAX);
 
        // now store all number as index of freq arr
        // so that we can improve searching
        HashSet<int> s = new HashSet<int>();
        for (int i = 0; i < arr.Length; i++)
            s.Add(arr[i]);
 
        // now check for each prime
        int ans = -1;
        for (int i = 0; i < prime.Count; i++)
        {
            if (!s.Contains((int)prime[i]))
            {
 
                ans = (int)(prime[i]);
                break;
            }
        }
        return (int)ans;
    }
 
    // Driver code
    static void Main()
    {
        int []arr = { 3, 0, 1, 2, 7 };
        int n = arr.Length;
 
        // find smallest prime
        // which is not present
        if (findSmallest(arr, n) == -1)
            Console.Write("No prime number missing");
        else
            Console.Write(findSmallest(arr, n));
    }
}
 
// This code is contributed by mits

Javascript




<script>
// Javascript implementation of the above approach
 
// this store all prime number
// upto 10^5
// this function find all prime
function findPrime(MAX)
{
    let pm = new Array(MAX + 1);
    pm.fill(true);
 
    // use sieve to find prime
    pm[0] = pm[1] = false;
    for (let i = 2; i <= MAX; i++)
        if (pm[i])
            for (let j = 2 * i; j <= MAX; j += i)
                pm[j] = false;
 
    // if number is prime then
    // store it in prime vector
    let prime = new Array();
    for (let i = 0; i <= MAX; i++)
        if (pm[i])
            prime.push(i);
 
    return prime;
}
 
// Function to find the smallest prime missing
function findSmallest(arr, n) {
    let MAX = arr.sort((A, B) => A - B).reverse()[0];
 
    // first of all find all prime
    let prime = findPrime(MAX);
 
    // now store all number as index of freq arr
    // so that we can improve searching
    let s = new Set();
    for (let i = 0; i < n; i++)
        s.add(arr[i]);
 
    // now check for each prime
    let ans = -1;
    for (let i = 0; i < prime.length; i++){
        if (!s.has(prime[i])) {
            ans = prime[i];
            break;
        }
    }
    return ans;
}
 
// Driver code
let arr = [3, 0, 1, 2, 7];
let n = arr.length;
 
// find smallest prime
// which is not present
if (findSmallest(arr, n) == -1)
    document.write("No prime number missing");
else
    document.write(findSmallest(arr, n));
 
// This code is contributed by gfgking.
</script>
Output: 
5

 

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