Nearest prime number in the array of every array element
Last Updated :
13 Feb, 2023
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Given an integer array arr[] consisting of N integers, the task is to find the nearest Prime Number in the array for every element in the array. If the array does not contain any prime number, then print -1.
Examples:
Input: arr[] = {1, 2, 3, 1, 6}
Output: 2 2 3 3 3
Explanation:
For the subarray {1, 2}, the nearest prime number is 2.
For the subarray {3, 1, 6}, the nearest prime number is 3.Input: arr[] = {8, 7, 12, 15, 3, 11}
Output: 7 7 7 3 3 11
Explanation:
For the subarray {8, 7, 12}, the nearest prime number is 7.
For the subarray {15, 3}, the nearest prime number is 3.
For the subarray {11}, the nearest prime number is 11 itself.
Approach:
Follow the steps below to solve the problem:
- Find the maximum element maxm in the array.
- Compute and store all prime numbers up to maxm using Sieve of Eratosthenes
- Traverse the array and store the indices of the prime numbers.
- If no prime numbers are present in the array, print -1 for all indices.
- Point curr to the first index consisting of a prime number.
- For every index up to curr, print the arr[primes[curr]] as the nearest prime number.
- For indices exceeding curr, compare the distance with primes[curr] and primes[curr + 1]. If primes[curr] is nearer, print arr[primes[curr]]. Otherwise, increment curr and print arr[primes[curr]].
- If curr is the last prime in the array, print arr[primes[curr]] for all indices onwards.
Below is the implementation of the above approach:
// C++ program to find nearest
// prime number in the array
// for all array elements
#include <bits/stdc++.h>
using namespace std;
#define max 10000000
// Create a boolean array and set all
// entries it as false. A value in
// prime[i] will be true if i is not a
// prime, else false
bool prime[max] = { false };
// Sieve of Eratosthenes function
void SieveOfEratosthenes(int maxm)
{
prime[0] = prime[1] = true;
for (int i = 2; i * i <= maxm; i++) {
// Update all multiples of i greater
// than or equal to the square of it
// numbers which are multiple of i and are
// less than i^2 are already been marked.
if (!prime[i]) {
for (int j = i * i; j <= maxm; j += i) {
prime[j] = true;
}
}
}
}
// Function to find nearest
// prime number for all elements
void print_nearest_prime(int arr[], int N)
{
int maxm = *max_element(arr, arr + N);
// Compute and store all prime
// numbers up to maxm
SieveOfEratosthenes(maxm);
vector<int> primes;
for (int i = 0; i < N; i++) {
// Store the indices of
// all primes
if (!prime[arr[i]])
primes.push_back(i);
}
// If no primes are present
// in the array
if (primes.size() == 0) {
for (int i = 0; i < N; i++) {
cout << -1 << " ";
}
return;
}
// Store the current prime
int curr = 0;
for (int i = 0; i < N; i++) {
// If the no further
// primes exist in the array
if (curr == primes.size() - 1
// For all indices less than
// that of the current prime
|| i <= primes[curr]) {
cout << arr[primes[curr]] << " ";
continue;
}
// If the current prime is
// nearer
if (abs(primes[curr] - i)
< abs(primes[curr + 1] - i)) {
cout << arr[primes[curr]] << " ";
}
// If the next prime is nearer
else {
// Make the next prime
// as the current
curr++;
cout << arr[primes[curr]] << " ";
}
}
}
// Driver Program
int main()
{
int N = 6;
int arr[] = { 8, 7, 12, 15, 3, 11 };
print_nearest_prime(arr, N);
return 0;
}
// Java program to find nearest
// prime number in the array
// for all array elements
import java.util.*;
class GFG{
static final int max = 10000000;
// Create a boolean array and set all
// entries it as false. A value in
// prime[i] will be true if i is not a
// prime, else false
static boolean []prime = new boolean[max];
// Sieve of Eratosthenes function
static void SieveOfEratosthenes(int maxm)
{
prime[0] = prime[1] = true;
for(int i = 2; i * i <= maxm; i++)
{
// Update all multiples of i greater
// than or equal to the square of it
// numbers which are multiple of i
// and are less than i^2 are already
// been marked.
if (!prime[i])
{
for(int j = i * i;
j <= maxm; j += i)
{
prime[j] = true;
}
}
}
}
// Function to find nearest
// prime number for all elements
static void print_nearest_prime(int arr[], int N)
{
int maxm = Arrays.stream(arr).max().getAsInt();
// Compute and store all prime
// numbers up to maxm
SieveOfEratosthenes(maxm);
Vector<Integer> primes = new Vector<Integer>();
for(int i = 0; i < N; i++)
{
// Store the indices of
// all primes
if (!prime[arr[i]])
primes.add(i);
}
// If no primes are present
// in the array
if (primes.size() == 0)
{
for(int i = 0; i < N; i++)
{
System.out.print(-1 + " ");
}
return;
}
// Store the current prime
int curr = 0;
for(int i = 0; i < N; i++)
{
// If the no further
// primes exist in the array
if (curr == primes.size() - 1 ||
// For all indices less than
// that of the current prime
i <= primes.get(curr))
{
System.out.print(
arr[primes.get(curr)] + " ");
continue;
}
// If the current prime is
// nearer
if (Math.abs(primes.get(curr) - i) <
Math.abs(primes.get(curr + 1) - i))
{
System.out.print(
arr[primes.get(curr)] + " ");
}
// If the next prime is nearer
else
{
// Make the next prime
// as the current
curr++;
System.out.print(
arr[primes.get(curr)] + " ");
}
}
}
// Driver code
public static void main(String[] args)
{
int N = 6;
int arr[] = { 8, 7, 12, 15, 3, 11 };
print_nearest_prime(arr, N);
}
}
// This code is contributed by PrinciRaj1992
# Python3 program to find nearest
# prime number in the array
# for all array elements
maxi = 10000000
# Create a boolean array and set all
# entries it as false. A value in
# prime[i] will be true if i is not a
# prime, else false
prime = [False] * (maxi)
# Sieve of Eratosthenes function
def SieveOfEratosthenes(maxm):
prime[0] = prime[1] = True
for i in range(2, maxm + 1):
if i * i > maxm:
break
# Update all multiples of i greater
# than or equal to the square of it
# numbers which are multiple of i and are
# less than i^2 are already been marked.
if (not prime[i]):
for j in range(i * i, maxm + 1, i):
prime[j] = True
# Function to find nearest
# prime number for all elements
def print_nearest_prime(arr, N):
maxm = max(arr)
# Compute and store all prime
# numbers up to maxm
SieveOfEratosthenes(maxm)
primes = []
for i in range(N):
# Store the indices of
# all primes
if (not prime[arr[i]]):
primes.append(i)
# If no primes are present
# in the array
if len(primes) == 0:
for i in range(N):
print(-1, end = " ")
return
# Store the current prime
curr = 0
for i in range(N):
# If the no further primes
# exist in the array
if (curr == len(primes) - 1 or
# For all indices less than
# that of the current prime
i <= primes[curr]):
print(arr[primes[curr]], end = " ")
continue
# If the current prime is
# nearer
if (abs(primes[curr] - i) <
abs(primes[curr + 1] - i)):
print(arr[primes[curr]], end = " ")
# If the next prime is nearer
else:
# Make the next prime
# as the current
curr += 1
print(arr[primes[curr]], end = " ")
# Driver code
if __name__ == '__main__':
N = 6
arr = [ 8, 7, 12, 15, 3, 11 ]
print_nearest_prime(arr, N)
# This code is contributed by mohit kumar 29
// C# program to find nearest
// prime number in the array
// for all array elements
using System;
using System.Linq;
using System.Collections.Generic;
class GFG{
static readonly int max = 10000000;
// Create a bool array and set all
// entries it as false. A value in
// prime[i] will be true if i is not a
// prime, else false
static bool []prime = new bool[max];
// Sieve of Eratosthenes function
static void SieveOfEratosthenes(int maxm)
{
prime[0] = prime[1] = true;
for(int i = 2; i * i <= maxm; i++)
{
// Update all multiples of i greater
// than or equal to the square of it
// numbers which are multiple of i
// and are less than i^2 are already
// been marked.
if (!prime[i])
{
for(int j = i * i;
j <= maxm; j += i)
{
prime[j] = true;
}
}
}
}
// Function to find nearest
// prime number for all elements
static void print_nearest_prime(int []arr,
int N)
{
int maxm = arr.Max();
// Compute and store all prime
// numbers up to maxm
SieveOfEratosthenes(maxm);
List<int> primes = new List<int>();
for(int i = 0; i < N; i++)
{
// Store the indices of
// all primes
if (!prime[arr[i]])
primes.Add(i);
}
// If no primes are present
// in the array
if (primes.Count == 0)
{
for(int i = 0; i < N; i++)
{
Console.Write(-1 + " ");
}
return;
}
// Store the current prime
int curr = 0;
for(int i = 0; i < N; i++)
{
// If the no further
// primes exist in the array
if (curr == primes.Count - 1 ||
// For all indices less than
// that of the current prime
i <= primes[curr])
{
Console.Write(
arr[primes[curr]] + " ");
continue;
}
// If the current prime is
// nearer
if (Math.Abs(primes[curr] - i) <
Math.Abs(primes[curr + 1] - i))
{
Console.Write(
arr[primes[curr]] + " ");
}
// If the next prime is nearer
else
{
// Make the next prime
// as the current
curr++;
Console.Write(
arr[primes[curr]] + " ");
}
}
}
// Driver code
public static void Main(String[] args)
{
int N = 6;
int []arr = { 8, 7, 12, 15, 3, 11 };
print_nearest_prime(arr, N);
}
}
// This code is contributed by PrinciRaj1992
<script>
// Javascript program to find nearest
// prime number in the array
// for all array elements
let max = 10000000;
// Create a boolean array and set all
// entries it as false. A value in
// prime[i] will be true if i is not a
// prime, else false
let prime = new Array(max);
// Sieve of Eratosthenes function
function SieveOfEratosthenes(maxm)
{
prime[0] = prime[1] = true;
for(let i = 2; i * i <= maxm; i++)
{
// Update all multiples of i greater
// than or equal to the square of it
// numbers which are multiple of i
// and are less than i^2 are already
// been marked.
if (!prime[i])
{
for(let j = i * i; j <= maxm; j += i)
{
prime[j] = true;
}
}
}
}
// Function to find nearest
// prime number for all elements
function print_nearest_prime(arr, N)
{
let maxm = Math.max(...arr);
// Compute and store all prime
// numbers up to maxm
SieveOfEratosthenes(maxm);
let primes = [];
for(let i = 0; i < N; i++)
{
// Store the indices of
// all primes
if (!prime[arr[i]])
primes.push(i);
}
// If no primes are present
// in the array
if (primes.length == 0)
{
for(let i = 0; i < N; i++)
{
document.write(-1 + " ");
}
return;
}
// Store the current prime
let curr = 0;
for(let i = 0; i < N; i++)
{
// If the no further
// primes exist in the array
if (curr == primes.length - 1 ||
// For all indices less than
// that of the current prime
i <= primes[curr])
{
document.write(arr[primes[curr]] + " ");
continue;
}
// If the current prime is
// nearer
if (Math.abs(primes[curr] - i) <
Math.abs(primes[curr + 1] - i))
{
document.write(
arr[primes[curr]] + " ");
}
// If the next prime is nearer
else
{
// Make the next prime
// as the current
curr++;
document.write(arr[primes[curr]] + " ");
}
}
}
// Driver code
let N = 6;
let arr = [ 8, 7, 12, 15, 3, 11 ];
print_nearest_prime(arr, N);
// This code is contributed by unknown2108
</script>
Output:
7 7 7 3 3 11
Time Complexity: O(maxm * (log(log(maxm))) + N)
Auxiliary Space: O(N)
