Sum of Maximum and Minimum prime factor of every number in the Array

Given an array arr[], the task is to find the sum of the maximum and the minimum prime factor of every number in the given array.
Examples:

Input: arr[] = {15} 
Output: 8 
The maximum and the minimum prime factors 
of 15 are 5 and 3 respectively.
Input: arr[] = {5, 10, 15, 20, 25, 30} 
Output: 10 7 8 7 10 7 

Approach: The idea is to use Sieve of Eratosthenes to precompute all the minimum and maximum prime factors of every number and store it in two arrays. After this precomputation, the sum of the minimum and the maximum prime factor can be found in constant time.
Below is the implementation of the above approach: 

CPP

// C++ implementation of the approach

#include <bits/stdc++.h>
using namespace std;

const int MAX = 100000;

// max_prime[i] represent maximum prime
// number that divides the number i
int max_prime[MAX];

// min_prime[i] represent minimum prime
// number that divides the number i
int min_prime[MAX];

// Function to store the minimum prime factor
// and the maximum prime factor in two arrays
void sieve(int n)
{
    for (int i = 2; i <= n; ++i) {

        // Check for prime number
        // if min_prime[i]>0,
        // then it is not a prime number
        if (min_prime[i] > 0) {
            continue;
        }

        // if i is a prime number
        // min_prime number that divide prime number
        // and max_prime number that divide prime number
        // is the number itself.
        min_prime[i] = i;
        max_prime[i] = i;

        int j = i + i;

        while (j <= n) {
            if (min_prime[j] == 0) {

                // If this number is being visited
                // for first time then this divisor
                // must be the smallest prime number
                // that divides this number
                min_prime[j] = i;
            }

            // Update prime number till
            // last prime number that divides this number

            // The last prime number that
            // divides this number will be maximum.
            max_prime[j] = i;
            j += i;
        }
    }
}

// Function to find the sum of the minimum
// and the maximum prime factors of every
// number from the given array
void findSum(int arr[], int n)
{

    // Pre-calculation
    sieve(MAX);

    // For every element of the given array
    for (int i = 0; i < n; i++) {

        // The sum of its smallest
        // and largest prime factor
        int sum = min_prime[arr[i]]
                  + max_prime[arr[i]];

        cout << sum << " ";
    }
}

// Driver code
int main()
{
    int arr[] = { 5, 10, 15, 20, 25, 30 };
    int n = sizeof(arr) / sizeof(int);

    findSum(arr, n);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

    static int MAX = 100000;

    // max_prime[i] represent maximum prime
    // number that divides the number i
    static int max_prime[] = new int[MAX + 1];

    // min_prime[i] represent minimum prime
    // number that divides the number i
    static int min_prime[] = new int[MAX + 1];

    // Function to store the minimum prime factor
    // and the maximum prime factor in two arrays
    static void sieve(int n)
    {
        for (int i = 2; i <= n; ++i)
        {

            // Check for prime number
            // if min_prime[i] > 0,
            // then it is not a prime number
            if (min_prime[i] > 0)
            {
                continue;
            }

            // if i is a prime number
            // min_prime number that divide prime number
            // and max_prime number that divide prime number
            // is the number itself.
            min_prime[i] = i;
            max_prime[i] = i;

            int j = i + i;

            while (j <= n)
            {
                if (min_prime[j] == 0)
                {

                    // If this number is being visited
                    // for first time then this divisor
                    // must be the smallest prime number
                    // that divides this number
                    min_prime[j] = i;
                }

                // Update prime number till
                // last prime number that divides this number

                // The last prime number that
                // divides this number will be maximum.
                max_prime[j] = i;
                j += i;
            }
        }
    }

    // Function to find the sum of the minimum
    // and the maximum prime factors of every
    // number from the given array
    static void findSum(int arr[], int n)
    {

        // Pre-calculation
        sieve(MAX);

        // For every element of the given array
        for (int i = 0; i < n; i++)
        {

            // The sum of its smallest
            // and largest prime factor
            int sum = min_prime[arr[i]]
                    + max_prime[arr[i]];

            System.out.print(sum + " ");
        }
    }

    // Driver code
    public static void main (String[] args)
    {
        int arr[] = { 5, 10, 15, 20, 25, 30 };
        int n = arr.length ;

        findSum(arr, n);

    }
}

// This code is contributed by AnkitRai01

Python

# Python3 implementation of the approach
MAX = 100000

# max_prime[i] represent maximum prime
# number that divides the number i
max_prime = [0]*(MAX + 1)

# min_prime[i] represent minimum prime
# number that divides the number i
min_prime = [0]*(MAX + 1)

# Function to store the minimum prime factor
# and the maximum prime factor in two arrays
def sieve(n):
    for i in range(2, n + 1):

        # Check for prime number
        # if min_prime[i]>0,
        # then it is not a prime number
        if (min_prime[i] > 0):
            continue

        # if i is a prime number
        # min_prime number that divide prime number
        # and max_prime number that divide prime number
        # is the number itself.
        min_prime[i] = i
        max_prime[i] = i

        j = i + i

        while (j <= n):
            if (min_prime[j] == 0):

                # If this number is being visited
                # for first time then this divisor
                # must be the smallest prime number
                # that divides this number
                min_prime[j] = i

            # Update prime number till
            # last prime number that divides this number

            # The last prime number that
            # divides this number will be maximum.
            max_prime[j] = i
            j += i

# Function to find the sum of the minimum
# and the maximum prime factors of every
# number from the given array
def findSum(arr, n):

    # Pre-calculation
    sieve(MAX)

    # For every element of the given array
    for i in range(n):

        # The sum of its smallest
        # and largest prime factor
        sum = min_prime[arr[i]] + max_prime[arr[i]]

        print(sum, end = " ")

# Driver code
arr = [5, 10, 15, 20, 25, 30]
n = len(arr)

findSum(arr, n)

# This code is contributed by mohit kumar 29

C#

// C# implementation of the approach
using System;

class GFG
{

    static int MAX = 100000;

    // max_prime[i] represent maximum prime
    // number that divides the number i
    static int []max_prime = new int[MAX + 1];

    // min_prime[i] represent minimum prime
    // number that divides the number i
    static int []min_prime = new int[MAX + 1];

    // Function to store the minimum prime factor
    // and the maximum prime factor in two arrays
    static void sieve(int n)
    {
        for (int i = 2; i <= n; ++i)
        {

            // Check for prime number
            // if min_prime[i] > 0,
            // then it is not a prime number
            if (min_prime[i] > 0)
            {
                continue;
            }

            // if i is a prime number
            // min_prime number that divide prime number
            // and max_prime number that divide prime number
            // is the number itself.
            min_prime[i] = i;
            max_prime[i] = i;

            int j = i + i;

            while (j <= n)
            {
                if (min_prime[j] == 0)
                {

                    // If this number is being visited
                    // for first time then this divisor
                    // must be the smallest prime number
                    // that divides this number
                    min_prime[j] = i;
                }

                // Update prime number till
                // last prime number that divides this number

                // The last prime number that
                // divides this number will be maximum.
                max_prime[j] = i;
                j += i;
            }
        }
    }

    // Function to find the sum of the minimum
    // and the maximum prime factors of every
    // number from the given array
    static void findSum(int []arr, int n)
    {

        // Pre-calculation
        sieve(MAX);

        // For every element of the given array
        for (int i = 0; i < n; i++)
        {

            // The sum of its smallest
            // and largest prime factor
            int sum = min_prime[arr[i]]
                    + max_prime[arr[i]];

            Console.Write(sum + " ");
        }
    }

    // Driver code
    public static void Main(String[] args)
    {
        int []arr = { 5, 10, 15, 20, 25, 30 };
        int n = arr.Length ;

        findSum(arr, n);

    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript implementation of the approach


var MAX = 100000;

// max_prime[i] represent maximum prime
// number that divides the number i
var max_prime = Array.from({length: MAX + 1}, (_, i) => 0);

// min_prime[i] represent minimum prime
// number that divides the number i
var min_prime = Array.from({length: MAX + 1}, (_, i) => 0);

// Function to store the minimum prime factor
// and the maximum prime factor in two arrays
function sieve(n)
{
    for (var i = 2; i <= n; ++i)
    {

        // Check for prime number
        // if min_prime[i] > 0,
        // then it is not a prime number
        if (min_prime[i] > 0)
        {
            continue;
        }

        // if i is a prime number
        // min_prime number that divide prime number
        // and max_prime number that divide prime number
        // is the number itself.
        min_prime[i] = i;
        max_prime[i] = i;

        var j = i + i;

        while (j <= n)
        {
            if (min_prime[j] == 0)
            {

                // If this number is being visited
                // for first time then this divisor
                // must be the smallest prime number
                // that divides this number
                min_prime[j] = i;
            }

            // Update prime number till
            // last prime number that divides this number

            // The last prime number that
            // divides this number will be maximum.
            max_prime[j] = i;
            j += i;
        }
    }
}

// Function to find the sum of the minimum
// and the maximum prime factors of every
// number from the given array
function findSum(arr , n)
{

    // Pre-calculation
    sieve(MAX);

    // For every element of the given array
    for (i = 0; i < n; i++)
    {

        // The sum of its smallest
        // and largest prime factor
        var sum = min_prime[arr[i]]
                + max_prime[arr[i]];

        document.write(sum + " ");
    }
}

// Driver code
var arr = [ 5, 10, 15, 20, 25, 30 ];
var n = arr.length ;

findSum(arr, n);


// This code contributed by shikhasingrajput

</script>

Output:

10 7 8 7 10 7

Time Complexity: O(n²)

Auxiliary Space: O(100000)