Largest Divisor for each element in an array other than 1 and the number itself

Given an array arr[] of N integers, the task is to find the largest divisor for each element in an array other than 1 and the number itself. If there is no such divisor, print -1.

Examples:

Input: arr[] = {5, 6, 7, 8, 9, 10} 
Output: -1 3 -1 4 3 5 
Divisors(5) = {1, 5} 
-> Since there is no divisor other than 1 and the number itself, therefore largest divisor = -1 
Divisors(6) = [1, 2, 3, 6] 
-> largest divisor other than 1 and the number itself = 3 
Divisors(7) = [1, 7] 
-> Since there is no divisor other than 1 and the number itself, therefore largest divisor = -1 
Divisors(8) = [1, 2, 4, 8] 
-> largest divisor other than 1 and the number itself = 4 
Divisors(9) = [1, 3, 9] 
-> largest divisor other than 1 and the number itself = 3 
Divisors(10) = [1, 2, 5, 10] 
-> largest divisor other than 1 and the number itself = 5

Input: arr[] = {15, 16, 17, 18, 19, 20, 21} 
Output: 5 8 -1 9 -1 10 7 

Naive approach: The idea is to iterate over all the array elements and find the largest divisor for each of the element using the approach discussed in this article. 
Time Complexity: O(N * ?N)

Efficient approach: A better solution is to precompute the maximum divisor of the numbers from 2 to 10⁵ and then just run a loop for array and print precomputed answer. 

• Use Sieve of Eratosthenes to mark the prime numbers and store the smallest prime divisor of each number.
• Now largest divisor for any number will be number / smallest_prime_divisor.
• Find the Largest divisor for each number using the precomputed answer.

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach

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

#define int long long
const int maxin = 100001;

// Divisors array to keep track
// of the maximum divisor
int divisors[maxin];

// Function to pre-compute the prime
// numbers and largest divisors
void Calc_Max_Div(int arr[], int n)
{

    // Visited array to keep
    // track of prime numbers
    bool vis[maxin];
    memset(vis, 1, maxin);

    // 0 and 1 are not prime numbers
    vis[0] = vis[1] = 0;

    // Initialising divisors[i] = i
    for (int i = 1; i <= maxin; i++)
        divisors[i] = i;

    // For all the numbers divisible by 2
    // the maximum divisor will be number / 2
    for (int i = 4; i <= maxin; i += 2) {
        vis[i] = 0;
        divisors[i] = i / 2;
    }
    for (int i = 3; i <= maxin; i += 2) {

        // If divisors[i] is not equal to i then
        // this means that divisors[i] contains
        // minimum prime divisor for the number
        if (divisors[i] != i) {

            // Update the answer to
            // i / smallest_prime_divisor[i]
            divisors[i] = i / divisors[i];
        }

        // Condition if i is a prime number
        if (vis[i] == 1) {
            for (int j = i * i; j < maxin; j += i) {
                vis[j] = 0;

                // If divisors[j] is equal to j then
                // this means that i is the first prime
                // divisor for j so we update divi[j] = i
                if (divisors[j] == j)
                    divisors[j] = i;
            }
        }
    }

    for (int i = 0; i < n; i++) {

        // If the current element is prime
        // then it has no divisors
        // other than 1 and itself
        if (divisors[arr[i]] == arr[i])
            cout << "-1 ";
        else
            cout << divisors[arr[i]] << " ";
    }
}

// Driver code
int32_t main()
{
    int arr[] = { 5, 6, 7, 8, 9, 10 };
    int n = sizeof(arr) / sizeof(int);

    Calc_Max_Div(arr, n);

    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
public class GFG
{

    final static int maxin = 10001;

    // Divisors array to keep track
    // of the maximum divisor
    static int divisors[] = new int[maxin + 1];

    // Function to pre-compute the prime
    // numbers and largest divisors
    static void Calc_Max_Div(int arr[], int n)
    {

        // Visited array to keep
        // track of prime numbers
        int vis[] = new int[maxin + 1];

        for(int i = 0;i <maxin+1 ; i++)
            vis[i] = 1;

        // 0 and 1 are not prime numbers
        vis[0] = vis[1] = 0;

        // Initialising divisors[i] = i
        for (int i = 1; i <= maxin; i++)
            divisors[i] = i;

        // For all the numbers divisible by 2
        // the maximum divisor will be number / 2
        for (int i = 4; i <= maxin; i += 2)
        {
            vis[i] = 0;
            divisors[i] = i / 2;
        }
        for (int i = 3; i <= maxin; i += 2)
        {

            // If divisors[i] is not equal to i then
            // this means that divisors[i] contains
            // minimum prime divisor for the number
            if (divisors[i] != i)
            {

                // Update the answer to
                // i / smallest_prime_divisor[i]
                divisors[i] = i / divisors[i];
            }

            // Condition if i is a prime number
            if (vis[i] == 1)
            {
                for (int j = i * i; j < maxin; j += i)
                {
                    vis[j] = 0;

                    // If divisors[j] is equal to j then
                    // this means that i is the first prime
                    // divisor for j so we update divi[j] = i
                    if (divisors[j] == j)
                        divisors[j] = i;
                }
            }
        }

        for (int i = 0; i < n; i++)
        {

            // If the current element is prime
            // then it has no divisors
            // other than 1 and itself
            if (divisors[arr[i]] == arr[i])
                    System.out.print("-1 ");
            else
                    System.out.print(divisors[arr[i]] + " ");
        }
    }

    // Driver code
    public static void main (String[] args)
    {
        int []arr = { 5, 6, 7, 8, 9, 10 };
        int n = arr.length;

        Calc_Max_Div(arr, n);
    }
}

// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
maxin = 100001;

# Divisors array to keep track
# of the maximum divisor
divisors = [0] * (maxin + 1);

# Function to pre-compute the prime
# numbers and largest divisors
def Calc_Max_Div(arr, n) :

    # Visited array to keep
    # track of prime numbers
    vis = [1] * (maxin + 1);

    # 0 and 1 are not prime numbers
    vis[0] = vis[1] = 0;

    # Initialising divisors[i] = i
    for i in range(1, maxin + 1) :
        divisors[i] = i;

    # For all the numbers divisible by 2
    # the maximum divisor will be number / 2
    for i in range(4 , maxin + 1, 2) :
        vis[i] = 0;
        divisors[i] = i // 2;

    for i in range(3, maxin + 1, 2) :

        # If divisors[i] is not equal to i then
        # this means that divisors[i] contains
        # minimum prime divisor for the number
        if (divisors[i] != i) :

            # Update the answer to
            # i / smallest_prime_divisor[i]
            divisors[i] = i // divisors[i];

        # Condition if i is a prime number
        if (vis[i] == 1) :
            for j in range( i * i, maxin, i) :
                vis[j] = 0;

                # If divisors[j] is equal to j then
                # this means that i is the first prime
                # divisor for j so we update divi[j] = i
                if (divisors[j] == j) :
                    divisors[j] = i;

    for i in range(n) :

        # If the current element is prime
        # then it has no divisors
        # other than 1 and itself
        if (divisors[arr[i]] == arr[i]) :
            print("-1 ", end = "");
        else :
            print(divisors[arr[i]], end = " ");

# Driver code
if __name__ == "__main__" :

    arr = [ 5, 6, 7, 8, 9, 10 ];
    n = len(arr);

    Calc_Max_Div(arr, n);

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG
{

    static int maxin = 10001;

    // Divisors array to keep track
    // of the maximum divisor
    static int []divisors = new int[maxin + 1];

    // Function to pre-compute the prime
    // numbers and largest divisors
    static void Calc_Max_Div(int []arr, int n)
    {

        // Visited array to keep
        // track of prime numbers
        int []vis = new int[maxin + 1];

        for(int i = 0; i < maxin + 1 ; i++)
            vis[i] = 1;

        // 0 and 1 are not prime numbers
        vis[0] = vis[1] = 0;

        // Initialising divisors[i] = i
        for (int i = 1; i <= maxin; i++)
            divisors[i] = i;

        // For all the numbers divisible by 2
        // the maximum divisor will be number / 2
        for (int i = 4; i <= maxin; i += 2)
        {
            vis[i] = 0;
            divisors[i] = i / 2;
        }
        for (int i = 3; i <= maxin; i += 2)
        {

            // If divisors[i] is not equal to i then
            // this means that divisors[i] contains
            // minimum prime divisor for the number
            if (divisors[i] != i)
            {

                // Update the answer to
                // i / smallest_prime_divisor[i]
                divisors[i] = i / divisors[i];
            }

            // Condition if i is a prime number
            if (vis[i] == 1)
            {
                for (int j = i * i; j < maxin; j += i)
                {
                    vis[j] = 0;

                    // If divisors[j] is equal to j then
                    // this means that i is the first prime
                    // divisor for j so we update divi[j] = i
                    if (divisors[j] == j)
                        divisors[j] = i;
                }
            }
        }

        for (int i = 0; i < n; i++)
        {

            // If the current element is prime
            // then it has no divisors
            // other than 1 and itself
            if (divisors[arr[i]] == arr[i])
                    Console.Write("-1 ");
            else
                    Console.Write(divisors[arr[i]] + " ");
        }
    }

    // Driver code
    public static void Main()
    {
        int []arr = { 5, 6, 7, 8, 9, 10 };
        int n = arr.Length;

        Calc_Max_Div(arr, n);
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>

// Javascript implementation of the approach
var maxin = 100001;

// Divisors array to keep track
// of the maximum divisor
var divisors = Array(maxin).fill(0);

// Function to pre-compute the prime
// numbers and largest divisors
function Calc_Max_Div(arr, n)
{

    // Visited array to keep
    // track of prime numbers
    var vis = Array(maxin).fill(1);

    // 0 and 1 are not prime numbers
    vis[0] = vis[1] = 0;

    var i,j;

    // Initialising divisors[i] = i
    for(i = 1; i <= maxin; i++)
        divisors[i] = i;

    // For all the numbers divisible by 2
    // the maximum divisor will be number / 2
    for(i = 4; i <= maxin; i += 2)
    {
        vis[i] = 0;
        divisors[i] = i / 2;
    }
    for(i = 3; i <= maxin; i += 2)
    {

        // If divisors[i] is not equal to i then
        // this means that divisors[i] contains
        // minimum prime divisor for the number
        if (divisors[i] != i)
        {

            // Update the answer to
            // i / smallest_prime_divisor[i]
            divisors[i] = i / divisors[i];
        }

        // Condition if i is a prime number
        if (vis[i] == 1)
        {
            for(j = i * i; j < maxin; j += i)
            {
                vis[j] = 0;

                // If divisors[j] is equal to j then
                // this means that i is the first prime
                // divisor for j so we update divi[j] = i
                if (divisors[j] == j)
                    divisors[j] = i;
            }
        }
    }

    for(i = 0; i < n; i++)
    {

        // If the current element is prime
        // then it has no divisors
        // other than 1 and itself
        if (divisors[arr[i]] == arr[i])
            document.write("-1 ");
        else
            document.write(divisors[arr[i]] + " ");
    }
}

// Driver code
var arr = [ 5, 6, 7, 8, 9, 10 ];
var n = arr.length;

Calc_Max_Div(arr, n);

// This code is contributed by bgangwar59

</script>

Output:

-1 3 -1 4 3 5

Time Complexity: O(N)
Auxiliary Space: O(100001)