Nearest element with at-least one common prime factor

Given an array arr[], find nearest element for every element such that there is at least one common prime factor. In output, we need to print position of closest element.

Example:

Input: arr[] = {2, 9, 4, 3, 13}
Output: 3 4 1 2 -1
Explanation : 
Closest element for 1st element is 3rd. 
=>Common prime factor of 1st and 3rd elements
  is 2.
Closest element for 2nd element is 4th.
=>Common prime factor of 2nd and 4th elements
  is 3.

Naive approach

Common prime factor will only exist if GCD of these two numbers will greater than 1. Simple brute force is to run the two loops one inside the another and iterate one by one from each index to both the sides simultaneously and find the gcd which is greater than 1. Whenever we found the answer then just break the loop and the print. If we reached the end of array after traversing both the sides then simply print -1.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to print nearest element with at least
// one common prime factor.
#include<bits/stdc++.h>
using namespace std;
  
void nearestGcd(int arr[], int n)
{
    // Loop covers the every element of arr[]
    for (int i=0; i<n; ++i)
    {
        int closest = -1;
  
        // Loop that covers from 0 to i-1 and i+1
        // to n-1 indexes simultaneously
        for (int j=i-1, k=i+1; j>0 || k<=n; --j, ++k)
        {
            if (j>=0 && __gcd(arr[i], arr[j]) > 1)
            {
                closest = j+1;
                break;
            }
            if (k<n && __gcd(arr[i], arr[k])>1)
            {
                closest = k+1;
                break;
            }
        }
  
        // print position of closest element
        cout << closest << " ";
    }
}
  
// Drive code
int main()
{
    int arr[] = {2, 9, 4, 3, 13};
    int n = sizeof(arr)/sizeof(arr[0]);
    nearestGcd(arr, n);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to print nearest element 
// with at least one common prime factor.
import java.util.*;
import java.lang.*;
  
class GFG 
{
static void nearestGcd(int []arr, int n) 
    // Loop covers the every
    // element of arr[] 
    for (int i = 0; i < n; ++i) 
    
        int closest = -1
  
        // Loop that covers from 0 to 
        // i-1 and i+1 to n-1 indexes 
        // simultaneously 
        for (int j = i - 1, k = i + 1
                 j > 0 || k <= n; --j, ++k) 
        
            if (j >= 0 && __gcd(arr[i], arr[j]) > 1
            
                closest = j + 1
                break
            
            if (k < n && __gcd(arr[i], arr[k]) > 1
            
                closest = k + 1
                break
            
        
  
        // print position of closest element 
        System.out.print(closest + " "); 
    
  
// Recursive function to return 
// gcd of a and b 
static int __gcd(int a, int b) 
    if (b == 0
        return a; 
    return __gcd(b, a % b); 
  
// Driver Code 
public static void main(String args[]) 
    int []arr = {2, 9, 4, 3, 13}; 
    int n = arr.length; 
    nearestGcd(arr, n); 
}
  
// This code is contributed 
// by Akanksha Rai

chevron_right


Python 3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python 3 program to print nearest 
# element with at least one common
# prime factor.
import math
  
def nearestGcd(arr, n):
  
    # Loop covers the every element of arr[]
    for i in range(n):
        closest = -1
  
        # Loop that covers from 0 to i-1 and 
        # i+1 to n-1 indexes simultaneously
        j = i - 1
        k = i + 1
        while j > 0 or k <= n:
            if (j >= 0 and 
                math.gcd(arr[i], arr[j]) > 1):
                closest = j + 1
                break
              
            if (k < n and
                math.gcd(arr[i], arr[k]) > 1):
                closest = k + 1
                break
            k += 1
            j -= 1
  
        # print position of closest element
        print(closest, end = " ")
  
# Drive code
if __name__=="__main__":
      
    arr = [2, 9, 4, 3, 13]
    n = len(arr)
    nearestGcd(arr, n)
  
# This code is contributed by ita_c

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to print nearest element 
// with at least one common prime factor.
using System;
  
class GFG 
{
static void nearestGcd(int []arr, int n) 
    // Loop covers the every
    // element of arr[] 
    for (int i = 0; i < n; ++i) 
    
        int closest = -1; 
  
        // Loop that covers from 0 to 
        // i-1 and i+1 to n-1 indexes 
        // simultaneously 
        for (int j = i - 1, k = i + 1; 
                 j > 0 || k <= n; --j, ++k) 
        
            if (j >= 0 && __gcd(arr[i], arr[j]) > 1) 
            
                closest = j + 1; 
                break
            
            if (k < n && __gcd(arr[i], arr[k]) > 1) 
            
                closest = k + 1; 
                break
            
        
  
        // print position of closest element 
        Console.Write(closest + " "); 
    
  
// Recursive function to return 
// gcd of a and b 
static int __gcd(int a, int b) 
{     
    if (b == 0) 
        return a; 
    return __gcd(b, a % b); 
  
// Drive code 
public static void Main() 
    int []arr = {2, 9, 4, 3, 13}; 
    int n = arr.Length; 
    nearestGcd(arr, n); 
}
  
// This code is contributed 
// by 29AjayKumar

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to print nearest element 
// with at least one common prime factor.
  
function nearestGcd($arr, $n
    // Loop covers the every
    // element of arr[] 
    for ($i = 0; $i < $n; ++$i
    
        $closest = -1; 
  
        // Loop that covers from 0 to 
        // i-1 and i+1 to n-1 indexes 
        // simultaneously 
        for ($j = $i - 1, $k = $i + 1; 
             $j > 0 || $k <= $n; --$j, ++$k
        
            if ($j >= 0 && __gcd($arr[$i], 
                                 $arr[$j]) > 1) 
            
                $closest = $j + 1; 
                break
            
            if ($k < $n && __gcd($arr[$i], 
                                 $arr[$k]) > 1) 
            
                $closest = $k + 1; 
                break
            
        
  
        // print position of closest element 
        echo $closest . " "
    
  
// Recursive function to return 
// gcd of a and b 
function __gcd($a, $b
    if ($b == 0) 
        return $a
    return __gcd($b, $a % $b); 
  
// Driver Code 
$arr = array(2, 9, 4, 3, 13); 
$n = sizeof($arr); 
nearestGcd($arr, $n); 
  
// This code is contributed 
// by Akanksha Rai
?>

chevron_right


Output: 3 4 1 2 -1

Time complexity: O(n2)
Auxiliary space: O(1)

Efficient Approach

We find prime factors of all array elements. To quickly find prime factors, we use Sieve of Eratosthenes. For every element, we consider all prime factors and keep track of closest element with common factor.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to print nearest element with at least
// one common prime factor.
#include <bits/stdc++.h>
using namespace std;
  
const int MAX = 100001;
const int INF = INT_MAX;
  
int primedivisor[MAX], dist[MAX], pos[MAX], divInd[MAX];
  
vector<int> divisors[MAX];
  
// Pre-computation of smallest prime divisor
// of all numbers
void sieveOfEratosthenes()
{
    for (int i=2; i*i < MAX; ++i)
    {
        if (!primedivisor[i])
            for (int j = i*i; j < MAX; j += i)
                primedivisor[j] = i;
    }
  
    // Prime number will have same divisor
    for (int i = 1; i < MAX; ++i)
        if (!primedivisor[i])
            primedivisor[i] = i;
}
  
// Function to calculate all divisors of
// input array
void findDivisors(int arr[], int n)
{
    for (int i=0; i<MAX; ++i)
        pos[i] = divInd[i] = -1, dist[i] = INF;
  
    for (int i=0; i<n; ++i)
    {
        int num = arr[i];
        while (num > 1)
        {
            int div = primedivisor[num];
            divisors[i].push_back(div);
            while (num % div == 0)
                num /= div;
        }
    }
}
  
void nearestGCD(int arr[], int n)
{
    // Pre-compute all the divisors of array
    // element by using prime factors
    findDivisors(arr, n);
  
    // Traverse all elements,
    for (int i=0; i<n; ++i)
    {
        // For every divisor of current element,
        // find closest element.
        for (auto &div: divisors[i])
        {
            // Visit divisor if not visited
            if (divInd[div] == -1)
                divInd[div] = i;
            else
            {
                // Fetch the index of visited divisor
                int ind = divInd[div];
  
                // Update the divisor index to current index
                divInd[div] = i;
  
                // Set the minimum distance
                if (dist[i] > abs(ind-i))
                {
                    // Set the min distance of current
                    // index 'i' to nearest one
                    dist[i] = abs(ind-i);
  
                    // Add 1 as indexing starts from 0
                    pos[i] = ind + 1;
                }
  
                if (dist[ind] > abs(ind-i))
                {
                    // Set the min distance of found index 'ind'
                    dist[ind] = abs(ind-i);
  
                    // Add 1 as indexing starts from 0
                    pos[ind] = i + 1;
                }
            }
        }
    }
}
  
// Driver code
int main()
{
    // Simple sieve to find smallest prime
    // divisor of number from 2 to MAX
    sieveOfEratosthenes();
  
    int arr[] = {2, 9, 4, 3, 13};
    int n = sizeof(arr)/sizeof(arr[0]);
  
    // function to calculate nearest distance
    // of every array elements
    nearestGCD(arr, n);
  
    // Print the nearest distance having GDC>1
    for (int i=0; i<n; ++i)
        cout << pos[i] << " ";
    return 0;
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to prnearest element with at least
# one common prime factor.
  
MAX = 100001
INF = 10**9
  
primedivisor = [0 for i in range(MAX)]
dist = [0 for i in range(MAX)]
pos = [0 for i in range(MAX)]
divInd = [0 for i in range(MAX)]
  
divisors = [[] for i in range(MAX)]
  
# Pre-computation of smallest prime divisor
# of all numbers
def sieveOfEratosthenes():
  
    for i in range(2,MAX):
        if i * i > MAX:
            break
  
        if (primedivisor[i] == 0):
            for j in range(2 * i, MAX, i):
                primedivisor[j] = i
  
    # Prime number will have same divisor
    for i in range(1, MAX):
        if (primedivisor[i] == 0):
            primedivisor[i] = i
  
# Function to calculate all divisors of
# input array
def findDivisors(arr, n):
  
    for i in range(MAX):
        pos[i] = divInd[i] = -1
        dist[i] = 10**9
    for i in range(n):
        num = arr[i]
        while (num > 1):
  
            div = primedivisor[num]
            divisors[i].append(div)
            while (num % div == 0):
                num //= div
  
  
def nearestGCD(arr, n):
    # Pre-compute all the divisors of array
    # element by using prime factors
    findDivisors(arr, n)
  
    # Traverse all elements,
    for i in range(n):
        # For every divisor of current element,
        # find closest element.
        for div in divisors[i]:
            # Visit divisor if not visited
            if (divInd[div] == -1):
                divInd[div] = i
            else:
  
                # Fetch the index of visited divisor
                ind = divInd[div]
  
                # Update the divisor index to current index
                divInd[div] = i
  
                # Set the minimum distance
                if (dist[i] > abs(ind-i)):
  
                    # Set the min distance of current
                    # index 'i' to nearest one
                    dist[i] = abs(ind-i)
  
                    # Add 1 as indexing starts from 0
                    pos[i] = ind + 1
  
  
                if (dist[ind] > abs(ind-i)):
  
                    # Set the min distance of found index 'ind'
                    dist[ind] = abs(ind-i)
  
                    # Add 1 as indexing starts from 0
                    pos[ind] = i + 1
  
  
# Driver code
  
# Simple sieve to find smallest prime
# divisor of number from 2 to MAX
sieveOfEratosthenes()
  
arr =[2, 9, 4, 3, 13]
n = len(arr)
  
# function to calculate nearest distance
# of every array elements
nearestGCD(arr, n)
  
# Prthe nearest distance having GDC>1
for i in range(n):
    print(pos[i],end=" ")
      
# This code is contributed by mohit kumar 29

chevron_right


Output: 
3 4 1 2 -1

Time complexity: O(MAX * log(log (MAX) ) )
Auxiliary space: O(MAX)

This article is contributed by Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



My Personal Notes arrow_drop_up



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.