Nearest element with at-least one common prime factor

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

The common prime factor will only exist if the 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 an array after traversing both the sides then simply print -1. 

C++

// 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 << " ";
    }
}

// Driver code
int main()
{
    int arr[] = {2, 9, 4, 3, 13};
    int n = sizeof(arr)/sizeof(arr[0]);
    nearestGcd(arr, n);
    return 0;
}

Java

// 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

Python 3

# 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 = " ")

# Driver code
if __name__=="__main__":

    arr = [2, 9, 4, 3, 13]
    n = len(arr)
    nearestGcd(arr, n)

# This code is contributed by ita_c

C#

// 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);
}

// Driver 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

PHP

<?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
?>

Javascript

<script>
// Javascript program to print nearest element
// with at least one common prime factor.

function nearestGcd(arr,n)
{
    // Loop covers the every
    // element of arr[]
    for (let i = 0; i < n; ++i)
    {
        let closest = -1;

        // Loop that covers from 0 to
        // i-1 and i+1 to n-1 indexes
        // simultaneously
        for (let 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
        document.write(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
let arr=[2, 9, 4, 3, 13];
let n = arr.length;
nearestGcd(arr, n);




// This code is contributed by unknown2108
</script>

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 the closest element with a common factor. 

C++

// 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;
}

Java

// Java program to print nearest element with at least
// one common prime factor.
import java.io.*;
import java.util.*;
class GFG
{
  static int MAX = 100001;
  static int INF = Integer.MAX_VALUE;
  static int[] primedivisor = new int [MAX];
  static int[] dist = new int [MAX];
  static int[] pos = new int [MAX];
  static int[] divInd = new int [MAX];

  static ArrayList<ArrayList<Integer>> divisors =
    new ArrayList<ArrayList<Integer>>();

  // Pre-computation of smallest prime divisor
  // of all numbers
  static void sieveOfEratosthenes()
  {
    for (int i = 2; i * i < MAX; ++i)
    {
      if (primedivisor[i] == 0)
      {
        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] == 0)
      {
        primedivisor[i] = i;
      }
    }

  }

  // Function to calculate all divisors of
  // input array
  static 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.get(i).add(div);
        while (num % div == 0)
        {
          num /= div;
        }
      }
    }

  }

  static 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(int div : divisors.get(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] > Math.abs(ind-i))
          {

            // Set the min distance of current
            // index 'i' to nearest one
            dist[i] = Math.abs(ind-i);

            // Add 1 as indexing starts from 0
            pos[i] = ind + 1;
          }

          if (dist[ind] > Math.abs(ind-i))
          {

            // Set the min distance of found index 'ind'
            dist[ind] = Math.abs(ind-i);

            // Add 1 as indexing starts from 0
            pos[ind] = i + 1;
          }
        }
      }
    }
  }

  // Driver code
  public static void main (String[] args)
  {

    for(int i = 0; i < MAX; i++)
    {
      divisors.add(new ArrayList<Integer>());
    }

    // Simple sieve to find smallest prime
    // divisor of number from 2 to MAX
    sieveOfEratosthenes();
    int arr[] = {2, 9, 4, 3, 13};
    int n = arr.length;

    // 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)
    {
      System.out.print(pos[i]+" ");
    }
  }
}

// This code is contributed by avanitrachhadiya2155

Python3

# Python3 program to print nearest 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)

# Print the nearest distance having GDC>1
for i in range(n):
    print(pos[i],end=" ")

# This code is contributed by mohit kumar 29

C#

// C# program to print nearest element with at least
// one common prime factor.
using System;
using System.Collections.Generic;

public class GFG{

  static int MAX = 100001;
  static int INF = Int32.MaxValue;
  static int[] primedivisor = new int [MAX];
  static int[] dist = new int [MAX];
  static int[] pos = new int [MAX];
  static int[] divInd = new int [MAX];

  static List<List<int>> divisors = new List<List<int>>();

  // Pre-computation of smallest prime divisor
  // of all numbers
  static void sieveOfEratosthenes()
  {
    for (int i = 2; i * i < MAX; ++i)
    {
      if (primedivisor[i] == 0)
      {
        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] == 0)
      {
        primedivisor[i] = i;
      }
    }

  }

  // Function to calculate all divisors of
  // input array
  static 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].Add(div);
        while (num % div == 0)
        {
          num /= div;
        }
      }
    }

  }

  static 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.
      foreach(int div in 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] > Math.Abs(ind-i))
          {

            // Set the min distance of current
            // index 'i' to nearest one
            dist[i] = Math.Abs(ind-i);

            // Add 1 as indexing starts from 0
            pos[i] = ind + 1;
          }

          if (dist[ind] > Math.Abs(ind-i))
          {

            // Set the min distance of found index 'ind'
            dist[ind] = Math.Abs(ind-i);

            // Add 1 as indexing starts from 0
            pos[ind] = i + 1;
          }
        }
      }
    }
  }

  // Driver code
  static public void Main ()
  {
    for(int i = 0; i < MAX; i++)
    {
      divisors.Add(new List<int>());
    }
    // Simple sieve to find smallest prime
    // divisor of number from 2 to MAX
    sieveOfEratosthenes();
    int[] arr = {2, 9, 4, 3, 13};
    int n = arr.Length;

    // 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)
    {
      Console.Write(pos[i]+" ");
    }
  }
}

// This code is contributed by rag2127

Javascript

<script>
// Javascript program to print nearest element with at least
// one common prime factor.

let MAX = 100001;
let INF = Number.MAX_VALUE;
let primedivisor = new Array(MAX);
let dist = new Array(MAX);
let pos = new Array(MAX);
let divInd = new Array(MAX);

for(let i=0;i<MAX;i++)
{
    primedivisor[i]=0;
    dist[i]=0;
    pos[i]=0;
    divInd[i]=0;
}

let divisors =[];

// Pre-computation of smallest prime divisor
  // of all numbers
function sieveOfEratosthenes()
{
    for (let i = 2; i * i < MAX; ++i)
    {
      if (primedivisor[i] == 0)
      {
        for (let j = i * i; j < MAX; j += i)
        {
          primedivisor[j] = i;
        }
      }
    }
    // Prime number will have same divisor
    for (let i = 1; i < MAX; ++i)
    {
      if (primedivisor[i] == 0)
      {
        primedivisor[i] = i;
      }
    }
}

// Function to calculate all divisors of
  // input array
function findDivisors(arr,n)
{
    for (let i=0; i<MAX; ++i)
    {
      pos[i] = divInd[i] = -1;
      dist[i] = INF;
    }
    for (let i = 0; i < n; ++i)
    {
      let num = arr[i];
      while (num > 1)
      {
        let div = primedivisor[num];
        divisors[i].push(div);
        while (num % div == 0)
        {
          num = Math.floor(num/div);
        }
      }
    }
}


function nearestGCD(arr,n)
{
    // Pre-compute all the divisors of array
    // element by using prime factors
    findDivisors(arr, n);

    // Traverse all elements,
    for (let i = 0; i < n; ++i)
    {

      // For every divisor of current element,
      // find closest element.
      for(let div=0;div<divisors[i].length;div++)
      {

        // Visit divisor if not visited
        if (divInd[divisors[i][div]] == -1)
        {
          divInd[divisors[i][div]] = i;
        }
        else
        {

          // Fetch the index of visited divisor
          let ind = divInd[divisors[i][div]];

          // Update the divisor index to current index
          divInd[divisors[i][div]] = i;

          // Set the minimum distance
          if (dist[i] > Math.abs(ind-i))
          {

            // Set the min distance of current
            // index 'i' to nearest one
            dist[i] = Math.abs(ind-i);

            // Add 1 as indexing starts from 0
            pos[i] = ind + 1;
          }

          if (dist[ind] > Math.abs(ind-i))
          {

            // Set the min distance of found index 'ind'
            dist[ind] = Math.abs(ind-i);

            // Add 1 as indexing starts from 0
            pos[ind] = i + 1;
          }
        }
      }
    }
}

// Driver code
for(let i = 0; i < MAX; i++)
{
divisors.push([]);
}

// Simple sieve to find smallest prime
// divisor of number from 2 to MAX
sieveOfEratosthenes();
let arr= [2, 9, 4, 3, 13];
let n = arr.length;

// function to calculate nearest distance
// of every array elements
nearestGCD(arr, n);

// Print the nearest distance having GDC>1
for (let i = 0; i < n; ++i)
{
document.write(pos[i]+" ");
}


// This code is contributed by patel2127

</script>

Output: 

3 4 1 2 -1

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