Maximum count of pairs in Array with GCD greater than 1 by reordering given Array

Given an array arr[] of size N. The task is to reorder arr[] and find the maximum number of GCD pairs which follows the conditions given below.

• Choose any two elements of array Ai and Aj of array where 0 <= i < j < N.
• Calculate GCD after multiplying Aj with 2 like (Ai, 2 * Aj) which is greater than 1. 

Examples

Input: arr[] = { 3, 6 . 5, 3}
Output: 4
Explanation: Reorder array like this : { 6, 5, 3, 3 } and below are the pairs formed from arr[].
P1 = GCD( 6, 5 * 2) => (6, 10) => 2 > 1
P2 = GCD( 6, 3 * 2) => (6, 6) => 6 > 1
P3 = GCD( 6, 3 * 2) => (6, 6) => 6 > 1
P4 = GCD( 3, 3 * 2) => (3, 6) => 3 > 1

Input: arr[] = { 1, 7 }
Output: 0
Explanation: If array is order like { 7, 1 } no pair can be formed
GCD(7, 1 * 2) = > (7, 2 ), GCD(1, 7 * 2) => (1, 14) == 1

Approach: If we observe that if even elements are in starting position then the pairs of (GCD > 1) is maximum because there is a condition to multiply the arr[j] * 2, and the ordering of odd elements does not matter its number of pair always same. Follow the steps below to solve the given problem.

• Initialize the variable idx with value 0.
• Traverse the array.
• If arr[i] is even then swap with arr[idx] and increment idx by 1.
• After traversing all the elements.
• Initialize the variable ans with 0.
• Use two loops first with i and second with j = i + 1.
• Now if gcd(arr[i], 2 * arr[j] * 2) > 1 increment the ans by 1.

Below is the implementation of the above approach.

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the maximum number of pairs
int maximumpairs(int arr[], int n)
{

    // Reorder array with even element first
    int idx = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] % 2 == 0) {
            swap(arr[i], arr[idx]);
            idx++;
        }
    }

    // Now count the ans
    int ans = 0;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            if (__gcd(arr[i], 2 * arr[j]) > 1) {
                ans++;
            }
        }
    }
    return ans;
}

// Driver Code
int main()
{

    // Initializations
    int arr[] = { 5, 3, 6, 3 };
    int N = sizeof(arr) / sizeof(int);

    // Function Call
    int ans = maximumpairs(arr, N);

    cout << ans;
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG {
  // find gcd
  static int gcd(int a, int b)
  {
    if (b == 0)
      return a;
    return gcd(b, a % b);
  }

  // Function to find the maximum number of pairs
  static int maximumpairs(int arr[], int n)
  {

    // Reorder array with even element first
    int idx = 0;
    for (int i = 0; i < n; i++) {
      if (arr[i] % 2 == 0) {
        //swap
        int temp = arr[i];
        arr[i] = arr[idx];
        arr[idx] = temp;
        idx++;
      }
    }

    // Now count the ans
    int ans = 0;
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        int a = arr[i];
        int b = 2 * arr[j];
        if (gcd(a, b) > 1) {
          ans++;
        }
      }
    }
    return ans;
  }

  public static void main (String[] args)
  {

    // Initializations
    int arr[] = { 5, 3, 6, 3 };
    int N = arr.length;

    // Function Call
    int ans = maximumpairs(arr, N);

    System.out.println(ans);
  }
}

// This code is contributed by hrithikgarg03188

Python3

# Python program for above approach

# Function to find GCD
def gcd(a, b):
    if(b == 0):
        return a
    else:
        return gcd(b, a % b)

# Function to find the maximum number of pairs
def maximumpairs(arr,n):

    # Reorder array with even element first
    idx = 0
    for i in range(0, n):
        if (arr[i] % 2 == 0):
            arr[i], arr[idx] = arr[idx], arr[i]
            idx = idx + 1

    # Now count the ans
    ans = 0
    for i in range(0,n):
        for j in range(i + 1,n):
            if (gcd(arr[i], 2*arr[j]) > 1):
                ans = ans + 1

    return ans

# Driver Code

# Initializations
arr = [ 5, 3, 6, 3 ]
N = len(arr)

# Function Call
ans = maximumpairs(arr, N)

print(ans)

# This code is contributed by Taranpreet

C#

// C# program for the above approach
using System;

class GFG {
  // find gcd
  static int gcd(int a, int b)
  {
    if (b == 0)
      return a;
    return gcd(b, a % b);
  }

  // Function to find the maximum number of pairs
  static int maximumpairs(int []arr, int n)
  {

    // Reorder array with even element first
    int idx = 0;
    for (int i = 0; i < n; i++) {
      if (arr[i] % 2 == 0) {
        //swap
        int temp = arr[i];
        arr[i] = arr[idx];
        arr[idx] = temp;
        idx++;
      }
    }

    // Now count the ans
    int ans = 0;
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        int a = arr[i];
        int b = 2 * arr[j];
        if (gcd(a, b) > 1) {
          ans++;
        }
      }
    }
    return ans;
  }

  public static void Main ()
  {

    // Initializations
    int []arr = { 5, 3, 6, 3 };
    int N = arr.Length;

    // Function Call
    int ans = maximumpairs(arr, N);

    Console.Write(ans);
  }
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
// Javascript program for the above approach

// find gcd
function gcd(a, b)
{
  if (b == 0)
    return a;
  return gcd(b, a % b);
}

// Function to find the maximum number of pairs
function maximumpairs(arr, n)
{

  // Reorder array with even element first
  let idx = 0;
  for (let i = 0; i < n; i++) {
    if (arr[i] % 2 == 0) {
      //swap
      let temp = arr[i];
      arr[i] = arr[idx];
      arr[idx] = temp;
      idx++;
    }
  }

  // Now count the ans
  let ans = 0;
  for (let i = 0; i < n; i++) {
    for (let j = i + 1; j < n; j++) {
      let a = arr[i];
      let b = 2 * arr[j];
      if (gcd(a, b) > 1) {
        ans++;
      }
    }
  }
  return ans;
}

  // Initializations
  let arr = [ 5, 3, 6, 3 ];
  let N = arr.length;

  // Function Call
  let ans = maximumpairs(arr, N);

  document.write(ans);

 // This code is contributed by Samim Hossain Mondal.
 </script>

Output

4

Time Complexity: O(N * N * logN) 
Auxiliary Space: O(1)