Minimize Array length by repeatedly replacing co-prime pairs with 1

Given an array arr[] consisting of N elements, the task is to minimize the array length by replacing any two coprime array elements with 1.
Examples:

Input: arr[] = {2, 3, 5} 
Output: 1 
Explanation:
Replace {2, 3} with 1 modifies the array to {1, 5}. 
Replace {1, 5} with 1 modifies the array to {1}.
Input: arr[] = {6, 9, 15} 
Output: 3 
Explanation: No coprime pairs exist in the array. Therefore, no reduction possible.

Naive Approach: The simplest approach is to iterate over the array and check for coprime pairs. If found replace it with 1 search for the next coprime pair and so on.

Time Complexity: O(N³  * log N) 
Auxiliary Space: O(1)

Efficient Approach: This approach is based on the fact:

1 is coprime with every number

The idea is to find if there is any co-prime pair present in the array or not. If found, then all the array elements can be reduced to 1 based on the above fact. Hence, if any co-prime pair is found, then, the required answer will be 1, else, the answer will be the initial size of the array.

Illustration:
For arr[] = {2, 4, 6, 8, 9}
Here, as there exists a coprime pair {2, 9}, replacing them by 1 modifies the array to {1, 4, 6, 8}. 
Since 1 is coprime with every number, the array can be reduced further in following steps: 
{1, 4, 6, 8} -> {1, 6, 8} -> {1, 8} -> {1} 
Hence, the array can be reduced to size 1.

Below is the implementation of the above approach:

C++

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

// Function to find the final array
// length by replacing coprime pair with 1
bool hasCoprimePair(vector<int>& arr, int n)
{

    // Iterate over all pairs of element
    for (int i = 0; i < n - 1; i++) {
        for (int j = i + 1; j < n; j++) {

            // Check if gcd is 1
            if (__gcd(arr[i], arr[j]) == 1) {
                return true;
            }
        }
    }

    // If no coprime pair
    // found return false
    return false;
}

// Driver code
int main()
{

    int n = 3;
    vector<int> arr = { 6, 9, 15 };

    // Check if atleast one coprime
    // pair exists in the array
    if (hasCoprimePair(arr, n)) {
        cout << 1 << endl;
    }

    // If no such pair exists
    else {
        cout << n << endl;
    }
}

Java

// Java Program for the above approach
import java.util.*;
class GFG{

// Recursive function to return
// gcd of a and b
static int __gcd(int a, int b)
{
    return b == 0? a:__gcd(b, a % b);
}

// Function to find the final array
// length by replacing coprime pair with 1
static boolean hasCoprimePair(int []arr, int n)
{

    // Iterate over all pairs of element
    for (int i = 0; i < n - 1; i++)
    {
        for (int j = i + 1; j < n; j++)
        {

            // Check if gcd is 1
            if ((__gcd(arr[i], arr[j])) == 1)
            {
                return true;
            }
        }
    }

    // If no coprime pair
    // found return false
    return false;
}

// Driver code
public static void main(String[] args)
{
    int n = 3;
    int []arr = { 6, 9, 15 };

    // Check if atleast one coprime
    // pair exists in the array
    if (hasCoprimePair(arr, n))
    {
        System.out.print(1 + "\n");
    }

    // If no such pair exists
    else
    {
        System.out.print(n + "\n");
    }
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 program for the above approach
import math

# Function to find the final array
# length by replacing coprime pair with 1
def hasCoprimePair(arr, n):

    # Iterate over all pairs of element
    for i in range(n - 1):
        for j in range(i + 1, n):

            # Check if gcd is 1
            if (math.gcd(arr[i], arr[j]) == 1):
                return True

    # If no coprime pair
    # found return false
    return False

# Driver code
if __name__ == "__main__":

    n = 3
    arr = [ 6, 9, 15 ]

    # Check if atleast one coprime
    # pair exists in the array
    if (hasCoprimePair(arr, n)):
        print(1)

    # If no such pair exists
    else:
        print(n)

# This code is contributed by chitranayal

C#

// C# Program for the above approach
using System;
class GFG{

// Recursive function to return
// gcd of a and b
static int __gcd(int a, int b)
{
    return b == 0 ? a : __gcd(b, a % b);
}

// Function to find the readonly array
// length by replacing coprime pair with 1
static bool hasCoprimePair(int []arr, int n)
{

    // Iterate over all pairs of element
    for (int i = 0; i < n - 1; i++)
    {
        for (int j = i + 1; j < n; j++)
        {

            // Check if gcd is 1
            if ((__gcd(arr[i],
                       arr[j])) == 1)
            {
                return true;
            }
        }
    }

    // If no coprime pair
    // found return false
    return false;
}

// Driver code
public static void Main(String[] args)
{
    int n = 3;
    int []arr = { 6, 9, 15 };

    // Check if atleast one coprime
    // pair exists in the array
    if (hasCoprimePair(arr, n))
    {
        Console.Write(1 + "\n");
    }

    // If no such pair exists
    else
    {
        Console.Write(n + "\n");
    }
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// Javascript Program for the above approach

    // Recursive function to return
    // gcd of a and b
    function __gcd(a , b) {
        return b == 0 ? a : __gcd(b, a % b);
    }

    // Function to find the final array
    // length by replacing coprime pair with 1
    function hasCoprimePair(arr , n) {

        // Iterate over all pairs of element
        for (i = 0; i < n - 1; i++) {
            for (j = i + 1; j < n; j++) {

                // Check if gcd is 1
                if ((__gcd(arr[i], arr[j])) == 1) {
                    return true;
                }
            }
        }

        // If no coprime pair
        // found return false
        return false;
    }

    // Driver code

        var n = 3;
        var arr = [ 6, 9, 15 ];

        // Check if atleast one coprime
        // pair exists in the array
        if (hasCoprimePair(arr, n)) {
            document.write(1 + "\n");
        }

        // If no such pair exists
        else {
            document.write(n + "\n");
        }
// This code contributed by gauravrajput1

</script>

Output:

3

Time Complexity: O(N² * log N) 
Auxiliary Space: O(1)