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Count pairs from an array having GCD equal to the minimum element in the pair
  • Difficulty Level : Expert
  • Last Updated : 06 May, 2021

Given an array arr[] consisting of N integers, the task is to find the number of pairs such that the GCD of any pair of array elements is the minimum element of that pair.

Examples:

Input: arr[ ] = {2, 3, 1, 2}
Output: 4
Explanation:
Below are the all possible pairs from the given array:

  1. (0, 1): The GCD of the pair formed by element at indices 0 and 2 is gcd(2, 1) = 1, which is equal to its minimum value of the pair {2, 1}.
  2. (0, 3): The GCD of the pair formed by element at indices 0 and 3 is gcd(2, 2) = 2, which is equal to its minimum value of the pair {2, 2}.
  3. (1, 2): The GCD of the pair formed by taking element at indices 1 and 2 is gcd(3, 1) = 1, which is equal to its minimum value of the pair {3, 1}.
  4. (2, 3): The GCD of the pair formed by taking element at indices 2 and 3 is gcd(1, 2) = 1, which is equal to its minimum value of the pair {1, 2}.

Therefore, there is a total of 4 possible pairs whose GCD is equal to their minimum element of the pair.

Input: arr[] = {4, 6}
Output: 0



Naive Approach: The simplest approach to solve the given problem is to generate all possible pairs from the given array and if there exists any pair whose GCD is equal to the minimum elements of that pair, then count that pair. After checking for all the pairs, print the value of count obtained as the result.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count pairs from an
// array having GCD equal to
// minimum element of that pair
int countPairs(int arr[], int N)
{
    // Stores the resultant count
    int count = 0;
 
    // Iterate over the range [0, N - 2]
    for (int i = 0; i < N - 1; i++) {
 
        // Iterate over the range [i + 1, N]
        for (int j = i + 1; j < N; j++) {
 
            // If arr[i] % arr[j] is 0
            // or arr[j] % arr[i] is 0
            if (arr[i] % arr[j] == 0
                || arr[j] % arr[i] == 0) {
 
                // Increment count by 1
                count++;
            }
        }
    }
 
    // Return the resultant count
    return count;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 3, 1, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << countPairs(arr, N);
 
    return 0;
}

Java




// java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
public class GFG {
 
    // Function to count pairs from an
    // array having GCD equal to
    // minimum element of that pair
    static int countPairs(int arr[], int N)
    {
       
        // Stores the resultant count
        int count = 0;
 
        // Iterate over the range [0, N - 2]
        for (int i = 0; i < N - 1; i++) {
 
            // Iterate over the range [i + 1, N]
            for (int j = i + 1; j < N; j++) {
 
                // If arr[i] % arr[j] is 0
                // or arr[j] % arr[i] is 0
                if (arr[i] % arr[j] == 0
                    || arr[j] % arr[i] == 0) {
 
                    // Increment count by 1
                    count++;
                }
            }
        }
 
        // Return the resultant count
        return count;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        int arr[] = { 2, 3, 1, 2 };
        int N = arr.length;
        System.out.print(countPairs(arr, N));
    }
}
 
// This code is contributed by Kingash.

Python3




# Python3 rogram for the above approach
 
# Function to count pairs from an
# array having GCD equal to
# minimum element of that pair
def countPairs(arr, N):
   
    # Stores the resultant count
    count = 0
 
    # Iterate over the range [0, N - 2]
    for i in range(N - 1):
       
        # Iterate over the range [i + 1, N]
        for j in range(i + 1, N):
           
            # If arr[i] % arr[j] is 0
            # or arr[j] % arr[i] is 0
            if (arr[i] % arr[j] == 0 or arr[j] % arr[i] == 0):
                 
                # Increment count by 1
                count += 1
                  
    # Return the resultant count
    return count
 
# Driver Code
if __name__ == '__main__':
    arr = [2, 3, 1, 2]
    N = len(arr)
    print (countPairs(arr, N))
 
# This code is contributed by mohit kumar 29.

C#




// C# program for the above approach
 
using System;
 
public class GFG {
 
    // Function to count pairs from an
    // array having GCD equal to
    // minimum element of that pair
    static int countPairs(int[] arr, int N)
    {
 
        // Stores the resultant count
        int count = 0;
 
        // Iterate over the range [0, N - 2]
        for (int i = 0; i < N - 1; i++) {
 
            // Iterate over the range [i + 1, N]
            for (int j = i + 1; j < N; j++) {
 
                // If arr[i] % arr[j] is 0
                // or arr[j] % arr[i] is 0
                if (arr[i] % arr[j] == 0
                    || arr[j] % arr[i] == 0) {
 
                    // Increment count by 1
                    count++;
                }
            }
        }
 
        // Return the resultant count
        return count;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
 
        int[] arr = { 2, 3, 1, 2 };
        int N = arr.Length;
        Console.WriteLine(countPairs(arr, N));
    }
}
 
// This code is contributed by ukasp.
Output: 
4

 

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized based on the following observations:

  • The minimum element of the pair should divide the maximum element of the pair, and it can be observed that element 1 can form a total of (N – 1) pairs.
  • Every element can form ^YC_2       pair with itself where Y is the count of an array element.
  • The idea is to traverse over the divisors of every array element and increment the count of pairs that can be formed by the frequencies of the divisors.

Follow the steps below to solve the problem:

  • Initialize a variable, say res, that stores the resultant count.
  • Initialize a map, say mp, that stores the count of every array element.
  • Traverse the array arr[] and increment the count of arr[i] in mp.
  • Iterate over the pairs of map mp and perform the following operations:
    • Store the value of array element in a variable say X and frequency of that number in a variable Y.
    • If the value of X is 1, then increment the value of res by (N – 1) and continue.
    • Increment the value of res by (Y*(Y – 1))/2.
    • Now, iterate over the divisors of the integer X using a variable j and increment the res by mp[j].
  • After completing the above steps, print the value of res as the total count obtained.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count pairs from an
// array having GCD equal to
// minimum element of that pair
int CountPairs(int arr[], int N)
{
    // Stores the resultant count
    int res = 0;
 
    // Stores the frequency of
    // each array element
    map<int, int> mp;
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
        mp[arr[i]]++;
    }
    // Iterate over the Map mp
    for (auto p : mp) {
 
        // Stores the array element
        int x = p.first;
 
        // Stores the count
        // of array element x
        int y = p.second;
 
        // If x is 1
        if (x == 1) {
 
            // Increment res by N-1
            res += N - 1;
            continue;
        }
 
        // Increment res by yC2
        res += (y * (y - 1)) / 2;
 
        // Iterate over the
        // range [2, sqrt(x)]
        for (int j = 2;
             j <= sqrt(x); j++) {
 
            // If x is divisble by j
            if (x % j == 0) {
 
                // Increment the value
                // of res by mp[j]
                res += mp[j];
 
                // If j is not equal to x/j
                if (j != x / j)
 
                    // Increment res
                    // by mp[x/j]
                    res += mp[x / j];
            }
        }
    }
 
    // Return the resultant count
    return res;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 3, 1, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << CountPairs(arr, N);
 
    return 0;
}
Output: 
4

 

Time Complexity: O(N*√M), where M is the maximum element of the array.
Auxiliary Space: O(1)

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