Number of pairs such that their HCF and LCM is equal

Given an array a[] of non-negative integers. The task is to count the number of pairs (i, j) in the array such that LCM(a[i], a[j]) = HCF(a[i], a[j]). 
Note: The pair (i, j) and (j, i) are considered same and i should not be equal to j.
Examples: 

Input : a[] = {3, 4, 3, 4, 5}
Output : 2
Pairs are (3, 3) and (4, 4)

Input  : a[] = {1, 1, 1}
Output : 3

Naive approach: Generate all possible pairs and count pairs with equal HCF and LCM. 

C++

// Naive C++ program to count number of pairs
// such that their hcf and lcm are equal
#include <bits/stdc++.h>
using namespace std;

// Function to return HCF of two numbers
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

// Function to return LCM of two numbers
int lcm(int a, int b)
{
    return (a * b) / gcd(a, b);
}

// Returns the number of valid pairs
int countPairs(int arr[], int n)
{
    int ans = 0; // initializing answer

    // Traversing the array. For each array
    // element, checking if it
    // follow the condition
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
            if (lcm(arr[i], arr[j]) == gcd(arr[i], arr[j]))
                ans++;
    return ans;
}

// Driver function
int main()
{
    int arr[] = { 1, 1, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countPairs(arr, n);
    return 0;
}

Java

// Naive Java program to count number of pairs
// such that their hcf and lcm are equal
import java.util.*;

class GFG{

// Function to return HCF of two numbers
static int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

// Function to return LCM of two numbers
static int lcm(int a, int b)
{
    return (a * b) / gcd(a, b);
}

// Returns the number of valid pairs
static int countPairs(int arr[], int n)
{
    int ans = 0; // initializing answer

    // Traversing the array. For each array
    // element, checking if it
    // follow the condition
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
            if (lcm(arr[i], arr[j]) == gcd(arr[i], arr[j]))
                ans++;
    return ans;
}

// Driver function
public static void main(String[] args)
{
    int arr[] = { 1, 1, 1 };
    int n = arr.length;
    System.out.print(countPairs(arr, n));
}
}

// This code is contributed by Rajput-Ji

Python3

# Naive Python program to count number of pairs
# such that their hcf and lcm are equal

# Function to return HCF of two numbers
def gcd(a, b):
    if (a == 0):
        return b;
    return gcd(b % a, a);

# Function to return LCM of two numbers
def lcm(a, b):
    return (a * b) / gcd(a, b);

# Returns the number of valid pairs
def countPairs(arr, n):
    ans = 0; # initializing answer

    # Traversing the array. For each array
    # element, checking if it
    # follow the condition
    for i in range(n):
        for j in range(i+1,n):
            if (lcm(arr[i], arr[j]) == gcd(arr[i], arr[j])):
                ans+=1;
    return ans;

# Driver function
if __name__ == '__main__':
    arr = [ 1, 1, 1 ];
    n = len(arr);
    print(countPairs(arr, n));

# This code is contributed by 29AjayKumar

C#

// Naive C# program to count number of pairs
// such that their hcf and lcm are equal
using System;

class GFG{

// Function to return HCF of two numbers
static int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

// Function to return LCM of two numbers
static int lcm(int a, int b)
{
    return (a * b) / gcd(a, b);
}

// Returns the number of valid pairs
static int countPairs(int []arr, int n)
{
    int ans = 0; // initializing answer

    // Traversing the array. For each array
    // element, checking if it
    // follow the condition
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
            if (lcm(arr[i], arr[j]) == gcd(arr[i], arr[j]))
                ans++;
    return ans;
}

// Driver function
public static void Main(String[] args)
{
    int []arr = { 1, 1, 1 };
    int n = arr.Length;
    Console.Write(countPairs(arr, n));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>


// Naive Javascript program to count number of pairs
// such that their hcf and lcm are equal

// Function to return HCF of two numbers
function gcd(a, b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

// Function to return LCM of two numbers
function lcm(a, b)
{
    return (a * b) / gcd(a, b);
}

// Returns the number of valid pairs
function countPairs(arr, n)
{
    var ans = 0; // initializing answer

    // Traversing the array. For each array
    // element, checking if it
    // follow the condition
    for (var i = 0; i < n; i++)
        for (var j = i + 1; j < n; j++)
            if (lcm(arr[i], arr[j]) == gcd(arr[i], arr[j]))
                ans++;
    return ans;
}

// Driver function
var arr = [ 1, 1, 1 ];
var n = arr.length;
document.write(countPairs(arr, n));

// This code is contributed by rutvik_56.
</script>

Output:

3

Time Complexity: O(n² * log(max(a, b)))

Auxiliary Space: O(log(max(a, b)))

Efficient Approach: If we observe carefully, it can be proved that the HCF and LCM of two numbers can be equal only when the numbers are also equal.
Proof: 

Let,
HCF(a[i], a[j]) = LCM(a[i], a[j]) = K

Since HCF(a[i], a[j]) = k,
a[i] = k*n1, a[j] = k*n2, for some natural numbers n1, n2

We know that,
HCF × LCM = Product of the two numbers

Therefore,
k*k = k*n1 × k*n2
or, n1*n2 = 1

Implies, n1 = n2 = 1, since n1, n2 are natural numbers.

Therefore,
a[i] = k*n1 = k, a[j] = k*n2 = k
i.e. the numbers must be equal.

So we have to count pairs with same elements in the pair.
It is observed that only the pairs of the form (arr[i], arr[j]) where arr[i] = arr[j] will satisfy the given condition. So, the problem now gets reduced to finding the number of pairs (arr[i], arr[j]) such that arr[i] = arr[j].
Below is the implementation of the above approach:

C++

// C++ program to count number of pairs
// such that their hcf and lcm are equal
#include <bits/stdc++.h>
using namespace std;

// Function to count number of pairs
// such that their hcf and lcm are equal
int countPairs(int a[], int n)
{
    // Store frequencies of array elements
    unordered_map<int, int> frequency;
    for (int i = 0; i < n; i++) {
        frequency[a[i]]++;
    }

    int count = 0;

    // Count of pairs (arr[i], arr[j])
    // where arr[i] = arr[j]
    for (auto x : frequency) {
        int f = x.second;
        count += f * (f - 1) / 2;
    }

    // Count of pairs (arr[i], arr[j]) where
    // arr[i] = arr[j],
    return count;
}

// Driver function
int main()
{
    int arr[] = { 1, 1, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countPairs(arr, n);
    return 0;
}

Java

// Java program to count number of pairs
// such that their hcf and lcm are equal

import java.util.*;

class GFG{

// Function to count number of pairs
// such that their hcf and lcm are equal
static int countPairs(int arr[], int n)
{
    // Store frequencies of array elements
    HashMap<Integer,Integer> frequency =
            new HashMap<Integer,Integer>();
    for (int i = 0; i < n; i++) {
        if(frequency.containsKey(arr[i])){
            frequency.put(arr[i], frequency.get(arr[i])+1);
        }else{
            frequency.put(arr[i], 1);
    }
    }

    int count = 0;

    // Count of pairs (arr[i], arr[j])
    // where arr[i] = arr[j]
    for (Map.Entry<Integer,Integer> x : frequency.entrySet()) {
        int f = x.getValue();
        count += f * (f - 1) / 2;
    }

    // Count of pairs (arr[i], arr[j]) where
    // arr[i] = arr[j],
    return count;
}

// Driver function
public static void main(String[] args)
{
    int arr[] = { 1, 1, 1 };
    int n = arr.length;
    System.out.print(countPairs(arr, n));
}
}

// This code contributed by sapnasingh4991

C#

// C# program to count number of pairs
// such that their hcf and lcm are equal
using System;
using System.Collections.Generic;

class GFG{

// Function to count number of pairs
// such that their hcf and lcm are equal
static int countPairs(int []a, int n)
{
    // Store frequencies of array elements
    Dictionary<int, int> frequency = new Dictionary<int,int>();
        for (int i = 0; i < n; i++)
        {
            if (frequency.ContainsKey(a[i]))
            {
                var val = frequency[a[i]];
                frequency.Remove(a[i]);
                frequency.Add(a[i], val + 1);
            }
            else
            {
                frequency.Add(a[i], 1);
            }
        }

    int count = 0;

    // Count of pairs (arr[i], arr[j])
    // where arr[i] = arr[j]
    foreach(KeyValuePair<int, int> entry in frequency) {
        int f = entry.Value;
        count += f * (f - 1) / 2;
    }

    // Count of pairs (arr[i], arr[j]) where
    // arr[i] = arr[j],
    return count;
}

// Driver function
public static void Main(String[] args)
{
    int []arr = { 1, 1, 1 };
    int n = arr.Length;
    Console.Write(countPairs(arr, n));
}
}

// This code is contributed by shivanisinghss2110

Python3

# Python 3 program to count number of pairs
# such that their hcf and lcm are equal
from collections import defaultdict

# Function to count number of pairs
# such that their hcf and lcm are equal
def countPairs(a, n):

    # Store frequencies of array elements
    frequency = defaultdict(int)
    for i in range(n) :
        frequency[a[i]] += 1


    count = 0

    # Count of pairs (arr[i], arr[j])
    # where arr[i] = arr[j]
    for x in frequency.keys():
        f = frequency[x]
        count += f * (f - 1) // 2

    # Count of pairs (arr[i], arr[j]) where
    # arr[i] = arr[j],
    return count

# Driver function
if __name__ == "__main__":

    arr = [ 1, 1, 1 ]
    n = len(arr)
    print(countPairs(arr, n))

# This code is contributed by chitranayal

Javascript

<script>

// JavaScript program to count number of pairs
// such that their hcf and lcm are equal

// Function to count number of pairs
// such that their hcf and lcm are equal
function countPairs(arr,n)
{
    // Store frequencies of array elements
    let frequency = new Map();
    for (let i = 0; i < n; i++) {
        if(frequency.has(arr[i])){
            frequency.set(arr[i], frequency.get(arr[i])+1);
        }else{
            frequency.set(arr[i], 1);
    }
    }

    let count = 0;

    // Count of pairs (arr[i], arr[j])
    // where arr[i] = arr[j]
    for (let [key, value] of frequency.entries()) {
        let f = value;
        count += f * (f - 1) / 2;
    }

    // Count of pairs (arr[i], arr[j]) where
    // arr[i] = arr[j],
    return count;
}

// Driver function
let arr=[1, 1, 1 ];
let n = arr.length;
document.write(countPairs(arr, n));



// This code is contributed by avanitrachhadiya2155

</script>

Output:

3

Time Complexity: O(n)

Auxiliary Space: O(n)