Count pairs in an array such that LCM(arr[i], arr[j]) > min(arr[i],arr[j])

Given an array arr[], the task is to find the count of pairs from the array such that LCM(arr[i], arr[j]) > min(arr[i], arr[j])
Note: Pairs (arr[i], arr[j]) and (arr[j], arr[i]) are considered identical and will be counted only once.

Examples:

Input: arr[] = {1, 1, 4, 9}
Output: 5
All valid pairs are (1, 4), (1, 9), (1, 4), (1, 9) and (4, 9)

Input: arr[] = {2, 4, 5, 2, 5, 7, 2, 8}
Output: 24

Approach: It is observed that only the pairs of the form (arr[i], arr[j]) where arr[i] = arr[j] won’t 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++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the count of valid pairs
int count_pairs(int n, int a[])
{
    // 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], i.e. Total pairs - pairs 
    // where arr[i] = arr[j]
    return ((n * (n - 1)) / 2) - count;
}
  
// Driver Code
int main()
{
    int arr[] = { 2, 4, 5, 2, 5, 7, 2, 8 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << count_pairs(n, arr);
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.HashMap;
import java.util.Map;
  
class GfG
{
  
    // Function to return the count of valid pairs 
    static int count_pairs(int n, int a[]) 
    
        // Store frequencies of array elements 
        HashMap<Integer, Integer> frequency = new HashMap<>(); 
        for (int i = 0; i < n; i++)
        
              
            if (!frequency.containsKey(a[i]))
                frequency.put(a[i], 0);
            frequency.put(a[i], frequency.get(a[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], i.e. Total pairs - pairs 
        // where arr[i] = arr[j] 
        return ((n * (n - 1)) / 2) - count; 
    }
  
    // Drive code
    public static void main(String []args)
    {
          
        int arr[] = { 2, 4, 5, 2, 5, 7, 2, 8 }; 
        int n = arr.length; 
        System.out.println(count_pairs(n, arr));
    }
}
      
// This code is contributed by Rituraj Jain

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Python3

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# Python3 implementation of the approach 
  
# Function to return the count 
# of valid pairs 
def count_pairs(n, a) : 
  
    # Store frequencies of array elements 
    frequency = dict.fromkeys(a, 0
    for i in range(n) :
        frequency[a[i]] += 1
  
    count = 0
  
    # Count of pairs (arr[i], arr[j]) 
    # where arr[i] = arr[j] 
    for f in frequency.values() : 
        count += f * (f - 1) // 2
      
    # Count of pairs (arr[i], arr[j]) where 
    # arr[i] != arr[j], i.e. Total pairs - pairs 
    # where arr[i] = arr[j] 
    return ((n * (n - 1)) // 2) - count
  
# Driver Code 
if __name__ == "__main__" :
      
    arr = [ 2, 4, 5, 2
            5, 7, 2, 8
    n = len(arr)
    print(count_pairs(n, arr))
  
# This code is contributed by Ryuga

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Output:

24


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Improved By : AnkitRai01, rituraj_jain