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Count of pairs in Array whose product is divisible by K

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  • Last Updated : 13 May, 2022
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Given an array A[] and positive integer K, the task is to count the total number of pairs in the array whose product is divisible by K

Examples :

Input: A[] = [1, 2, 3, 4, 5], K = 2
Output: 7
Explanation: The 7 pairs of indices whose corresponding products are divisible by 2 are
(0, 1), (0, 3), (1, 2), (1, 3), (1, 4), (2, 3), and (3, 4).
Other pairs such as (0, 2) and (2, 4) have products 3 and 15 respectively, which are not divisible by 2. 

Input: A[] = [1, 2, 3, 4], K = 5
Output: 0
Explanation: There does not exist any pair of indices whose corresponding product is divisible by 5.

 

Naive approach: For finding the counts of all pairs we can  simply do a nested loop iteration and for each of element we can check all remaining elements whether their product is divisible by given key or not.

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

Efficient approach: The problem can be solved efficiently using hashing based on the following observation:

  • For checking the divisibility of product with K, better to deal with the GCD of number with K. This will remove all other factors from consideration. As, the number of divisors of K, would be very small when compared with the length of original array size.
  • If GCD(a, K) * GCD(b, K) is divisible by key, then a * b should also be divisible by key.

Follow the steps mentioned below to solve the problem:

  • Create a map which will store the GCD(A[i], K) as key and its occurrence as value.
  • For each element of the array, check all elements of map, whether map’s element is divisible by X (X =  quotient when K is divide by GCD(A[i], K))
    •  if yes add the occurrence of that element from map to answer. 
    • Also, keep incrementing the occurrence for each element’s GCD with key.
  • Return the final count.

Below is the implementation of the above approach

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Program to count pairs whose product
// is divisible by key
long long countPairs(vector<int>& A, int
                                         key)
{
    long long ans = 0;
    unordered_map<int, int> mp;
 
    for (auto ele : A) {
        // Calculate gcd of nums[i] and
        // key
        long long gcd = __gcd(key, ele);
        long long x = key / gcd;
 
        // Iterate over all possible gcds
        for (auto it : mp)
            if (it.first % x == 0)
                // Add count to answer
                ans += it.second;
        // Add gcd to map
        mp[gcd]++;
    }
 
    return ans;
}
 
// Driver code
int main()
{
    vector<int> A = { 1, 2, 3, 4, 5 };
    int key = 2;
 
    cout << countPairs(A, key) << endl;
    return 0;
}


Java




// JAVA program for the above approach
import java.util.*;
class GFG {
 
  public static int agcd(int a, int b)
  {
    if (b == 0)
      return a;
    return agcd(b, a % b);
  }
 
  // Program to count pairs whose product
  // is divisible by key
  public static long countPairs(int[] A, int key)
  {
    long ans = 0;
    HashMap<Long, Integer> mp = new HashMap<>();
 
    for (int i = 0; i < A.length; ++i)
    {
 
      // Calculate gcd of nums[i] and
      // key
      long gcd = agcd(key, A[i]);
      long x = key / gcd;
 
      // Iterate over all possible gcds
      for (Map.Entry<Long, Integer> it :
           mp.entrySet())
        if (it.getKey() % x == 0)
          // Add count to answer
          ans += it.getValue();
      // Add gcd to map
      if (mp.containsKey(gcd)) {
        mp.put(gcd, mp.get(gcd) + 1);
      }
      else {
        mp.put(gcd, 1);
      }
      // mp[gcd]++;
    }
 
    return ans;
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int A[] = new int[] { 1, 2, 3, 4, 5 };
    int key = 2;
 
    System.out.println(countPairs(A, key));
  }
}
 
// This code is contributed by Taranpreet


Python3




# Python3 Program to count pairs whose product
# is divisible by key
import math
 
def countPairs(A,key):
    ans = 0
    mp = {}
 
    for ele in A:
 
        # Calculate gcd of nums[i] and
        # key
        gcd = math.gcd(ele, key)
        x = key // gcd
 
        # Iterate over all possible gcds
        for Key,value in mp.items():
            if (Key % x == 0):
                # Add count to answer
                ans += value
        # Add gcd to map
        if(gcd in mp):
            mp[gcd] = mp[gcd] + 1
        else:
            mp[gcd] = 1
 
    return ans
 
# Driver code
 
A = [ 1, 2, 3, 4, 5 ]
key = 2
 
print(countPairs(A, key))
 
# This code is contributed by shinjanpatra


C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG {
 
  static int agcd(int a, int b)
  {
    if (b == 0)
      return a;
    return agcd(b, a % b);
  }
 
  // Program to count pairs whose product
  // is divisible by key
  static long countPairs(int[] A, int key)
  {
    long ans = 0;
    Dictionary<long, int> mp
      = new Dictionary<long, int>();
 
    for (int i = 0; i < A.Length; ++i) {
 
      // Calculate gcd of nums[i] and
      // key
      long gcd = agcd(key, A[i]);
      long x = key / gcd;
 
      // Iterate over all possible gcds
      foreach(KeyValuePair<long, int> it in mp)
      {
 
        if (it.Key % x == 0)
          // Add count to answer
          ans += it.Value;
      }
      // Add gcd to map
      if (mp.ContainsKey(gcd)) {
        mp[gcd] = mp[gcd] + 1;
      }
      else {
        mp.Add(gcd, 1);
      }
      // mp[gcd]++;
    }
 
    return ans;
  }
 
  // Driver code
  public static void Main()
  {
    int[] A = { 1, 2, 3, 4, 5 };
    int key = 2;
 
    Console.WriteLine(countPairs(A, key));
  }
}
 
// This code is contributed by Samim Hossain Mondal.


Javascript




<script>
 
// Program to count pairs whose product
// is divisible by key
 
function _gcd(x,y){
    if(!y)return x;
    return _gcd(y,x%y);
}
 
function countPairs(A,key){
 
    let ans = 0
    let mp = new Map()
 
    for(let ele of A){
 
        // Calculate gcd of nums[i] and
        // key
        let gcd = _gcd(ele,key)
        let x = Math.floor(key / gcd)
 
        // Iterate over all possible gcds
        for(let [Key,value] of mp){
            if (Key % x == 0)
                // Add count to answer
                ans += value
        }
        // Add gcd to map
        if(mp.has(gcd))
            mp.set(gcd,mp.get(gcd)+1)
        else
            mp.set(gcd,1)
    }
 
    return ans
}
 
// Driver code
 
let A = [ 1, 2, 3, 4, 5 ]
let key = 2
 
document.write(countPairs(A, key))
 
// code is contributed by shinjanpatra
 
</script>


Output

7

Time Complexity: O(N*K1/2)
Space Complexity: O(K1/2)


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