Count of triplets in an Array such that A[i] * A[j] = A[k] and i < j < k

Namachila November 19, 2025

0 0 7 minutes read

Given an array A[ ] consisting of N positive integers, the task is to find the number of triplets A[i], A[j] & A[k] in the array such that i < j < k and A[i] * A[j] = A[k].

Examples:

Input: N = 5, A[ ] = {2, 3, 4, 6, 12}
Output: 3
Explanation:
The valid triplets from the given array are:
(A[0], A[1], A[3]) = (2, 3, 6) where (2*3 = 6)
(A[0], A[3], A[4]) = (2, 6, 12) where (2*6 = 12)
(A[1], A[2], A[4]) = (3, 4, 12) where (3*4 = 12)
Hence, a total of 3 triplets exists which satisfies the given condition.

Input: N = 3, A[ ] = {1, 1, 1}
Output: 1
Explanation:
The only valid triplet is (A[0], A[1], A[2]) = (1, 1, 1)

Naive Approach:
The simplest approach to solve the problem is to generate all possible triplets and for each triplet, check if it satisfies the required condition. If found to be true, increase the count of triplets. After complete traversal of the array and generating all possible triplets, print the final count.

C++

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Returns total number of
// valid triplets possible
int countTriplets(int A[], int N)
{
    // Stores the count
    int ans = 0;
 
    for (int i = 0; i < N; i++) {
        for (int j = i + 1; j < N; j++) {
            for (int k = j + 1; k < N; k++) {
                if (A[i] * A[j] == A[k])
                    ans++;
            }
        }
    }
 
    // Return the final count
    return ans;
}
 
// Driver Code
int main()
{
    int N = 5;
    int A[] = { 2, 3, 4, 6, 12 };
 
    cout << countTriplets(A, N);
 
    return 0;
}

Java

/*package whatever //do not write package name here */
import java.io.*;
 
public class GFG {
 
  // Returns total number of
  // valid triplets possible
  static int countTriplets(int A[], int N)
  {
    // Stores the count
    int ans = 0;
 
    for (int i = 0; i < N; i++) {
      for (int j = i + 1; j < N; j++) {
        for (int k = j + 1; k < N; k++) {
          if (A[i] * A[j] == A[k])
            ans++;
        }
      }
    }
 
    // Return the final count
    return ans;
  }
 
  public static void main (String[] args) {
 
    int N = 5;
    int A[] = { 2, 3, 4, 6, 12 };
 
    System.out.println(countTriplets(A, N));
  }
}
 
// This code is contributed by aadityaburujwale.

Python3

# Python3 Program to implement
# the above approach
 
# Returns total number of
# valid triplets possible
def countTriplets( A, N):
   
    # Stores the count
    ans = 0;
 
    for i in range(0, N):
        for j in range(i + 1, N):
            for k in range(j + 1, N):
                if (A[i] * A[j] == A[k]):
                    ans += 1;
         
    # Return the final count
    return ans;
 
# Driver Code
N = 5;
A = [ 2, 3, 4, 6, 12 ];
print(countTriplets(A, N));
 
# This code is contributed by poojaagrawal2.

C#

// Include namespace system
using System;
 
 
public class GFG
{
    // Returns total number of
    // valid triplets possible
    public static int countTriplets(int[] A, int N)
    {
        // Stores the count
        var ans = 0;
        for (int i = 0; i < N; i++)
        {
            for (int j = i + 1; j < N; j++)
            {
                for (int k = j + 1; k < N; k++)
                {
                    if (A[i] * A[j] == A[k])
                    {
                        ans++;
                    }
                }
            }
        }
        // Return the final count
        return ans;
    }
    public static void Main(String[] args)
    {
        var N = 5;
        int[] A = {2, 3, 4, 6, 12};
        Console.WriteLine(countTriplets(A, N));
    }
}

Javascript

// Javascript Program to implement
// the above approach
 
// Returns total number of
// valid triplets possible
function countTriplets(A, N)
{
    // Stores the count
    let ans = 0;
 
    for (let i = 0; i < N; i++) {
        for (let j = i + 1; j < N; j++) {
            for (let k = j + 1; k < N; k++) {
                if (A[i] * A[j] == A[k])
                    ans++;
            }
        }
    }
 
    // Return the final count
    return ans;
}
 
// Driver Code
let N = 5;
let A = [ 2, 3, 4, 6, 12 ];
 
console.log(countTriplets(A, N));

Output

Time Complexity: O(N³)
Auxiliary Space: O(1)

Efficient Approach:
The above approach can be optimized using Two Pointers and HashMap.
Follow the steps below to solve the problem:

Initialize a Map to store frequencies of array elements.
Iterate over the array in reverse, i.e. loop with a variable j in the range [N – 2, 1].
For every j, increase the count of A[j + 1] in the map. Iterate over the range [0, j – 1] using variable i and check if A[i] * A[j] is present in the map or not.
If A[i] * A[j] is found in the map, increase the count of triplets by the frequency of A[i] * A[j] stored in the map.
After complete traversal of the array, print the final count.

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Returns total number of
// valid triplets possible
int countTriplets(int A[], int N)
{
    // Stores the count
    int ans = 0;
 
    // Map to store frequency
    // of array elements
    map<int, int> map;
 
    for (int j = N - 2; j >= 1; j--) {
 
        // Increment the frequency
        // of A[j+1] as it can be
        // a valid A[k]
        map[A[j + 1]]++;
 
        for (int i = 0; i < j; i++) {
 
            int target = A[i] * A[j];
 
            // If target exists in the map
            if (map.find(target)
                != map.end())
                ans += map[target];
        }
    }
 
    // Return the final count
    return ans;
}
 
// Driver Code
int main()
{
    int N = 5;
    int A[] = { 2, 3, 4, 6, 12 };
 
    cout << countTriplets(A, N);
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Returns total number of
// valid triplets possible
static int countTriplets(int A[], int N)
{
     
    // Stores the count
    int ans = 0;
 
    // Map to store frequency
    // of array elements
    HashMap<Integer,
            Integer> map = new HashMap<Integer,
                                       Integer>();
                                        
    for(int j = N - 2; j >= 1; j--) 
    {
 
        // Increment the frequency
        // of A[j+1] as it can be
        // a valid A[k]
        if(map.containsKey(A[j + 1]))
            map.put(A[j + 1], map.get(A[j + 1]) + 1);
        else
            map.put(A[j + 1], 1);
 
        for(int i = 0; i < j; i++) 
        {
            int target = A[i] * A[j];
 
            // If target exists in the map
            if (map.containsKey(target))
                ans += map.get(target);
        }
    }
 
    // Return the final count
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 5;
    int A[] = { 2, 3, 4, 6, 12 };
 
    System.out.print(countTriplets(A, N));
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 program for the above approach
from collections import defaultdict
 
# Returns total number of
# valid triplets possible
def countTriplets(A, N):
 
    # Stores the count
    ans = 0
 
    # Map to store frequency
    # of array elements
    map = defaultdict(lambda: 0)
 
    for j in range(N - 2, 0, -1):
 
        # Increment the frequency
        # of A[j+1] as it can be
        # a valid A[k]
        map[A[j + 1]] += 1
 
        for i in range(j):
            target = A[i] * A[j]
 
            # If target exists in the map
            if(target in map.keys()):
                ans += map[target]
 
    # Return the final count 
    return ans
 
# Driver code
if __name__ == '__main__':
 
    N = 5
    A = [ 2, 3, 4, 6, 12 ]
 
    print(countTriplets(A, N))
 
# This code is contributed by Shivam Singh

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
 
// Returns total number of
// valid triplets possible
static int countTriplets(int []A, int N)
{
     
    // Stores the count
    int ans = 0;
 
    // Map to store frequency
    // of array elements
    Dictionary<int,
               int> map = new Dictionary<int,
                                         int>();
                                        
    for(int j = N - 2; j >= 1; j--) 
    {
 
        // Increment the frequency
        // of A[j+1] as it can be
        // a valid A[k]
        if(map.ContainsKey(A[j + 1]))
            map[A[j + 1]] = map[A[j + 1]] + 1;
        else
            map.Add(A[j + 1], 1);
 
        for(int i = 0; i < j; i++) 
        {
            int target = A[i] * A[j];
 
            // If target exists in the map
            if (map.ContainsKey(target))
                ans += map[target];
        }
    }
 
    // Return the readonly count
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 5;
    int []A = { 2, 3, 4, 6, 12 };
 
    Console.Write(countTriplets(A, N));
}
}
 
// This code is contributed by sapnasingh4991

Javascript

<script>
 
// Javascript program to implement
// the above approach
  
// Returns total number of
// valid triplets possible
function countTriplets(A, N)
{
      
    // Stores the count
    let ans = 0;
  
    // Map to store frequency
    // of array elements
    let map = new Map();
                                         
    for(let j = N - 2; j >= 1; j--)
    {
  
        // Increment the frequency
        // of A[j+1] as it can be
        // a valid A[k]
        if(map.has(A[j + 1]))
            map.set(A[j + 1], map.get(A[j + 1]) + 1);
        else
            map.set(A[j + 1], 1);
  
        for(let i = 0; i < j; i++)
        {
            let target = A[i] * A[j];
  
            // If target exists in the map
            if (map.has(target))
                ans += map.get(target);
        }
    }
  
    // Return the final count
    return ans;
}
  
// Driver code
    let N = 5;
    let A = [ 2, 3, 4, 6, 12 ];
  
    document.write(countTriplets(A, N));
 
// This code is contributed by souravghosh0416.
</script>

Output

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

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