Count triplets with specific property

Last Updated : 29 Nov, 2023

Given an array arr[] consisting of integer elements, find the number of triplets having distinct indices i, j, k where i < j < k, such that arr[i]*arr[j] = arr[j] + arr[k].

Examples:

Input: arr[] = {4, 4, 3, 12, 24, 36}
Output: 4
Explanation: There are 4 such pairs (4, 4, 12), (3, 12, 24), (3, 12, 24), (4, 12, 36) such that arr[i]*arr[j]=arr[j]+arr[k]

Input: arr[] = {3, 16, 32, 2, 8, 12, 14, 28}
Output: 2
Explanation: There are 2 such pairs (3, 16, 32), (3, 14, 28) such that arr[i]*arr[j] = arr[j] + arr[k]

Naive Approach: To solve the problem follow the below steps:

Iterate array arr[] from left to right such that for i^th element loop is iterated from left to right for each jth element, including iteration for the kth element to form a triplet, holding the condition i<j<k.
If the condition arr[i]*arr[j] = arr[j]+arr[k] satisfies then the counter is incremented by 1. The final count is the number of triplets and then print the final count as output.

Below is the implementation of the above approach:

C++

// C++ program for the naive approach
#include <bits/stdc++.h>
using namespace std;

// Function to count triplets with the
// given condition arr[i]*arr[j]=arr[j]+arr[k]
int countTriplets(int arr[], int n)
{
    int cnt = 0;
    for (int i = 0; i < n - 2; i++) {
        for (int j = i + 1; j < n - 1; j++) {
            for (int k = j + 1; k < n; k++) {

                // If it satisfy the given
                // conditions then increase
                // counter
                if (arr[i] * arr[j] == arr[j] + arr[k]) {
                    cnt += 1;
                }
            }
        }
    }
    return cnt;
}

// Driver Code
int main()
{

    // Given array arr[]
    int arr[] = { 4, 4, 3, 12, 24, 36 };

    // Function Call
    cout << countTriplets(arr, 6) << endl;
    return 0;
}

Java

public class CountTriplets {
    // Function to count triplets with the given condition
    static int countTriplets(int[] arr, int n) {
        int cnt = 0;
        for (int i = 0; i < n - 2; i++) {
            for (int j = i + 1; j < n - 1; j++) {
                for (int k = j + 1; k < n; k++) {
                    // If it satisfies the given conditions, then increase counter
                    if (arr[i] * arr[j] == arr[j] + arr[k]) {
                        cnt++;
                    }
                }
            }
        }
        return cnt;
    }
//  Driver Code
    public static void main(String[] args) {
        
        int[] arr = {4, 4, 3, 12, 24, 36};
        
        System.out.println(countTriplets(arr, 6));
    }
}

Python3

# python program for the naive approach

def countTriplets(arr, n):
    cnt = 0
    for i in range(n - 2):
        for j in range(i + 1, n - 1):
            for k in range(j + 1, n):
                # If it satisfies the given conditions, then increase the counter
                if arr[i] * arr[j] == arr[j] + arr[k]:
                    cnt += 1
    return cnt

# Driver Code
if __name__ == "__main__":
    # Given array arr[]
    arr = [4, 4, 3, 12, 24, 36]

    # Function Call
    print(countTriplets(arr, 6))

using System;

class Program
{
    // Function to count triplets with the given condition arr[i]*arr[j]=arr[j]+arr[k]
    static int CountTriplets(int[] arr)
    {
        int n = arr.Length;
        int count = 0;
        
        for (int i = 0; i < n - 2; i++)
        {
            for (int j = i + 1; j < n - 1; j++)
            {
                for (int k = j + 1; k < n; k++)
                {
                    // If it satisfies the given conditions, increase the counter
                    if (arr[i] * arr[j] == arr[j] + arr[k])
                    {
                        count++;
                    }
                }
            }
        }
        return count;
    }

    static void Main()
    {
        // Given array arr[]
        int[] arr = { 4, 4, 3, 12, 24, 36 };

        // Function Call
        Console.WriteLine(CountTriplets(arr));
    }
}

JavaScript

// Javascript program for the naive approach

// Function to count triplets with the
// given condition arr[i]*arr[j]===arr[j]+arr[k]
function countTriplets(arr) {
    let cnt = 0;
    for (let i = 0; i < arr.length - 2; i++) {
        for (let j = i + 1; j < arr.length - 1; j++) {
            for (let k = j + 1; k < arr.length; k++) {
                // If it satisfy the given
                // conditions then increase
                // counter
                if (arr[i] * arr[j] === arr[j] + arr[k]) {
                    cnt += 1;
                }
            }
        }
    }
    return cnt;
}
// Driver Code

// Given array arr[]
const arr = [4, 4, 3, 12, 24, 36];

// Function Call
console.log(countTriplets(arr));

Output

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

Efficient Approach: To solve the problem follow the below idea:

Given condition: arr[i]*arr[j] = arr[j]+arr[k], we can move arr[j] to the other side of this equation.

arr[i] = (arr[j]+arr[k])/arr[j]
arr[i] = 1 + arr[k]/arr[j]

Follow the steps to solve the problem:

We will run nested for loops to find pair (j, k) such that j<k.
If arr[j] is not zero and arr[k] is divisible by arr[j] then add the count of arr[i] to the total count.
After the completion of the loop for the jth element, the jth element is now considered as the ith element, and the count of arr[i] is incremented by 1. Now, Move to step 1 if the loop iteration hasn't ended yet.
After the completion of loops, The final count is the number of triplets holding condition arr[i]*arr[j]=arr[i]+arr[k], and then print the final count as output.

Below is the implementation of the above approach:

C++

// C++ program for the efficient approach
#include <bits/stdc++.h>
using namespace std;

// Function to count triplets with the given
// condition arr[i]*arr[j]=arr[j]+arr[k]
int countTriplets(int arr[], int n)
{
    int cnt = 0;
    unordered_map<int, int> countARRi;
    for (int j = 0; j < n - 1; j++) {
        if (arr[j] != 0) {
            for (int k = j + 1; k < n; k++) {

                // If it satisfy that arr[k]
                // is divisible by arr[j] then
                // increment counter by
                // countArri[1+arr[k]/arr[j]];
                // here countArri is a map which
                // stores count of arr[i]
                // computed previously.
                if (arr[k] % arr[j] == 0) {
                    cnt += countARRi[1 + arr[k] / arr[j]];
                }
            }
        }
        countARRi[arr[j]]++;
    }
    return cnt;
}

// Driver Code
int main()
{

    // Given array arr[]
    int arr[] = { 3, 16, 32, 2, 8, 12, 14, 28 };

    // Function Call
    cout << countTriplets(arr, 8) << endl;
    return 0;
}

Java

import java.util.HashMap;
import java.util.Map;

public class CountTriplets {

    // Function to count triplets with the given condition
    static int countTriplets(int[] arr, int n) {
        int cnt = 0;
        Map<Integer, Integer> countARRi = new HashMap<>();

        for (int j = 0; j < n - 1; j++) {
            if (arr[j] != 0) {
                for (int k = j + 1; k < n; k++) {
                    // If it satisfies that arr[k] is divisible by arr[j]
                    // then increment the counter by countARRi.get(1 + arr[k] / arr[j]);
                    if (arr[k] % arr[j] == 0) {
                        cnt += countARRi.getOrDefault(1 + arr[k] / arr[j], 0);
                    }
                }
            }
            countARRi.put(arr[j], countARRi.getOrDefault(arr[j], 0) + 1);
        }
        return cnt;
    }

    // Driver code
    public static void main(String[] args) {
        // Given array arr[]
        int[] arr = {3, 16, 32, 2, 8, 12, 14, 28};
        
        // Function Call
        System.out.println(countTriplets(arr, 8));
    }
}

Python3

import collections

# Function to count triplets with the given
# condition arr[i]*arr[j]=arr[j]+arr[k]
def countTriplets(arr, n):
    cnt = 0
    countARRi = collections.defaultdict(int)
    for j in range(n - 1):
        if arr[j] != 0:
            for k in range(j + 1, n):
              
                  # If it satisfy that arr[k]
                # is divisible by arr[j] then
                # increment counter by
                # countArri[1+arr[k]/arr[j]];
                # here countArri is a map which
                # stores count of arr[i]
                # computed previously.
                if arr[k] % arr[j] == 0:
                    cnt += countARRi[1 + arr[k] // arr[j]]
        countARRi[arr[j]] += 1
    return cnt

# Driver Code
arr = [3, 16, 32, 2, 8, 12, 14, 28]
print(countTriplets(arr, 8))

using System;
using System.Collections.Generic;

class GFG
{
    // Function to count triplets with 
  // given condition arr[i]*arr[j]=arr[j]+arr[k]
    static int CountTriplets(int[] arr, int n)
    {
        int cnt = 0;
        Dictionary<int, int> countARRi = new Dictionary<int, int>();
        for (int j = 0; j < n - 1; j++)
        {
            if (arr[j] != 0)
            {
                for (int k = j + 1; k < n; k++)
                {
                    // If it satisfies that arr[k] is divisible by arr[j].
                    if (arr[k] % arr[j] == 0)
                    {
                        if (countARRi.ContainsKey(1 + arr[k] / arr[j]))
                        {
                            cnt += countARRi[1 + arr[k] / arr[j]];
                        }
                    }
                }
            }
            if (countARRi.ContainsKey(arr[j]))
            {
                countARRi[arr[j]]++;
            }
            else
            {
                countARRi.Add(arr[j], 1);
            }
        }
        return cnt;
    }
    static void Main(string[] args)
    {
        // Given array arr[]
        int[] arr = { 3, 16, 32, 2, 8, 12, 14, 28 };
        // The  Function Call
        Console.WriteLine(CountTriplets(arr, 8));
    }
}

JavaScript

// Function to count triplets with the given condition
function countTriplets(arr) {
    let cnt = 0;
    const countARRi = new Map();

    for (let j = 0; j < arr.length - 1; j++) {
        if (arr[j] !== 0) {
            for (let k = j + 1; k < arr.length; k++) {
                // If it satisfies that arr[k] is divisible by arr[j]
                if (arr[k] % arr[j] === 0) {
                    // Increment the counter by countARRi[1 + arr[k] / arr[j]]
                    // Here countARRi is a map that stores the count of arr[i] computed previously.
                    cnt += countARRi.get(1 + arr[k] / arr[j]) || 0;
                }
            }
        }
        countARRi.set(arr[j], (countARRi.get(arr[j]) || 0) + 1);
    }
    return cnt;
}

// Driver Code
const arr = [3, 16, 32, 2, 8, 12, 14, 28];

// Function Call
console.log(countTriplets(arr));

Output

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

Analysis of Algorithms

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