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# Number of quadruples where the first three terms are in AP and last three terms are in GP

Given an array arr[] of N integers. The task is to find the number of index quadruples (i, j, k, l) such that a[i], a[j] and a[k] are in AP and a[j], a[k] and a[l] are in GP. All the quadruples have to be distinct.
Examples:

Input: arr[] = {2, 6, 4, 9, 2}
Output:
Indexes of elements in the quadruples are (0, 2, 1, 3) and (4, 2, 1, 3) and corresponding quadruples are (2, 4, 6, 9) and (2, 4, 6, 9)
Input: arr[] = {1, 1, 1, 1}
Output: 24

A naive approach is to solve the above problem using four nested loops. Check for the first three elements if they are in AP or not and then check whether the last three elements are in GP or not. If both the conditions satisfy, then they increase the count by 1.
Time Complexity: O(n4
An efficient approach is to use combinatorics to solve the above problem. Initially keep a count of the number of occurrences of every array element. Run two nested loops, and consider both elements to be the second and third numbers. Hence the first element will be a[j] – (a[k] – a[j]) and the fourth element will be a[k] * a[k] / a[j] if it is an integer value. Hence the number of quadruples using these two indexes j and k will be a count of the first number * count of fourth number with the second and third element being fixed.
Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach#include using namespace std; // Function to return the count of quadruplesint countQuadruples(int a[], int n){     // Hash table to count the number of occurrences    unordered_map mpp;     // Traverse and increment the count    for (int i = 0; i < n; i++)        mpp[a[i]]++;     int count = 0;     // Run two nested loop for second and third element    for (int j = 0; j < n; j++) {        for (int k = 0; k < n; k++) {             // If they are same            if (j == k)                continue;             // Initially decrease the count            mpp[a[j]]--;            mpp[a[k]]--;             // Find the first element using common difference            int first = a[j] - (a[k] - a[j]);             // Find the fourth element using GP            // y^2 = x * z property            int fourth = (a[k] * a[k]) / a[j];             // If it is an integer            if ((a[k] * a[k]) % a[j] == 0) {                 // If not equal                if (a[j] != a[k])                    count += mpp[first] * mpp[fourth];                 // Same elements                else                    count += mpp[first] * (mpp[fourth] - 1);            }             // Later increase the value for            // future calculations            mpp[a[j]]++;            mpp[a[k]]++;        }    }    return count;} // Driver codeint main(){    int a[] = { 2, 6, 4, 9, 2 };    int n = sizeof(a) / sizeof(a[0]);     cout << countQuadruples(a, n);     return 0;}

## Java

 // Java implementation of the approachimport java.util.*; class GFG{     // Function to return the count of quadruples    static int countQuadruples(int a[], int n)    {         // Hash table to count the number of occurrences        HashMap mp = new HashMap();         // Traverse and increment the count        for (int i = 0; i < n; i++)            if (mp.containsKey(a[i]))            {                mp.put(a[i], mp.get(a[i]) + 1);            }            else            {                mp.put(a[i], 1);            }         int count = 0;         // Run two nested loop for second and third element        for (int j = 0; j < n; j++)        {            for (int k = 0; k < n; k++)            {                 // If they are same                if (j == k)                    continue;                 // Initially decrease the count                mp.put(a[j], mp.get(a[j]) - 1);                mp.put(a[k], mp.get(a[k]) - 1);                 // Find the first element using common difference                int first = a[j] - (a[k] - a[j]);                 // Find the fourth element using GP                // y^2 = x * z property                int fourth = (a[k] * a[k]) / a[j];                 // If it is an integer                if ((a[k] * a[k]) % a[j] == 0)                {                     // If not equal                    if (a[j] != a[k])                    {                        if (mp.containsKey(first) && mp.containsKey(fourth))                            count += mp.get(first) * mp.get(fourth);                    }                                         // Same elements                    else if (mp.containsKey(first) && mp.containsKey(fourth))                        count += mp.get(first) * (mp.get(fourth) - 1);                }                 // Later increase the value for                // future calculations                if (mp.containsKey(a[j]))                {                    mp.put(a[j], mp.get(a[j]) + 1);                }                else                {                    mp.put(a[j], 1);                }                if (mp.containsKey(a[k]))                {                    mp.put(a[k], mp.get(a[k]) + 1);                }                else                {                    mp.put(a[k], 1);                }            }        }        return count;    }     // Driver code    public static void main(String[] args)    {        int a[] = { 2, 6, 4, 9, 2 };        int n = a.length;         System.out.print(countQuadruples(a, n));    }} // This code is contributed by 29AjayKumar

## Python3

 # Python3 implementation of the approach # Function to return the count of quadruplesdef countQuadruples(a, n) :     # Hash table to count the number    # of occurrences    mpp = dict.fromkeys(a, 0);     # Traverse and increment the count    for i in range(n) :        mpp[a[i]] += 1;     count = 0;     # Run two nested loop for second    # and third element    for j in range(n) :        for k in range(n) :             # If they are same            if (j == k) :                continue;             # Initially decrease the count            mpp[a[j]] -= 1;            mpp[a[k]] -= 1;             # Find the first element using            # common difference            first = a[j] - (a[k] - a[j]);                         if first not in mpp :                mpp[first] = 0;                             # Find the fourth element using            # GP y^2 = x * z property            fourth = (a[k] * a[k]) // a[j];                         if fourth not in mpp :                mpp[fourth] = 0;                             # If it is an integer            if ((a[k] * a[k]) % a[j] == 0) :                 # If not equal                if (a[j] != a[k]) :                    count += mpp[first] * mpp[fourth];                 # Same elements                else :                    count += (mpp[first] *                             (mpp[fourth] - 1));                         # Later increase the value for            # future calculations            mpp[a[j]] += 1;            mpp[a[k]] += 1;                 return count; # Driver codeif __name__ == "__main__" :     a = [ 2, 6, 4, 9, 2 ];    n = len(a) ;     print(countQuadruples(a, n)); # This code is contributed by Ryuga

## C#

 // C# implementation of the approachusing System;using System.Collections.Generic; class GFG{     // Function to return the count of quadruples    static int countQuadruples(int []a, int n)    {         // Hash table to count the number of occurrences        Dictionary mp = new Dictionary();         // Traverse and increment the count        for (int i = 0; i < n; i++)            if (mp.ContainsKey(a[i]))            {                mp[a[i]] = mp[a[i]] + 1;            }            else            {                mp.Add(a[i], 1);            }         int count = 0;         // Run two nested loop for second and third element        for (int j = 0; j < n; j++)        {            for (int k = 0; k < n; k++)            {                 // If they are same                if (j == k)                    continue;                 // Initially decrease the count                mp[a[j]] = mp[a[j]] - 1;                mp[a[k]] = mp[a[k]] - 1;                 // Find the first element using common difference                int first = a[j] - (a[k] - a[j]);                 // Find the fourth element using GP                // y^2 = x * z property                int fourth = (a[k] * a[k]) / a[j];                 // If it is an integer                if ((a[k] * a[k]) % a[j] == 0)                {                     // If not equal                    if (a[j] != a[k])                    {                        if (mp.ContainsKey(first) && mp.ContainsKey(fourth))                            count += mp[first] * mp[fourth];                    }                                         // Same elements                    else if (mp.ContainsKey(first) && mp.ContainsKey(fourth))                        count += mp[first] * (mp[fourth] - 1);                }                 // Later increase the value for                // future calculations                if (mp.ContainsKey(a[j]))                {                    mp[a[j]] = mp[a[j]] + 1;                }                else                {                    mp.Add(a[j], 1);                }                if (mp.ContainsKey(a[k]))                {                    mp[a[k]] = mp[a[k]] + 1;                }                else                {                    mp.Add(a[k], 1);                }            }        }        return count;    }     // Driver code    public static void Main(String[] args)    {        int []a = { 2, 6, 4, 9, 2 };        int n = a.Length;         Console.Write(countQuadruples(a, n));    }} // This code is contributed by 29AjayKumar

## Javascript



Output:

2

Time Complexity: O(N2), as we are using nested loops to traverse N*N times. Where N is the number of elements in the array.
Auxiliary Space: O(N), as are using extra space for the map.

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