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: 2
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 number. 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 this two index j and k will be count of first number * count of fourth number with the second and third element being fixed.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the count of quadruples
int countQuadruples(int a[], int n)
{
  
    // Hash table to count the number of occurrences
    unordered_map<int, int> 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 code
int main()
{
    int a[] = { 2, 6, 4, 9, 2 };
    int n = sizeof(a) / sizeof(a[0]);
  
    cout << countQuadruples(a, n);
  
    return 0;
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach 
  
# Function to return the count of quadruples 
def 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 code 
if __name__ == "__main__" :
  
    a = [ 2, 6, 4, 9, 2 ]; 
    n = len(a) ; 
  
    print(countQuadruples(a, n)); 
  
# This code is contributed by Ryuga

chevron_right


Output:

2

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



My Personal Notes arrow_drop_up

Striver(underscore)79 at Codechef and codeforces D

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : AnkitRai01