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 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 <bits/stdc++.h>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<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 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<Integer, Integer> mp = new HashMap<Integer, Integer>();
// 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 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 |
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<int, int> mp = new Dictionary<int, int>();
// 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
<script>// JavaScript implementation of the approach // Function to return the count of quadruples
function countQuadruples(a, n)
{
// Hash table to count the
// number of occurrences
let mp = new Map();
// Traverse and increment the count
for (let i = 0; i < n; i++)
if (mp.has(a[i]))
{
mp.set(a[i], mp.get(a[i]) + 1);
}
else
{
mp.set(a[i], 1);
}
let count = 0;
// Run two nested loop for second
// and third element
for (let j = 0; j < n; j++)
{
for (let k = 0; k < n; k++)
{
// If they are same
if (j == k)
continue;
// Initially decrease the count
mp.set(a[j], mp.get(a[j]) - 1);
mp.set(a[k], mp.get(a[k]) - 1);
// Find the first element using
// common difference
let first = a[j] - (a[k] - a[j]);
// Find the fourth element using GP
// y^2 = x * z property
let 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.has(first) && mp.has(fourth))
count +=
mp.get(first) * mp.get(fourth);
}
// Same elements
else if (mp.has(first) && mp.has(fourth))
count +=
mp.get(first) * (mp.get(fourth) - 1);
}
// Later increase the value for
// future calculations
if (mp.has(a[j]))
{
mp.set(a[j], mp.get(a[j]) + 1);
}
else
{
mp.set(a[j], 1);
}
if (mp.has(a[k]))
{
mp.set(a[k], mp.get(a[k]) + 1);
}
else
{
mp.set(a[k], 1);
}
}
}
return count;
}
// Driver code
let a = [ 2, 6, 4, 9, 2 ];
let n = a.length;
document.write(countQuadruples(a, n));
</script> |
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.