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Number of GP (Geometric Progression) subsequences of size 3
  • Difficulty Level : Hard
  • Last Updated : 20 Nov, 2020

Given n elements and a ratio r, find the number of G.P. subsequences with length 3. A subsequence is considered GP with length 3 with ration r.

Examples:

Input : arr[] = {1, 1, 2, 2, 4}
            r = 2
Output : 4 
Explanation: Any of the two 1s can be chosen 
as the first element, the second element can 
be any of the two 2s, and the third element 
of the subsequence must be equal to 4.
             
Input : arr[] = {1, 1, 2, 2, 4}
            r = 3
Output : 0

A naive approach is to use three nested for loops and check for every subsequence with length 3 and keep a count of the subsequences. The complexity is O(n3).

An efficient approach is to solve the problem for the fixed middle element of progression. This means that if we fix element a[i] as middle, then it must be multiple of r, and a[i]/r and a[i]*r must be present. We count the number of occurrences of a[i]/r and a[i]*r and then multiply the counts. To do this, we can use the concept of hashing where we store the count of all possible elements in two hash maps, one indicating the number of elements on the left and the other indicating the number of elements to the right.

Below is the implementation of the above approach



C++




// C++ program to count GP subsequences of size 3.
#include <bits/stdc++.h>
using namespace std;
 
// Returns count of G.P. subseqeunces
// with length 3 and common ratio r
long long subsequences(int a[], int n, int r)
{
    // hashing to maintain left and right array
    // elements to the main count
    unordered_map<int, int> left, right;
 
    // stores the answer
    long long ans = 0;
 
    // traverse through the elements
    for (int i = 0; i < n; i++)
        right[a[i]]++; // keep the count in the hash
 
    // traverse through all elements
    // and find out the number of elements as k1*k2
    for (int i = 0; i < n; i++) {
 
        // keep the count of left and right elements
        // left is a[i]/r and right a[i]*r
        long long c1 = 0, c2 = 0;
 
        // if the current element is divisible by k,
        // count elements in left hash.
        if (a[i] % r == 0)
            c1 = left[a[i] / r];
 
        // decrease the count in right hash
        right[a[i]]--;
 
        // number of right elements
        c2 = right[a[i] * r];
 
        // calculate the answer
        ans += c1 * c2;
 
        left[a[i]]++; // left count of a[i]
    }
 
    // returns answer
    return ans;
}
 
// driver program
int main()
{
    int a[] = { 1, 2, 6, 2, 3, 6, 9, 18, 3, 9 };
    int n = sizeof(a) / sizeof(a[0]);
    int r = 3;
    cout << subsequences(a, n, r);
    return 0;
}

Java




// Java program to count GP subsequences
// of size 3.
import java.util.*;
import java.lang.*;
 
class GFG{
 
// Returns count of G.P. subseqeunces
// with length 3 and common ratio r
static long subsequences(int a[], int n, int r)
{
     
    // Hashing to maintain left and right array
    // elements to the main count
    Map<Integer, Integer> left = new HashMap<>(),
                         right = new HashMap<>();
 
    // Stores the answer
    long ans = 0;
 
    // Traverse through the elements
    for(int i = 0; i < n; i++)
     
        // Keep the count in the hash
        right.put(a[i],
        right.getOrDefault(a[i], 0) + 1);
 
    // Traverse through all elements
    // and find out the number of
    // elements as k1*k2
    for(int i = 0; i < n; i++)
    {
         
        // Keep the count of left and right
        // elements left is a[i]/r and
        // right a[i]*r
        long c1 = 0, c2 = 0;
 
        // If the current element is divisible
        // by k, count elements in left hash.
        if (a[i] % r == 0)
            c1 = left.getOrDefault(a[i] / r, 0);
 
        // Decrease the count in right hash
        right.put(a[i],
        right.getOrDefault(a[i], 0) - 1);
 
        // Number of right elements
        c2 = right.getOrDefault(a[i] * r, 0);
 
        // Calculate the answer
        ans += c1 * c2;
         
        // left count of a[i]
        left.put(a[i],
        left.getOrDefault(a[i], 0) + 1);
    }
     
    // Returns answer
    return ans;
}
 
// Driver Code
public static void main (String[] args)
{
    int a[] = { 1, 2, 6, 2, 3,
                6, 9, 18, 3, 9 };
    int n = a.length;
    int r = 3;
     
    System.out.println(subsequences(a, n, r));
}
}
 
// This code is contributed by offbeat

Python3




# Python3 program to count GP subsequences
# of size 3.
from collections import defaultdict
 
# Returns count of G.P. subseqeunces
# with length 3 and common ratio r
def subsequences(a, n, r):
 
    # hashing to maintain left and right
    # array elements to the main count
    left = defaultdict(lambda:0)
    right = defaultdict(lambda:0)
 
    # stores the answer
    ans = 0
 
    # traverse through the elements
    for i in range(0, n):
        right[a[i]] += 1 # keep the count in the hash
 
    # traverse through all elements and
    # find out the number of elements as k1*k2
    for i in range(0, n):
 
        # keep the count of left and right elements
        # left is a[i]/r and right a[i]*r
        c1, c2 = 0, 0
 
        # if the current element is divisible
        # by k, count elements in left hash.
        if a[i] % r == 0:
            c1 = left[a[i] // r]
 
        # decrease the count in right hash
        right[a[i]] -= 1
 
        # number of right elements
        c2 = right[a[i] * r]
 
        # calculate the answer
        ans += c1 * c2
 
        left[a[i]] += 1 # left count of a[i]
 
    return ans
 
# Driver Code
if __name__ == "__main__":
 
    a = [1, 2, 6, 2, 3, 6, 9, 18, 3, 9]
    n = len(a)
    r = 3
    print(subsequences(a, n, r))
 
# This code is contributed by
# Rituraj Jain

C#




// C# program to count GP
// subsequences of size 3.
using System;
using System.Collections.Generic;
class GFG{
 
// Returns count of G.P. subseqeunces
// with length 3 and common ratio r
static long subsequences(int []a,
                         int n, int r)
{    
  // Hashing to maintain left and
  // right array elements to the
  // main count
  Dictionary<int,
             int> left =
             new Dictionary<int,
                            int>(),
  right = new Dictionary<int,
                         int>();
 
  // Stores the answer
  long ans = -1;
 
  // Traverse through the
  // elements
  for(int i = 0; i < n; i++)
 
    // Keep the count in the hash
    if (right.ContainsKey(a[i]))
      right[a[i]] = right[a[i]] + 1;
  else
    right.Add(a[i], 1);
 
  // Traverse through all elements
  // and find out the number of
  // elements as k1*k2
  for(int i = 0; i < n; i++)
  {
    // Keep the count of left and
    // right elements left is a[i]/r
    // and right a[i]*r
    long c1 = 0, c2 = 0;
 
    // If the current element is
    // divisible by k, count elements
    // in left hash.
    if (a[i] % r == 0)
      if (left.ContainsKey(a[i] / r))
        c1 =  right[a[i] / r];
    else
 
      c1 =  0;
 
    // Decrease the count in right
    // hash
    if (right.ContainsKey(a[i]))
      right[a[i]] = right[a[i]];
    else
      right.Add(a[i], -1);
 
    // Number of right elements
    if (right.ContainsKey(a[i] * r))
      c2 = right[a[i] * r];
    else
      c2 = 0;
 
    // Calculate the answer
    ans += (c1 * c2);
 
    // left count of a[i]
    if (left.ContainsKey(a[i]))
      left[a[i]] = 0;
    else
      left.Add(a[i], 1);
  }
 
  // Returns answer
  return ans - 1;
}
 
// Driver Code
public static void Main(String[] args)
{
  int []a = {1, 2, 6, 2, 3,
             6, 9, 18, 3, 9};
  int n = a.Length;
  int r = 3;
  Console.WriteLine(subsequences(a,
                                 n, r));
}
}
 
// This code is contributed by Princi Singh
Output: 
6













Time complexity: O(n)

The above solution does not handle the case when r is 1 : For example, for input = {1,1,1,1,1}, there are 10 possible for G.P. subsequences of length 3, which can be calculated by using 5C3. Such a procedure should be implemented for all cases where r = 1. Below is the modified code to handle this.

C++




// C++ program to count GP subsequences of size 3.
#include <bits/stdc++.h>
using namespace std;
 
// to calculate nCr
// DP approach
int binomialCoeff(int n, int k) {
  int C[k + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1; // nC0 is 1
  for (int i = 1; i <= n; i++) {
 
    // Compute next row of pascal triangle using
    // the previous row
    for (int j = min(i, k); j > 0; j--)
      C[j] = C[j] + C[j - 1];
  }
  return C[k];
}
 
// Returns count of G.P. subseqeunces
// with length 3 and common ratio r
long long subsequences(int a[], int n, int r)
{
    // hashing to maintain left and right array
    // elements to the main count
    unordered_map<int, int> left, right;
 
    // stores the answer
    long long ans = 0;
 
    // traverse through the elements
    for (int i = 0; i < n; i++)
        right[a[i]]++; // keep the count in the hash
 
    // IF RATIO IS ONE
    if (r == 1){
 
        // traverse the count in hash
        for (auto i : right) {
 
             // calculating nC3, where 'n' is
             // the number of times each number is
             // repeated in the input
             ans += binomialCoeff(i.second, 3);
        }
 
        return ans;
    }
 
    // traverse through all elements
    // and find out the number of elements as k1*k2
    for (int i = 0; i < n; i++) {
 
        // keep the count of left and right elements
        // left is a[i]/r and right a[i]*r
        long long c1 = 0, c2 = 0;
 
        // if the current element is divisible by k,
        // count elements in left hash.
        if (a[i] % r == 0)
            c1 = left[a[i] / r];
 
        // decrease the count in right hash
        right[a[i]]--;
 
        // number of right elements
        c2 = right[a[i] * r];
 
        // calculate the answer
        ans += c1 * c2;
 
        left[a[i]]++; // left count of a[i]
    }
 
    // returns answer
    return ans;
}
 
// driver program
int main()
{
    int a[] = { 1, 2, 6, 2, 3, 6, 9, 18, 3, 9 };
    int n = sizeof(a) / sizeof(a[0]);
    int r = 3;
    cout << subsequences(a, n, r);
    return 0;
}

Java




// Java program to count GP
// subsequences of size 3.
import java.util.*;
 
class GFG{
 
// To calculate nCr
// DP approach
static int binomialCoeff(int n, int k)
{
    int []C = new int[k + 1];
     
    C[0] = 1; // nC0 is 1
    for(int i = 1; i <= n; i++)
    {
         
        // Compute next row of pascal
        // triangle using the previous row
        for(int j = Math.min(i, k); j > 0; j--)
            C[j] = C[j] + C[j - 1];
    }
    return C[k];
}
 
// Returns count of G.P. subseqeunces
// with length 3 and common ratio r
static long subsequences(int a[], int n, int r)
{
     
    // Hashing to maintain left and right array
    // elements to the main count
    HashMap<Integer, Integer> left = new HashMap<>();
    HashMap<Integer, Integer> right = new HashMap<>();
 
    // Stores the answer
    long ans = 0;
     
    // Traverse through the elements
    for(int i = 0; i < n; i++)
        if (right.containsKey(a[i]))
        {
            right.put(a[i], right.get(a[i]) + 1);
        }
        else
        {
            right.put(a[i], 1);
        }
 
    // IF RATIO IS ONE
    if (r == 1)
    {
         
        // Traverse the count in hash
        for(Map.Entry<Integer, Integer> i : right.entrySet())
        {
             
            // Calculating nC3, where 'n' is
            // the number of times each number is
            // repeated in the input
            ans += binomialCoeff(i.getValue(), 3);
        }
        return ans;
    }
 
    // Traverse through all elements and
    // find out the number of elements as k1*k2
    for(int i = 0; i < n; i++)
    {
         
        // Keep the count of left and right
        // elements left is a[i]/r and
        // right a[i]*r
        long c1 = 0, c2 = 0;
 
        // If the current element is divisible
        // by k, count elements in left hash.
        if (a[i] % r == 0)
            if (left.containsKey(a[i] / r))
                c1 = left.get(a[i] / r);
 
        // Decrease the count in right hash
        if (right.containsKey(a[i]))
        {
            right.put(a[i], right.get(a[i]) - 1);
        }
        else
        {
            right.put(a[i], -1);
        }
         
        // Number of right elements
        if (right.containsKey(a[i] * r))
            c2 = right.get(a[i] * r);
 
        // Calculate the answer
        ans += c1 * c2;
 
        if (left.containsKey(a[i]))
        {
            left.put(a[i], left.get(a[i]) + 1);
        }
        else
        {
            left.put(a[i], 1);
        }// left count of a[i]
    }
 
    // Returns answer
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int a[] = { 1, 2, 6, 2, 3,
                6, 9, 18, 3, 9 };
    int n = a.length;
    int r = 3;
     
    System.out.print(subsequences(a, n, r));
}
}
 
// This code is contributed by Amit Katiyar

Python3




# Python3 program to count
# GP subsequences of size 3.
from collections import defaultdict
 
# To calculate nCr
# DP approach
def binomialCoeff(n, k):
   
  C = [0] * (k + 1)
  
  # nC0 is 1
  C[0] = 1 
   
  for  i in range (1, n + 1):
 
    # Compute next row of pascal
    # triangle using the previous row
    for j in range (min(i, k), -1, -1):
      C[j] = C[j] + C[j - 1]
  return C[k]
 
# Returns count of G.P. subseqeunces
# with length 3 and common ratio r
def subsequences(a, n, r):
 
    # hashing to maintain left
    # and right array elements
    # to the main count
    left = defaultdict (int)
    right = defaultdict (int)
 
    # Stores the answer
    ans = 0
 
    # Traverse through
    # the elements
    for i in range (n):
       
        # Keep the count
        # in the hash
        right[a[i]] += 1
 
    # IF RATIO IS ONE
    if (r == 1):
 
        # Traverse the count
        # in hash
        for  i in right:
 
             # calculating nC3, where 'n' is
             # the number of times each number is
             # repeated in the input
             ans += binomialCoeff(right[i], 3)
       
        return ans
 
    # traverse through all elements
    # and find out the number
    # of elements as k1*k2
    for i in range (n):
 
        # Keep the count of left
        # and right elements left
        # is a[i]/r and right a[i]*r
        c1 = 0
        c2 = 0;
 
        # if the current element
        # is divisible by k, count
        # elements in left hash.
        if (a[i] % r == 0):
            c1 = left[a[i] // r]
 
        # Decrease the count
        # in right hash
        right[a[i]] -= 1
 
        # Number of right elements
        c2 = right[a[i] * r]
 
        # Calculate the answer
        ans += c1 * c2
 
        # left count of a[i]
        left[a[i]] += 1
    
    # returns answer
    return ans
 
# Driver code
if __name__ == "__main__":
   
    a = [1, 2, 6, 2, 3,
         6, 9, 18, 3, 9]
    n = len(a)
    r = 3
    print ( subsequences(a, n, r))
    
# This code is contributed by Chitranayal

C#




// C# program to count GP
// subsequences of size 3.
using System;
using System.Collections.Generic;
class GFG{
 
// To calculate nCr
// DP approach
static int binomialCoeff(int n,
                         int k)
{
  int []C = new int[k + 1];
 
  // nC0 is 1
  C[0] = 1;
  for(int i = 1; i <= n; i++)
  {
    // Compute next row of pascal
    // triangle using the previous
    // row
    for(int j = Math.Min(i, k);
            j > 0; j--)
      C[j] = C[j] + C[j - 1];
  }
  return C[k];
}
 
// Returns count of G.P. subseqeunces
// with length 3 and common ratio r
static long subsequences(int []a,
                         int n, int r)
{   
  // Hashing to maintain left and
  // right array elements to the
  // main count
  Dictionary<int,
             int> left =
             new Dictionary<int,
                            int>();
  Dictionary<int,
             int> right =
             new Dictionary<int,
                            int>();       
 
  // Stores the answer
  long ans = 0;
 
  // Traverse through the elements
  for(int i = 0; i < n; i++)
    if (right.ContainsKey(a[i]))
    {
      right[a[i]]++;
    }
  else
  {
    right.Add(a[i], 1);
  }
 
  // IF RATIO IS ONE
  if (r == 1)
  {
    // Traverse the count in hash
    foreach(KeyValuePair<int,
                         int> i in right)
    {
 
      // Calculating nC3, where 'n' is
      // the number of times each number is
      // repeated in the input
      ans += binomialCoeff(i.Value, 3);
    }
    return ans;
  }
 
  // Traverse through all elements
  // and find out the number of
  // elements as k1*k2
  for(int i = 0; i < n; i++)
  {
    // Keep the count of left and
    // right elements left is a[i]/r
    // and right a[i]*r
    long c1 = 0, c2 = 0;
 
    // If the current element is
    // divisible by k, count elements
    // in left hash.
    if (a[i] % r == 0)
      if (left.ContainsKey(a[i] / r))
        c1 = left[a[i] / r];
 
    // Decrease the count in right
    // hash
    if (right.ContainsKey(a[i]))
    {
      right[a[i]]--;       
    }
    else
    {
      right.Add(a[i], -1);
    }
 
    // Number of right elements
    if (right.ContainsKey(a[i] * r))
      c2 = right[a[i] * r];
 
    // Calculate the answer
    ans += c1 * c2;
 
    if (left.ContainsKey(a[i]))
    {
      left[a[i]]++;
    }
    else
    {
      left.Add(a[i], 1);
    }// left count of a[i]
  }
 
  // Returns answer
  return ans;
}
 
// Driver code
public static void Main(String[] args)
{
  int []a = {1, 2, 6, 2, 3,
             6, 9, 18, 3, 9};
  int n = a.GetLength(0);
  int r = 3;
  Console.Write(subsequences(a, n, r));
}
}
 
// This code is contributed by shikhasingrajput
Output: 
6













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