# Number of GP (Geometric Progression) subsequences of size 3

Last Updated : 24 Mar, 2023

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 using namespace std;    // Returns count of G.P. subsequences // 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 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. subsequences  // 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 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. subsequences  # 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. subsequences  // 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 left =               new Dictionary(),   right = new Dictionary();       // 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

## Javascript



Output:

6

Time Complexity: O(n), where n represents the size of the given array.

Auxiliary Space: O(n), where n represents the size of the given array.

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 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. subsequences // 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 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. subsequences // 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 left = new HashMap<>();     HashMap 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 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. subsequences # 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. subsequences // 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 left =               new Dictionary();   Dictionary right =               new Dictionary();              // 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 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

## Javascript

 // JavaScript program to count GP  // subsequences of size 3.       // To calculate nCr // DP approach function binomialCoeff(n, k)  {     let C = new Array(k + 1);            C[0] = 1; // nC0 is 1     for(var i = 1; i <= n; i++)     {                    // Compute next row of pascal          // triangle using the previous row         for(var j = Math.min(i, k); j > 0; j--)             C[j] = C[j] + C[j - 1];     }     return C[k]; }    // Returns count of G.P. subsequences // with length 3 and common ratio r function subsequences(a, n, r) {            // Hashing to maintain left and right array     // elements to the main count     let left = {};     let right = {};        // Stores the answer     let ans = 0;            // Traverse through the elements     for(var i = 0; i < n; i++)         if (right.hasOwnProperty(a[i]))         {             right[a[i]] = right[a[i]] + 1;         }         else         {             right[a[i]] = 1;         }        // IF RATIO IS ONE     if (r == 1)     {                    // Traverse the count in hash         for(var i of 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(var i = 0; i < n; i++)      {                    // Keep the count of left and right         // elements left is a[i]/r and          // right a[i]*r         let c1 = 0, c2 = 0;            // If the current element is divisible         // by k, count elements in left hash.         if (a[i] % r == 0)             if (left.hasOwnProperty(Math.floor(a[i] / r)))                 c1 = left[(Math.floor(a[i] / r))];            // Decrease the count in right hash         if (right.hasOwnProperty(a[i]))         {             right[a[i]] = right[a[i]] - 1;         }         else         {             right[a[i]] = -1;         }                    // Number of right elements          if (right.hasOwnProperty(a[i] * r))             c2 = right[a[i] * r];            // Calculate the answer         ans += c1 * c2;            if (left.hasOwnProperty(a[i]))         {             left[a[i]] = left[a[i]] + 1;         }         else         {             left[a[i]] = 1;         }// left count of a[i]     }        // Returns answer     return ans; }    // Driver code let a = [ 1, 2, 6, 2, 3, 6, 9, 18, 3, 9 ]; let n = a.length; let r = 3;        console.log(subsequences(a, n, r));       // This code is contributed by phasing17

Output:

6

Time Complexity: O(n), where n represents the size of the given array.

Auxiliary Space: O(n), where n represents the size of the given array.

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