大小为3的GP(几何级数)子序列的数量

给定n个元素和一个比率r，求出长度为3的GP子序列的数量。一个子序列被视为长度为3且比率为r的GP。

例子：

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

幼稚的方法是使用三个嵌套的for循环，并检查每个长度为3的子序列，并保留这些子序列的计数。复杂度为O(n ³ )。

一种有效的方法是解决进度的固定中间元素的问题。这意味着如果将元素a [i]固定为中间，则它必须是r的倍数，并且必须存在a [i] / r和a [i] * r。我们计算a [i] / r和a [i] * r的出现次数，然后将其相乘。为此，我们可以使用哈希的概念，将所有可能元素的计数存储在两个哈希图中，一个表示左侧的元素数量，另一个表示右侧的元素数量。

下面是上述方法的实现

C++

// C++ program to count GP subsequences of size 3.
#include 
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 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 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 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

输出：

6

时间复杂度： O(n)

上述解决方案不适用于r为1的情况：例如，对于input = {1,1,1,1,1}，长度为3的GP子序列可能有10个，可以使用⁵ C来计算₃ 。应该在r = 1的所有情况下实施这样的过程。下面是修改后的代码来处理此问题。

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. 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 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 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. 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 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

输出：

6