Dados n elementos y una razón r, encuentre el número de subsecuencias GP con longitud 3. Una subsecuencia se considera GP con longitud 3 con razón r.
Ejemplos:
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
Un enfoque ingenuo es usar tres bucles for anidados y verificar cada subsecuencia con una longitud de 3 y llevar un conteo de las subsecuencias. La complejidad es O(n 3 ).
Un enfoque eficiente es resolver el problema para el elemento medio fijo de progresión. Esto significa que si fijamos el elemento a[i] como medio, entonces debe ser múltiplo de r, y a[i]/r y a[i]*r deben estar presentes. Contamos el número de ocurrencias de a[i]/r y a[i]*r y luego multiplicamos las cuentas. Para hacer esto, podemos usar el concepto de hash donde almacenamos el conteo de todos los elementos posibles en dos mapas hash, uno que indica el número de elementos a la izquierda y el otro que indica el número de elementos a la derecha.
A continuación se muestra la implementación del enfoque anterior.
C++
// C++ program to count GP subsequences of size 3.
#include <bits/stdc++.h>
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<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. 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<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. 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<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
Javascript
<script>
// JavaScript program to count GP subsequences of size 3.
// 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 = new Map(), right = new Map();
// stores the answer
let ans = 0;
// traverse through the elements
for (let i = 0; i < n; i++){
// keep the count in the hash
if(right.has(a[i])){
right.set(a[i],right.get(a[i])+1);
}
else right.set(a[i],1);
}
// traverse through all elements
// and find out the number of elements as k1*k2
for (let 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)
c1 = left.has(a[i] / r)?left.get(a[i] / r):0;
// decrease the count in right hash
right.set(a[i],right.get(a[i])-1);
// number of right elements
c2 = right.has(a[i] * r)?right.get(a[i] * r):0;
// calculate the answer
ans += c1 * c2;
// left count of a[i]
if(left.has(a[i])){
left.set(a[i],left.get(a[i])+1);
}
else left.set(a[i],1);
}
// returns answer
return ans;
}
// driver program
let a = [ 1, 2, 6, 2, 3, 6, 9, 18, 3, 9 ];
let n = a.length;
let r = 3;
document.write(subsequences(a, n, r));
// This code is contributed by shinjanpatra
</script>
Producción:
6
Complejidad de tiempo: O(n), donde n representa el tamaño de la array dada.
Espacio auxiliar: O(n), donde n representa el tamaño de la array dada.
La solución anterior no maneja el caso cuando r es 1: por ejemplo, para input = {1,1,1,1,1}, hay 10 subsecuencias GP posibles de longitud 3, que se pueden calcular usando 5 C 3 . Dicho procedimiento debe implementarse para todos los casos en los que r = 1. A continuación se muestra el código modificado para manejar esto.
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. 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<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. 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<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. 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<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
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
Producción:
6
Complejidad de tiempo: O(n), donde n representa el tamaño de la array dada.
Espacio auxiliar: O(n), donde n representa el tamaño de la array dada.