Dado un conjunto de números, encuentre la L ongitud de la Progresión G eometrix Más Larga ( LLGP ) en él. La razón común de GP debe ser un número entero. Ejemplos:
set[] = {5, 7, 10, 15, 20, 29}
output = 3
The longest geometric progression is {5, 10, 20}
set[] = {3, 9, 27, 81}
output = 4
Este problema es similar al problema de progresión aritmética más larga . Podemos resolver este problema usando Programación Dinámica.
Primero ordenamos el conjunto dado. Usamos una tabla auxiliar L[n][n] para almacenar los resultados de los subproblemas. Una entrada L[i][j] en esta tabla almacena LLGP con set[i] y set[j] como los dos primeros elementos de GP y j > i. La tabla se llena de abajo a la derecha a arriba a la izquierda. Para llenar la tabla, primero se fija j (segundo elemento en GP). i y k se buscan para un j fijo. Si i y k se encuentran de tal manera que i, j, k forman un GP, entonces el valor de L[i][j] se establece como L[j][k] + 1. Tenga en cuenta que el valor de L[j] [k] debe haberse rellenado antes, ya que el ciclo atraviesa de las columnas de derecha a izquierda.
A continuación se muestra la implementación del algoritmo de Programación Dinámica.
C++
// C++ program to find length
// of the longest geometric
// progression in a given set
#include <iostream>
#include <algorithm>
using namespace std;
// Returns length of the
// longest GP subset of set[]
int lenOfLongestGP(int set[], int n)
{
// Base cases
if (n < 2)
return n;
if (n == 2)
return (set[1] % set[0] == 0) ? 2 : 1;
// Let us sort the set first
sort(set, set+n);
// An entry L[i][j] in this
// table stores LLGP with
// set[i] and set[j] as first
// two elements of GP
// and j > i.
int L[n][n];
// Initialize result (A single element
// is always a GP)
int llgp = 1;
// Initialize values of last column
for (int i = 0; i < n - 1; ++i) {
if (set[n-1] % set[i] == 0)
{
L[i][n-1] = 2;
if (2 > llgp)
llgp = 2;
}
else
{
L[i][n-1] = 1;
}
}
L[n-1][n-1] = 1;
// Consider every element as
// second element of GP
for (int j = n - 2; j >= 1; --j)
{
// Search for i and k for j
int i = j - 1, k = j+1;
while (i>=0 && k <= n-1)
{
// Two cases when i, j and k don't form
// a GP.
if (set[i] * set[k] < set[j]*set[j])
{
++k;
}
else if (set[i] * set[k] > set[j]*set[j])
{
if (set[j] % set[i] == 0)
{
L[i][j] = 2;
}
else
{
L[i][j] = 1;
}
--i;
}
// i, j and k form GP, LLGP with i and j as
// first two elements is equal to LLGP with
// j and k as first two elements plus 1.
// L[j][k] must have been filled before as
// we run the loop from right side
else
{
if (set[j] % set[i] == 0)
{
L[i][j] = L[j][k] + 1;
// Update overall LLGP
if (L[i][j] > llgp)
llgp = L[i][j];
} else {
L[i][j] = 1;
}
// Change i and k to fill more L[i][j]
// values for current j
--i;
++k;
}
}
// If the loop was stopped due to k becoming
// more than n-1, set the remaining entries
// in column j as 1 or 2 based on divisibility
// of set[j] by set[i]
while (i >= 0)
{
if (set[j] % set[i] == 0)
{
L[i][j] = 2;
if (2 > llgp)
llgp = 2;
}
else
L[i][j] = 1;
--i;
}
}
// Return result
return llgp;
}
// Driver code
int main()
{
int set1[] = {1, 3, 9, 27, 81, 243};
int n1 = sizeof(set1)/sizeof(set1[0]);
cout << lenOfLongestGP(set1, n1) << "\n";
int set2[] = {1, 3, 4, 9, 7, 27};
int n2 = sizeof(set2)/sizeof(set2[0]);
cout << lenOfLongestGP(set2, n2) << "\n";
int set3[] = {2, 3, 5, 7, 11, 13};
int n3 = sizeof(set3)/sizeof(set3[0]);
cout << lenOfLongestGP(set3, n3) << "\n";
return 0;
}
Java
// Java program to find length
// of the longest geometric
// progression in a given set
import java.util.*;
class GFG {
// Returns length of the longest GP subset of set[]
static int lenOfLongestGP(int set[], int n)
{
// Base cases
if (n < 2) {
return n;
}
if (n == 2) {
return (set[1] % set[0] == 0 ? 2 : 1);
}
// Let us sort the set first
Arrays.sort(set);
// An entry L[i][j] in this table
// stores LLGP with set[i] and set[j]
// as first two elements of GP
// and j > i.
int L[][] = new int[n][n];
// Initialize result (A single
// element is always a GP)
int llgp = 1;
// Initialize values of last column
for (int i = 0; i < n - 1; ++i) {
if (set[n - 1] % set[i] == 0) {
L[i][n - 1] = 2;
if (2 > llgp)
llgp = 2;
}
else {
L[i][n - 1] = 1;
}
}
L[n - 1][n - 1] = 1;
// Consider every element as second element of GP
for (int j = n - 2; j >= 1; --j) {
// Search for i and k for j
int i = j - 1, k = j + 1;
while (i >= 0 && k <= n - 1) {
// Two cases when i, j and k
// don't form a GP.
if (set[i] * set[k] < set[j] * set[j]) {
++k;
}
else if (set[i] * set[k]
> set[j] * set[j]) {
if (set[j] % set[i] == 0) {
L[i][j] = 2;
if (2 > llgp)
llgp = 2;
}
else {
L[i][j] = 1;
}
--i;
}
// i, j and k form GP, LLGP with i and j as
// first two elements is equal to LLGP with
// j and k as first two elements plus 1.
// L[j][k] must have been filled before as
// we run the loop from right side
else {
if (set[j] % set[i] == 0) {
L[i][j] = L[j][k] + 1;
// Update overall LLGP
if (L[i][j] > llgp) {
llgp = L[i][j];
}
}
else {
L[i][j] = 1;
}
// Change i and k to fill more L[i][j]
// values for current j
--i;
++k;
}
}
// If the loop was stopped due to k becoming
// more than n-1, set the remaining entries
// in column j as 1 or 2 based on divisibility
// of set[j] by set[i]
while (i >= 0) {
if (set[j] % set[i] == 0) {
L[i][j] = 2;
if (2 > llgp)
llgp = 2;
}
else {
L[i][j] = 1;
}
--i;
}
}
// Return result
return llgp;
}
// Driver code
public static void main(String[] args)
{
int set1[] = { 1, 3, 9, 27, 81, 243 };
int n1 = set1.length;
System.out.print(lenOfLongestGP(set1, n1) + "\n");
int set2[] = { 1, 3, 4, 9, 7, 27 };
int n2 = set2.length;
System.out.print(lenOfLongestGP(set2, n2) + "\n");
int set3[] = { 2, 3, 5, 7, 11, 13 };
int n3 = set3.length;
System.out.print(lenOfLongestGP(set3, n3) + "\n");
}
}
/* This code contributed by PrinciRaj1992 */
Python3
# Python3 program to find length # of the longest geometric # progression in a given sett # Returns length of the longest GP # subset of sett[] def lenOfLongestGP(sett, n): # Base cases if n < 2: return n if n == 2: return 2 if (sett[1] % sett[0] == 0) else 1 # let us sort the sett first sett.sort() # An entry L[i][j] in this # table stores LLGP with # sett[i] and sett[j] as first # two elements of GP # and j > i. L = [[0 for i in range(n)] for i in range(n)] # Initialize result (A single # element is always a GP) llgp = 1 # Initialize values of last column for i in range(0, n-1): if sett[n-1] % sett[i] == 0: L[i][n-1] = 2 if 2 > llgp: llgp = 2 else: L[i][n-1] = 1 L[n-1][n-1] = 1 # Consider every element as second element of GP for j in range(n-2, 0, -1): # Search for i and k for j i = j - 1 k = j + 1 while i >= 0 and k <= n - 1: # Two cases when i, j and k don't form # a GP. if sett[i] * sett[k] < sett[j] * sett[j]: k += 1 else if sett[i] * sett[k] > sett[j] * sett[j]: if sett[j] % sett[i] == 0: L[i][j] = 2 else: L[i][j] = 1 i -= 1 # i, j and k form GP, LLGP with i and j as # first two elements is equal to LLGP with # j and k as first two elements plus 1. # L[j][k] must have been filled before as # we run the loop from right side else: if sett[j] % sett[i] == 0: L[i][j] = L[j][k] + 1 # Update overall LLGP if L[i][j] > llgp: llgp = L[i][j] else: L[i][j] = 1 # Change i and k to fill more L[i][j] # values for current j i -= 1 k += 1 # If the loop was stopped due to k becoming # more than n-1, set the remaining entries # in column j as 1 or 2 based on divisibility # of sett[j] by sett[i] while i >= 0: if sett[j] % sett[i] == 0: L[i][j] = 2 else: L[i][j] = 1 i -= 1 return llgp # Driver code if __name__ == '__main__': set1 = [1, 3, 9, 27, 81, 243] n1 = len(set1) print(lenOfLongestGP(set1, n1)) set2 = [1, 3, 4, 9, 7, 27] n2 = len(set2) print(lenOfLongestGP(set2, n2)) set3 = [2, 3, 5, 7, 11, 13] n3 = len(set3) print(lenOfLongestGP(set3, n3)) # this code is contributed by sahilshelangia
C#
// C# program to find length
// of the longest geometric
// progression in a given Set
using System;
class GFG
{
// Returns length of the
// longest GP subset of Set[]
static int lenOfLongestGP(int []Set, int n)
{
// Base cases
if (n < 2)
{
return n;
}
if (n == 2)
{
return (Set[1] % Set[0] == 0 ? 2 : 1);
}
// Let us sort the Set first
Array.Sort(Set);
// An entry L[i,j] in this table
// stores LLGP with Set[i] and Set[j]
// as first two elements of GP
// and j > i.
int [,]L = new int[n, n];
// Initialize result (A single
// element is always a GP)
int llgp = 1;
// Initialize values of last column
for (int i = 0; i < n - 1; ++i)
{
if (Set[n - 1] % Set[i] == 0)
{
L[i, n - 1] = 2;
if (2 > llgp)
llgp = 2;
}
else
{
L[i, n - 1] = 1;
}
}
L[n - 1, n - 1] = 1;
// Consider every element
// as second element of GP
for (int j = n - 2; j >= 1; --j)
{
// Search for i and k for j
int i = j - 1, k = j + 1;
while (i >= 0 && k <= n - 1)
{
// Two cases when i, j and k
// don't form a GP.
if (Set[i] * Set[k] < Set[j] * Set[j])
{
++k;
}
else if (Set[i] * Set[k] > Set[j] * Set[j])
{
if (Set[j] % Set[i] == 0)
{
L[i,j] = 2;
if (2 > llgp)
llgp = 2;
}
else
{
L[i,j] = 1;
}
--i;
}
// i, j and k form GP, LLGP with i and j as
// first two elements is equal to LLGP with
// j and k as first two elements plus 1.
// L[j,k] must have been filled before as
// we run the loop from right side
else
{
if (Set[j] % Set[i] == 0)
{
L[i, j] = L[j, k] + 1;
// Update overall LLGP
if (L[i, j] > llgp)
{
llgp = L[i, j];
}
}
else
{
L[i, j] = 1;
}
// Change i and k to fill more L[i,j]
// values for current j
--i;
++k;
}
}
// If the loop was stopped due to k becoming
// more than n-1, set the remaining entries
// in column j as 1 or 2 based on divisibility
// of Set[j] by Set[i]
while (i >= 0)
{
if (Set[j] % Set[i] == 0)
{
L[i, j] = 2;
if (2 > llgp)
llgp = 2;
}
else
{
L[i, j] = 1;
}
--i;
}
}
// Return result
return llgp;
}
// Driver code
public static void Main(String[] args)
{
int []set1 = {1, 3, 9, 27, 81, 243};
int n1 = set1.Length;
Console.Write(lenOfLongestGP(set1, n1) + "\n");
int []set2 = {1, 3, 4, 9, 7, 27};
int n2 = set2.Length;
Console.Write(lenOfLongestGP(set2, n2) + "\n");
int []set3 = {2, 3, 5, 7, 11, 13};
int n3 = set3.Length;
Console.Write(lenOfLongestGP(set3, n3) + "\n");
}
}
// This code has been contributed by 29AjayKumar
Javascript
<script>
// Javascript program to find length
// of the longest geometric
// progression in a given set
// Returns length of the longest GP subset of set[]
function lenOfLongestGP(set, n)
{
// Base cases
if (n < 2) {
return n;
}
if (n == 2) {
return (set[1] % set[0] == 0 ? 2 : 1);
}
// Let us sort the set first
set.sort(function(a, b){return a - b});
// An entry L[i][j] in this table
// stores LLGP with set[i] and set[j]
// as first two elements of GP
// and j > i.
let L = new Array(n);
for (let i = 0; i < n; ++i)
{
L[i] = new Array(n);
for (let j = 0; j < n; ++j)
{
L[i][j] = 0;
}
}
// Initialize result (A single
// element is always a GP)
let llgp = 1;
// Initialize values of last column
for (let i = 0; i < n - 1; ++i) {
if (set[n - 1] % set[i] == 0) {
L[i][n - 1] = 2;
if (2 > llgp)
llgp = 2;
}
else {
L[i][n - 1] = 1;
}
}
L[n - 1][n - 1] = 1;
// Consider every element as second element of GP
for (let j = n - 2; j >= 1; --j) {
// Search for i and k for j
let i = j - 1, k = j + 1;
while (i >= 0 && k <= n - 1) {
// Two cases when i, j and k
// don't form a GP.
if (set[i] * set[k] < set[j] * set[j]) {
++k;
}
else if (set[i] * set[k]
> set[j] * set[j]) {
if (set[j] % set[i] == 0) {
L[i][j] = 2;
if (2 > llgp)
llgp = 2;
}
else {
L[i][j] = 1;
}
--i;
}
// i, j and k form GP, LLGP with i and j as
// first two elements is equal to LLGP with
// j and k as first two elements plus 1.
// L[j][k] must have been filled before as
// we run the loop from right side
else {
if (set[j] % set[i] == 0) {
L[i][j] = L[j][k] + 1;
// Update overall LLGP
if (L[i][j] > llgp) {
llgp = L[i][j];
}
}
else {
L[i][j] = 1;
}
// Change i and k to fill more L[i][j]
// values for current j
--i;
++k;
}
}
// If the loop was stopped due to k becoming
// more than n-1, set the remaining entries
// in column j as 1 or 2 based on divisibility
// of set[j] by set[i]
while (i >= 0) {
if (set[j] % set[i] == 0) {
L[i][j] = 2;
if (2 > llgp)
llgp = 2;
}
else {
L[i][j] = 1;
}
--i;
}
}
// Return result
return llgp;
}
let set1 = [ 1, 3, 9, 27, 81, 243 ];
let n1 = set1.length;
document.write(lenOfLongestGP(set1, n1) + "</br>");
let set2 = [ 1, 3, 4, 9, 7, 27 ];
let n2 = set2.length;
document.write(lenOfLongestGP(set2, n2) + "</br>");
let set3 = [ 2, 3, 5, 7, 11, 13 ];
let n3 = set3.length;
document.write(lenOfLongestGP(set3, n3) + "</br>");
// This code is contributed by rameshtravel07.
</script>
Producción:
6 4 1
Complejidad de tiempo: O(n 2 )
Espacio auxiliar: O(n 2 )
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA