Dado un número N , la tarea es verificar si el número N dado es un cubo perfecto o no.
Ejemplos:
Entrada: N = 216
Salida: Sí
Explicación:
Como 216 = 6*6*6. Por lo tanto, la raíz cúbica de 216 es 6.
Entrada: N = 100
Salida: No
Método 1: enfoque ingenuo
La idea es verificar para cada número del 1 al N si el cubo de cualquiera de estos números es igual a N . Si es así, entonces ese número es la raíz cúbica de N y N es un cubo perfecto.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to check whether the given
// number N is the perfect cube or not
.
#include <bits/stdc++.h>
using namespace std;
// Function to check if a number
// is a perfect Cube or not
void perfectCube(int N)
{
int cube;
// Iterate from 1-N
for (int i; i <= N; i++) {
// Find the cube of
// every number
cube = i * i * i;
// Check if cube equals
// N or not
if (cube == N) {
cout << "Yes";
return;
}
else if (cube > N) {
cout << "NO";
return;
}
}
}
// Driver code
int main()
{
int N = 216;
// Function call
perfectCube(N);
return 0;
}
Java
// Java program to check whether the given
// number N is the perfect cube or not
class GFG {
// Function to check if a number
// is a perfect Cube or not
static void perfectCube(int N)
{
int cube;
// Iterate from 1-N
for (int i = 0; i <= N; i++) {
// Find the cube of
// every number
cube = i * i * i;
// Check if cube equals
// N or not
if (cube == N) {
System.out.println("Yes");
return;
}
else if (cube > N) {
System.out.println("NO");
return;
}
}
}
// Driver code
public static void main (String[] args)
{
int N = 216;
// Function call
perfectCube(N);
}
}
// This code is contributed by AnkitRai01
Python3
# Python3 program to check whether the given
# number N is the perfect cube or not
# Function to check if a number
# is a perfect Cube or not
def perfectCube(N) :
cube = 0;
# Iterate from 1-N
for i in range(N + 1) :
# Find the cube of
# every number
cube = i * i * i;
# Check if cube equals
# N or not
if (cube == N) :
print("Yes");
return;
elif (cube > N) :
print("NO");
return;
# Driver code
if __name__ == "__main__" :
N = 216;
# Function call
perfectCube(N);
# This code is contributed by Yash_R
C#
// C# program to check whether the given
// number N is the perfect cube or not
using System;
class GFG {
// Function to check if a number
// is a perfect Cube or not
static void perfectCube(int N)
{
int cube;
// Iterate from 1-N
for (int i = 0; i <= N; i++) {
// Find the cube of
// every number
cube = i * i * i;
// Check if cube equals
// N or not
if (cube == N) {
Console.WriteLine("Yes");
return;
}
else if (cube > N) {
Console.WriteLine("NO");
return;
}
}
}
// Driver code
public static void Main (string[] args)
{
int N = 216;
// Function call
perfectCube(N);
}
}
// This code is contributed by AnkitRai01
Javascript
<script>
// JavaScript program to check whether the given
// number N is the perfect cube or not
// Function to check if a number
// is a perfect Cube or not
function perfectCube(N)
{
let cube;
// Iterate from 1-N
for(let i = 0; i <= N; i++)
{
// Find the cube of
// every number
cube = i * i * i;
// Check if cube equals
// N or not
if (cube === N)
{
document.write("Yes");
return;
}
else if (cube > N)
{
document.write("NO");
return;
}
}
}
// Driver code
let N = 216;
// Function call
perfectCube(N);
// This code is contributed by Manoj.
</script>
Producción:
Yes
Complejidad de tiempo: O(N)
Espacio Auxiliar: O(1)
Método 2: Usar la función incorporada
La idea es usar la función incorporada ( cbrt() ) para encontrar la raíz cúbica de un número que devuelve el valor mínimo de la raíz cúbica del número N. Si el cubo de este número es igual a N , entonces N es un cubo perfecto; de lo contrario , N no es un cubo perfecto.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to check whether the given
// number N is the perfect cube or not
#include <bits/stdc++.h>
using namespace std;
// Function to check if a number is
// a perfect Cube using inbuilt function
void perfectCube(int N)
{
int cube_root;
cube_root = round(cbrt(N));
// If cube of cube_root is equals to N,
// then print Yes Else print No
if (cube_root * cube_root * cube_root == N) {
cout << "Yes";
return;
}
else {
cout << "NO";
return;
}
}
// Driver's code
int main()
{
int N = 125;
// Function call to check
// N is cube or not
perfectCube(N);
return 0;
}
Java
// Java program to check whether the given
// number N is the perfect cube or not
public class GFG {
// Function to check if a number is
// a perfect Cube using inbuilt function
static void perfectCube(int N)
{
int cube_root;
cube_root = (int)Math.round(Math.cbrt(N));
// If cube of cube_root is equals to N,
// then print Yes Else print No
if (cube_root * cube_root * cube_root == N) {
System.out.println("Yes");
return;
}
else {
System.out.println("NO");
return;
}
}
// Driver's code
public static void main (String[] args)
{
int N = 125;
// Function call to check
// N is cube or not
perfectCube(N);
}
}
// This code is contributed by AnkitRai01
Python3
# Python program to check whether the given
# number N is the perfect cube or not
# Function to check if a number is
# a perfect Cube using inbuilt function
def perfectCube(N) :
cube_root = round(N**(1/3));
# If cube of cube_root is equals to N,
# then print Yes Else print No
if cube_root * cube_root * cube_root == N :
print("Yes");
return;
else :
print("NO");
return;
# Driver's code
if __name__ == "__main__" :
N = 125;
# Function call to check
# N is cube or not
perfectCube(N);
# This code is contributed by AnkitRai01
C#
// C# program to check whether the given
// number N is the perfect cube or not
using System;
class GFG {
// Function to check if a number is
// a perfect Cube using inbuilt function
static void perfectCube(int N)
{
int cube_root;
cube_root = (int)Math.Round(Math.Cbrt(N));
// If cube of cube_root is equals to N,
// then print Yes Else print No
if (cube_root * cube_root * cube_root == N) {
Console.WriteLine("Yes");
return;
}
else {
Console.WriteLine("NO");
return;
}
}
// Driver's code
public static void Main (string[] args)
{
int N = 125;
// Function call to check
// N is cube or not
perfectCube(N);
}
}
// This code is contributed by AnkitRai01
Javascript
<script>
// Javascript program to check whether the given
// number N is the perfect cube or not
// Function to check if a number is
// a perfect Cube using inbuilt function
function perfectCube(N)
{
let cube_root;
cube_root = Math.round(Math.cbrt(N));
// If cube of cube_root is equals to N,
// then print Yes Else print No
if ((cube_root * cube_root * cube_root) == N) {
document.write("Yes");
return;
}
else {
document.write("NO");
return;
}
}
// Driver's code
let N = 125;
// Function call to check
// N is cube or not
perfectCube(N);
// This code is contributed by subhammahato348.
</script>
Producción:
Yes
Complejidad del tiempo: O(cbrt(N))
Espacio Auxiliar: O(1)
Método 3: Uso de factores primos
- Encuentre todos los factores primos del número N dado utilizando el enfoque de este artículo.
- Almacene la frecuencia de todos los factores primos obtenidos anteriormente en un Hash Map .
- Recorra el mapa hash y si la frecuencia de todos los factores primos no es un múltiplo de 3 , entonces el número N dado no es un cubo perfecto.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to check if a number
// is a perfect cube using prime factors
#include<bits/stdc++.h>
using namespace std;
// Inserts the prime factor in HashMap
// if not present
// if present updates it's frequency
map<int, int> insertPF(map<int, int> primeFact,
int fact)
{
if (primeFact.find(fact) != primeFact.end())
{
primeFact[fact]++;
}
else
{
primeFact[fact] = 1;
}
return primeFact;
}
// A utility function to find all
// prime factors of a given number N
map<int, int> primeFactors (int n)
{
map<int, int> primeFact;
// Insert the number of 2s
// that divide n
while (n % 2 == 0)
{
primeFact = insertPF(primeFact, 2);
n /= 2;
}
// n must be odd at this point
// So we can skip one element
for(int i = 3; i <= sqrt(n); i += 2)
{
// while i divides n, insert
// i and divide n
while (n % i == 0)
{
primeFact = insertPF(primeFact, i);
n /= i;
}
}
// This condition is to handle
// the case when n is a prime
// number greater than 2
if (n > 2)
primeFact = insertPF(primeFact, n);
return primeFact;
}
// Function to check if a
// number is perfect cube
string perfectCube (int n)
{
map<int, int> primeFact;
primeFact = primeFactors(n);
// Iteration in Map
for(auto x : primeFact)
{
if (x.second % 3 != 0)
return "No";
}
return "Yes";
}
// Driver Code
int main()
{
int N = 216;
// Function to check if N is
// perfect cube or not
cout << perfectCube(N);
return 0;
}
// This code is contributed by himanshu77
Java
// Java program to check if a number
// is a perfect cube using prime factors
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
// Inserts the prime factor in the Hash Map
// if not present
// If present updates it's frequency
public static HashMap<Integer, Integer>
insertPF(HashMap<Integer, Integer> primeFact,
int fact)
{
if (primeFact.containsKey(fact)) {
int freq;
freq = primeFact.get(fact);
primeFact.replace(fact, ++freq);
}
else {
primeFact.put(fact, 1);
}
return primeFact;
}
// A utility function to find all
// prime factors of a given number N
public static HashMap<Integer, Integer>
primeFactors(int n)
{
HashMap<Integer, Integer> primeFact
= new HashMap<>();
// Insert the number of 2s
// that divide n
while (n % 2 == 0) {
primeFact = insertPF(primeFact, 2);
n /= 2;
}
// n must be odd at this point.
// So we can skip one element
for (int i = 3; i <= Math.sqrt(n);
i += 2) {
// While i divides n, insert i
// and divide n
while (n % i == 0) {
primeFact = insertPF(primeFact, i);
n /= i;
}
}
// This condition is to handle
// the case when n is a prime
// number greater than 2
if (n > 2)
primeFact = insertPF(primeFact, n);
return primeFact;
}
// Function to check if a number
// is a perfect cube
public static String perfectCube(int n)
{
HashMap<Integer, Integer> primeFact;
primeFact = primeFactors(n);
// Using values() for iteration
// over keys
for (int freq : primeFact.values()) {
if (freq % 3 != 0)
return "No";
}
return "Yes";
}
// Driver Code
public static void main(String[] args)
{
int N = 216;
// Function to check if N is
// perfect cube or not
System.out.println(perfectCube(N));
}
}
Python3
# Python3 program to check if a number
# is a perfect cube using prime factors
import math
# Inserts the prime factor in HashMap
# if not present
# if present updates it's frequency
def insertPF(primeFact, fact) :
if (fact in primeFact) :
primeFact[fact] += 1
else :
primeFact[fact] = 1
return primeFact
# A utility function to find all
# prime factors of a given number N
def primeFactors (n) :
primeFact = {}
# Insert the number of 2s
# that divide n
while (n % 2 == 0) :
primeFact = insertPF(primeFact, 2)
n = n // 2
# n must be odd at this point
# So we can skip one element
for i in range(3, int(math.sqrt(n)) + 1, 2) :
# while i divides n, insert
# i and divide n
while (n % i == 0) :
primeFact = insertPF(primeFact, i)
n = n // i
# This condition is to handle
# the case when n is a prime
# number greater than 2
if (n > 2) :
primeFact = insertPF(primeFact, n)
return primeFact
# Function to check if a
# number is perfect cube
def perfectCube (n) :
primeFact = {}
primeFact = primeFactors(n)
# Iteration in Map
for x in primeFact :
if (primeFact[x] % 3 != 0) :
return "No"
return "Yes"
N = 216
# Function to check if N is
# perfect cube or not
print(perfectCube(N))
# This code is contributed by divyeshrabadiya07.
C#
// C# program to check if a number
// is a perfect cube using prime factors
using System;
using System.Collections.Generic;
public class GFG {
// Inserts the prime factor in the Hash Map
// if not present
// If present updates it's frequency
public static Dictionary<int, int>
insertPF(Dictionary<int, int> primeFact,
int fact)
{
if (primeFact.ContainsKey(fact)) {
int freq;
freq = primeFact[fact];
primeFact[fact] = ++freq;
}
else {
primeFact.Add(fact, 1);
}
return primeFact;
}
// A utility function to find all
// prime factors of a given number N
public static Dictionary<int, int>
primeFactors(int n)
{
Dictionary<int, int> primeFact
= new Dictionary<int, int>();
// Insert the number of 2s
// that divide n
while (n % 2 == 0) {
primeFact = insertPF(primeFact, 2);
n /= 2;
}
// n must be odd at this point.
// So we can skip one element
for (int i = 3; i <= Math.Sqrt(n);
i += 2) {
// While i divides n, insert i
// and divide n
while (n % i == 0) {
primeFact = insertPF(primeFact, i);
n /= i;
}
}
// This condition is to handle
// the case when n is a prime
// number greater than 2
if (n > 2)
primeFact = insertPF(primeFact, n);
return primeFact;
}
// Function to check if a number
// is a perfect cube
public static String perfectCube(int n)
{
Dictionary<int, int> primeFact;
primeFact = primeFactors(n);
// Using values() for iteration
// over keys
foreach (int freq in primeFact.Values) {
if (freq % 3 != 0)
return "No";
}
return "Yes";
}
// Driver Code
public static void Main(String[] args)
{
int N = 216;
// Function to check if N is
// perfect cube or not
Console.WriteLine(perfectCube(N));
}
}
// This code is contributed by sapnasingh4991
Javascript
<script>
// Javascript program to check if a number
// is a perfect cube using prime factors
// Inserts the prime factor in HashMap
// if not present
// if present updates it's frequency
function insertPF(primeFact, fact)
{
if (primeFact.has(fact))
{
primeFact.set(fact, primeFact.get(fact)+1);
}
else
{
primeFact.set(fact, 1);
}
return primeFact;
}
// A utility function to find all
// prime factors of a given number N
function primeFactors (n)
{
var primeFact = new Map();
// Insert the number of 2s
// that divide n
while (n % 2 == 0)
{
primeFact = insertPF(primeFact, 2);
n = parseInt(n/2);
}
// n must be odd at this point
// So we can skip one element
for(var i = 3; i <= Math.sqrt(n); i += 2)
{
// while i divides n, insert
// i and divide n
while (n % i == 0)
{
primeFact = insertPF(primeFact, i);
n =parseInt(n/i);
}
}
// This condition is to handle
// the case when n is a prime
// number greater than 2
if (n > 2)
primeFact = insertPF(primeFact, n);
return primeFact;
}
// Function to check if a
// number is perfect cube
function perfectCube (n)
{
var primeFact = new Map();
primeFact = primeFactors(n);
// Iteration in Map
primeFact.forEach((value, key) => {
if (value % 3 != 0)
return "No";
});
return "Yes";
}
// Driver Code
var N = 216;
// Function to check if N is
// perfect cube or not
document.write( perfectCube(N));
// This code is contributed by rrrtnx.
</script>
Producción:
Yes
Complejidad del tiempo: O(sqrt(n))
Espacio auxiliar: O(sqrt(n))