Dada una array arr[] , la tarea es determinar el número de elementos de la array que no son divisibles por ningún otro elemento de la array dada.
Ejemplos:
Entrada: arr[] = {86, 45, 18, 4, 8, 28, 19, 33, 2}
Salida: 4
Explicación:
Los elementos {2, 19, 33, 45} no son divisibles por ningún otro elemento de array .Entrada: arr[] = {3, 3, 3}
Salida: 0
Enfoque ingenuo: el enfoque ingenuo es iterar sobre toda la array y contar la cantidad de elementos que no son divisibles por ningún otro elemento en la array dada e imprimir la cuenta.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
// Function to count the number of
// elements of array which are not
// divisible by any other element
// in the array arr[]
int count(int a[], int n)
{
int countElements = 0;
// Iterate over the array
for (int i = 0; i < n; i++) {
bool flag = true;
for (int j = 0; j < n; j++) {
// Check if the element
// is itself or not
if (i == j)
continue;
// Check for divisibility
if (a[i] % a[j] == 0) {
flag = false;
break;
}
}
if (flag == true)
++countElements;
}
// Return the final result
return countElements;
}
// Driver Code
int main()
{
// Given array
int arr[] = { 86, 45, 18, 4, 8,
28, 19, 33, 2 };
int n = sizeof(arr) / sizeof(int);
// Function Call
cout << count(arr, n);
return 0;
}
Java
// Java program for the above approach
class GFG{
// Function to count the number of
// elements of array which are not
// divisible by any other element
// in the array arr[]
static int count(int a[], int n)
{
int countElements = 0;
// Iterate over the array
for(int i = 0; i < n; i++)
{
boolean flag = true;
for(int j = 0; j < n; j++)
{
// Check if the element
// is itself or not
if (i == j)
continue;
// Check for divisibility
if (a[i] % a[j] == 0)
{
flag = false;
break;
}
}
if (flag == true)
++countElements;
}
// Return the final result
return countElements;
}
// Driver Code
public static void main(String[] args)
{
// Given array
int arr[] = { 86, 45, 18, 4, 8,
28, 19, 33, 2 };
int n = arr.length;
// Function call
System.out.print(count(arr, n));
}
}
// This code is contributed by Amit Katiyar
Python3
# Python3 program for # the above approach # Function to count the number of # elements of array which are not # divisible by any other element # in the array arr[] def count(a, n): countElements = 0 # Iterate over the array for i in range (n): flag = True for j in range (n): # Check if the element # is itself or not if (i == j): continue # Check for divisibility if (a[i] % a[j] == 0): flag = False break if (flag == True): countElements += 1 # Return the final result return countElements # Driver Code if __name__ == "__main__": # Given array arr = [86, 45, 18, 4, 8, 28, 19, 33, 2] n = len(arr) # Function Call print( count(arr, n)) # This code is contributed by Chitranayal
C#
// C# program for the
// above approach
using System;
class GFG{
// Function to count the
// number of elements of
// array which are not
// divisible by any other
// element in the array []arr
static int count(int []a,
int n)
{
int countElements = 0;
// Iterate over the array
for(int i = 0; i < n; i++)
{
bool flag = true;
for(int j = 0; j < n; j++)
{
// Check if the element
// is itself or not
if (i == j)
continue;
// Check for divisibility
if (a[i] % a[j] == 0)
{
flag = false;
break;
}
}
if (flag == true)
++countElements;
}
// Return the readonly result
return countElements;
}
// Driver Code
public static void Main(String[] args)
{
// Given array
int []arr = {86, 45, 18, 4, 8,
28, 19, 33, 2};
int n = arr.Length;
// Function call
Console.Write(count(arr, n));
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// Javascript program for the above approach
// Function to count the number of
// elements of array which are not
// divisible by any other element
// in the array arr[]
function count(a, n)
{
let countElements = 0;
// Iterate over the array
for (let i = 0; i < n; i++) {
let flag = true;
for (let j = 0; j < n; j++) {
// Check if the element
// is itself or not
if (i == j)
continue;
// Check for divisibility
if (a[i] % a[j] == 0) {
flag = false;
break;
}
}
if (flag == true)
++countElements;
}
// Return the final result
return countElements;
}
// Driver Code
// Given array
let arr = [ 86, 45, 18, 4, 8,
28, 19, 33, 2 ];
let n = arr.length;
// Function Call
document.write(count(arr, n));
</script>
Producción
4
Complejidad de Tiempo: O(N 2 )
Espacio Auxiliar: O(1)
Enfoque Eficiente: Para optimizar el enfoque anterior, utilizaremos el concepto de Tamiz de Eratóstenes . A continuación se muestran los pasos:
- Inicialice una array booleana (digamos v[] ) de tamaño igual al elemento máximo presente en la array + 1 con verdadero en cada índice.
- Recorra la array dada arr[] y cambie el valor en el índice de múltiplos de los elementos actuales como falso en la array v[] .
- Cree un Hashmap y almacene la frecuencia de cada elemento en él.
- Para cada elemento (por ejemplo, elemento_actual ) en la array, si v[elemento_actual] es verdadero, entonces ese elemento no es divisible por ningún otro elemento en la array dada e incrementa el recuento del elemento actual.
- Imprima el valor final de la cuenta después de los pasos anteriores.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count the number of
// elements of array which are not
// divisible by any other element
// of same array
int countEle(int a[], int n)
{
// Length for boolean array
int len = 0;
// Hash map for storing the
// element and it's frequency
unordered_map<int, int> hmap;
for (int i = 0; i < n; i++) {
// Update the maximum element
len = max(len, a[i]);
hmap[a[i]]++;
}
// Boolean array of size
// of the max element + 1
bool v[len + 1];
for (int i = 0; i <= len; i++) {
v[i] = true;
}
// Marking the multiples as false
for (int i = 0; i < n; i++) {
if (v[a[i]] == false)
continue;
for (int j = 2 * a[i];
j <= len; j += a[i]) {
v[j] = false;
}
}
// To store the final count
int count = 0;
// Traverse boolean array
for (int i = 1; i <= len; i++) {
// Check if i is not divisible by
// any other array elements and
// appears in the array only once
if (v[i] == true
&& hmap.count(i) == 1
&& hmap[i] == 1) {
count += 1;
}
}
// Return the final Count
return count;
}
// Driver Code
int main()
{
// Given array
int arr[] = { 86, 45, 18, 4, 8,
28, 19, 33, 2 };
int n = sizeof(arr) / sizeof(int);
// Function Call
cout << countEle(arr, n);
return 0;
}
Java
// Java program for
// the above approach
import java.util.*;
class GFG{
// Function to count the
// number of elements of
// array which are not
// divisible by any other
// element of same array
static int countEle(int a[],
int n)
{
// Length for boolean array
int len = 0;
// Hash map for storing the
// element and it's frequency
HashMap<Integer,
Integer> hmap = new HashMap<>();
for (int i = 0; i < n; i++)
{
// Update the maximum element
len = Math.max(len, a[i]);
if(hmap.containsKey(a[i]))
{
hmap.put(a[i],
hmap.get(a[i]) + 1);
}
else
{
hmap.put(a[i], 1);
}
}
// Boolean array of size
// of the max element + 1
boolean []v = new boolean[len + 1];
for (int i = 0; i <= len; i++)
{
v[i] = true;
}
// Marking the multiples
// as false
for (int i = 0; i < n; i++)
{
if (v[a[i]] == false)
continue;
for (int j = 2 * a[i];
j <= len; j += a[i])
{
v[j] = false;
}
}
// To store the
// final count
int count = 0;
// Traverse boolean array
for (int i = 1; i <= len; i++)
{
// Check if i is not divisible by
// any other array elements and
// appears in the array only once
if (v[i] == true &&
hmap.containsKey(i) &&
hmap.get(i) == 1 &&
hmap.get(i) == 1)
{
count += 1;
}
}
// Return the final
// Count
return count;
}
// Driver Code
public static void main(String[] args)
{
// Given array
int arr[] = {86, 45, 18, 4, 8,
28, 19, 33, 2};
int n = arr.length;
// Function Call
System.out.print(countEle(arr, n));
}
}
// This code is contributed by Princi Singh
Python3
# Python3 program for the above approach
# Function to count the number of
# elements of array which are not
# divisible by any other element
# of same array
def countEle(a, n):
# Length for boolean array
len = 0
# Hash map for storing the
# element and it's frequency
hmap = {}
for i in range(n):
# Update the maximum element
len = max(len, a[i])
hmap[a[i]] = hmap.get(a[i], 0) + 1
# Boolean array of size
# of the max element + 1
v = [True for i in range(len + 1)]
# Marking the multiples as false
for i in range(n):
if (v[a[i]] == False):
continue
for j in range(2 * a[i], len + 1, a[i]):
v[j] = False
# To store the final count
count = 0
# Traverse boolean array
for i in range(1, len + 1):
# Check if i is not divisible by
# any other array elements and
# appears in the array only once
if (v[i] == True and (i in hmap) and hmap[i] == 1):
count += 1
# Return the final Count
return count
# Driver Code
if __name__ == '__main__':
# Given array
arr = [86, 45, 18, 4, 8, 28, 19, 33, 2]
n = len(arr)
# Function Call
print(countEle(arr, n))
# This code is contributed by mohit kumar 29
C#
// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to count the
// number of elements of
// array which are not
// divisible by any other
// element of same array
static int countEle(int []a,
int n)
{
// Length for bool array
int len = 0;
// Hash map for storing the
// element and it's frequency
Dictionary<int,
int> hmap =
new Dictionary<int,
int>();
for (int i = 0; i < n; i++)
{
// Update the maximum
// element
len = Math.Max(len, a[i]);
if(hmap.ContainsKey(a[i]))
{
hmap[a[i]]++;
}
else
{
hmap.Add(a[i], 1);
}
}
// Boolean array of size
// of the max element + 1
bool []v = new bool[len + 1];
for (int i = 0; i <= len; i++)
{
v[i] = true;
}
// Marking the multiples
// as false
for (int i = 0; i < n; i++)
{
if (v[a[i]] == false)
continue;
for (int j = 2 * a[i];
j <= len; j += a[i])
{
v[j] = false;
}
}
// To store the
// readonly count
int count = 0;
// Traverse bool array
for (int i = 1; i <= len; i++)
{
// Check if i is not divisible by
// any other array elements and
// appears in the array only once
if (v[i] == true &&
hmap.ContainsKey(i) &&
hmap[i] == 1 &&
hmap[i] == 1)
{
count += 1;
}
}
// Return the readonly
// Count
return count;
}
// Driver Code
public static void Main(String[] args)
{
// Given array
int []arr = {86, 45, 18, 4, 8,
28, 19, 33, 2};
int n = arr.Length;
// Function Call
Console.Write(countEle(arr, n));
}
}
// This code is contributed by gauravrajput1
Javascript
<script>
// Javascript program for
// the above approach
// Function to count the
// number of elements of
// array which are not
// divisible by any other
// element of same array
function countEle(a, n)
{
// Length for boolean array
let len = 0;
// Hash map for storing the
// element and it's frequency
let hmap = new Map();
for (let i = 0; i < n; i++)
{
// Update the maximum element
len = Math.max(len, a[i]);
if(hmap.has(a[i]))
{
hmap.set(a[i],
hmap.get(a[i]) + 1);
}
else
{
hmap.set(a[i], 1);
}
}
// Boolean array of size
// of the max element + 1
let v = Array.from({length: len + 1}, (_, i) => 0);
for (let i = 0; i <= len; i++)
{
v[i] = true;
}
// Marking the multiples
// as false
for (let i = 0; i < n; i++)
{
if (v[a[i]] == false)
continue;
for (let j = 2 * a[i];
j <= len; j += a[i])
{
v[j] = false;
}
}
// To store the
// final count
let count = 0;
// Traverse boolean array
for (let i = 1; i <= len; i++)
{
// Check if i is not divisible by
// any other array elements and
// appears in the array only once
if (v[i] == true &&
hmap.has(i) &&
hmap.get(i) == 1 &&
hmap.get(i) == 1)
{
count += 1;
}
}
// Return the final
// Count
return count;
}
// Driver code
// Given array
let arr = [86, 45, 18, 4, 8,
28, 19, 33, 2];
let n = arr.length;
// Function Call
document.write(countEle(arr, n));
// This code is contributed by souravghosh0416.
</script>
Producción
4
Complejidad de tiempo: O(N*log(M)) donde N es el número de elementos en la array dada y M es el elemento máximo en la array dada.
Espacio auxiliar: O(M + N) donde N es el número de elementos en el arreglo dado y M es el elemento máximo en el arreglo dado.
Publicación traducida automáticamente
Artículo escrito por mansimar_anand y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA