Dada una array arr[] que consta de N enteros positivos, la tarea es encontrar el número de elementos de la array cuya mayor potencia de 2 menor o igual a ese número está presente en la array .
Ejemplos:
Entrada: arr[] = {3, 4, 6, 9}
Salida: 2
Explicación:
Hay 2 elementos de array (4 y 6), cuya potencia más alta de 2 es menor que (es decir, 4) están presentes en la array.Entrada: arr[] = {3, 9, 10, 8, 1}
Salida: 3
Enfoque ingenuo: el problema dado se puede resolver contando aquellos elementos cuya potencia más alta de 2 existe en la array dada y que se pueden encontrar recorriendo la array nuevamente. Después de verificar todos los elementos, imprima el recuento total obtenido.
Complejidad de Tiempo: O(N 2 )
Espacio Auxiliar: O(1)
Enfoque eficiente: el enfoque anterior también se puede optimizar mediante el uso de unordered_map para mantener el recuento de elementos visitados y actualizar el recuento de elementos resultantes en consecuencia. Siga los pasos a continuación para resolver el problema dado:
- Inicialice una variable, digamos count , que almacena el recuento de elementos cuya potencia más alta de 2 menor o igual que está presente en la array .
- Inicialice un mapa M y almacene la frecuencia de los elementos del arreglo .
- Recorra la array dada y para cada elemento, si la frecuencia de la potencia más alta de 2 de arr[i] que no excede el elemento existe en el mapa, entonces incremente el valor de count en 1.
- Después de completar los pasos anteriores, imprima el valor de conteo como el conteo resultante de elementos.
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 array elements
// whose highest power of 2 is less
// than or equal to that number is
// present in the given array
int countElement(int arr[], int N)
{
// Stores the resultant count
// of array elements
int count = 0;
// Stores frequency of
// visited array elements
unordered_map<int, int> m;
// Traverse the array
for (int i = 0; i < N; i++) {
m[arr[i]]++;
}
for (int i = 0; i < N; i++) {
// Calculate log base 2
// of the element arr[i]
int lg = log2(arr[i]);
// Highest power of 2 whose
// value is at most arr[i]
int p = pow(2, lg);
// Increment the count by 1
if (m[p]) {
count++;
}
}
// Return the resultant count
return count;
}
// Driver Code
int main()
{
int arr[] = { 3, 4, 6, 9 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << countElement(arr, N);
return 0;
}
Java
// Java program for the above approach
import java.util.HashMap;
import java.io.*;
class GFG{
static int log2(int N)
{
// Calculate log2 N indirectly
// using log() method
int result = (int)(Math.log(N) /
Math.log(2));
return result;
}
// Function to count array elements
// whose highest power of 2 is less
// than or equal to that number is
// present in the given array
static int countElement(int arr[], int N)
{
// Stores the resultant count
// of array elements
int count = 0;
// Stores frequency of
// visited array elements
HashMap<Integer, Integer> m = new HashMap<>();
// Traverse the array
for(int i = 0; i < N; i++)
{
if (m.containsKey(arr[i]))
{
m.put(arr[i], m.get(arr[i]) + 1);
}
else
{
m.put(arr[i], 1);
}
}
for(int i = 0; i < N; i++)
{
// Calculate log base 2
// of the element arr[i]
int lg = log2(arr[i]);
// Highest power of 2 whose
// value is at most arr[i]
int p = (int)Math.pow(2, lg);
// Increment the count by 1
if (m.containsKey(p))
{
count++;
}
}
// Return the resultant count
return count;
}
// Driver Code
public static void main (String[] args)
{
int arr[] = { 3, 4, 6, 9 };
int N = arr.length;
System.out.println(countElement(arr, N));
}
}
// This code is contributed by Potta Lokesh
Python3
# Python program for the above approach
from math import log2
# Function to count array elements
# whose highest power of 2 is less
# than or equal to that number is
# present in the given array
def countElement(arr, N):
# Stores the resultant count
# of array elements
count = 0
# Stores frequency of
# visited array elements
m = {}
# Traverse the array
for i in range(N):
m[arr[i]] = m.get(arr[i], 0) + 1
for i in range(N):
# Calculate log base 2
# of the element arr[i]
lg = int(log2(arr[i]))
# Highest power of 2 whose
# value is at most arr[i]
p = pow(2, lg)
# Increment the count by 1
if (p in m):
count += 1
# Return the resultant count
return count
# Driver Code
if __name__ == '__main__':
arr= [3, 4, 6, 9]
N = len(arr)
print (countElement(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 array elements
// whose highest power of 2 is less
// than or equal to that number is
// present in the given array
static int countElement(int []arr, int N)
{
// Stores the resultant count
// of array elements
int count = 1;
// Stores frequency of
// visited array elements
Dictionary<int,int> m = new Dictionary<int,int>();
// Traverse the array
for (int i = 0; i < N; i++) {
if(m.ContainsKey(arr[i]))
m[arr[i]]++;
else
m.Add(arr[i],1);
}
for(int i = 0; i < N; i++) {
// Calculate log base 2
// of the element arr[i]
int lg = (int)Math.Log(arr[i]);
// Highest power of 2 whose
// value is at most arr[i]
int p = (int)Math.Pow(2, lg);
// Increment the count by 1
if (m.ContainsKey(p)) {
count++;
}
}
// Return the resultant count
return count;
}
// Driver Code
public static void Main()
{
int []arr = { 3, 4, 6, 9 };
int N = arr.Length;
Console.Write(countElement(arr, N));
}
}
// This code is contributed by bgangwar59.
Javascript
<script>
// Javascript program for the above approach
// Function to count array elements
// whose highest power of 2 is less
// than or equal to that number is
// present in the given array
function countElement(arr, N) {
// Stores the resultant count
// of array elements
let count = 0;
// Stores frequency of
// visited array elements
let m = new Map();
// Traverse the array
for (let i = 0; i < N; i++) {
if (m.has(arr[i])) {
m.set(arr[i], m.get(arr[i]) + 1)
} else {
m.set(arr[i], 1)
}
}
for (let i = 0; i < N; i++) {
// Calculate log base 2
// of the element arr[i]
let lg = Math.floor(Math.log2(arr[i]));
// Highest power of 2 whose
// value is at most arr[i]
let p = Math.pow(2, lg);
// Increment the count by 1
if (m.get(p)) {
count++;
}
}
// Return the resultant count
return count;
}
// Driver Code
let arr = [3, 4, 6, 9];
let N = arr.length
document.write(countElement(arr, N));
// This code is contributed by _saurabh_jaiswal.
</script>
Producción:
2
Complejidad temporal: O(N)
Espacio auxiliar: O(N)
Publicación traducida automáticamente
Artículo escrito por subhammahato348 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA