Cuente los bits establecidos totales en una array

Dada una array arr , la tarea es contar el número total de bits establecidos en todos los números de esa array arr .

Ejemplo:

Entrada: arr[] = {1, 2, 5, 7}
Salida: 7
Explicación: El número de bits establecidos en {1, 2, 5, 7} son {1, 1, 2, 3} respectivamente

Entrada: arr[] = {0, 4, 9, 8}
Salida: 4

Enfoque: siga los pasos a continuación para resolver este problema:

1. Cree una variable cnt para almacenar la respuesta e inicialícela con 0 .
2. Recorra cada elemento de la array arr .
3. Ahora, para cada elemento, digamos x , ejecuta un bucle mientras sea mayor que 0 .
4. Extraiga el último bit de x usando (x&1) y luego desplace x a la derecha un solo bit.
5. Devuelve cnt como la respuesta a este problema.

A continuación se muestra la implementación del enfoque anterior:

// C++ code for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the total number of set bits
// in an array of integers
int totalSetBits(vector<int>& arr)
{
    int cnt = 0;
    for (auto x : arr) {
 
        // While x is greater than 0
        while (x > 0) {
 
            // Adding last bit to cnt
            cnt += (x & 1);
 
            // Right shifting x by a single bit
            x >>= 1;
        }
    }
    return cnt;
}
 
// Driver Code
int main()
{
    vector<int> arr = { 1, 2, 5, 7 };
    cout << totalSetBits(arr);
}

// Java code for the above approach
import java.util.*;
 
class GFG{
 
// Function to count the total number of set bits
// in an array of integers
static int totalSetBits(int[] arr)
{
    int cnt = 0;
    for (int x : arr) {
 
        // While x is greater than 0
        while (x > 0) {
 
            // Adding last bit to cnt
            cnt += (x & 1);
 
            // Right shifting x by a single bit
            x >>= 1;
        }
    }
    return cnt;
}
 
// Driver Code
public static void main(String[] args)
{
    int[] arr = { 1, 2, 5, 7 };
    System.out.print(totalSetBits(arr));
}
}
 
// This code is contributed by shikhasingrajput

# python code for the above approach
 
# Function to count the total number of set bits
# in an array of integers
def totalSetBits(arr):
 
    cnt = 0
    for x in arr:
 
        # While x is greater than 0
        while (x > 0):
 
            # Adding last bit to cnt
            cnt += (x & 1)
 
            # Right shifting x by a single bit
            x >>= 1
    return cnt
 
# Driver Code
if __name__ == "__main__":
 
    arr = [1, 2, 5, 7]
    print(totalSetBits(arr))
 
# This code is contributed by rakeshsahni

// C# code for the above approach
using System;
 
class GFG {
 
    // Function to count the total number of set bits
    // in an array of integers
    static int totalSetBits(int[] arr)
    {
        int cnt = 0;
        for (int x = 0; x < arr.Length; x++) {
 
            // While x is greater than 0
            while (arr[x] > 0) {
 
                // Adding last bit to cnt
                cnt += (arr[x] & 1);
 
                // Right shifting x by a single bit
                arr[x] >>= 1;
            }
        }
        return cnt;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        int[] arr = { 1, 2, 5, 7 };
        Console.WriteLine(totalSetBits(arr));
    }
}
 
// This code is contributed by ukasp.

<script>
 
// JavaScript code for the above approach
 
// Function to count the total number of set bits
// in an array of integers
function totalSetBits(arr)
{
    let cnt = 0;
    for(let x of arr)
    {
         
        // While x is greater than 0
        while (x > 0)
        {
             
            // Adding last bit to cnt
            cnt += (x & 1);
 
            // Right shifting x by a single bit
            x >>= 1;
        }
    }
    return cnt;
}
 
// Driver Code
let arr = [ 1, 2, 5, 7 ];
 
document.write(totalSetBits(arr));
 
// This code is contributed by Potta Lokesh
 
</script>

7

Complejidad temporal: O(N)
Espacio auxiliar: O(1)