Minimizar la suma de la array aplicando la operación XOR en todos los elementos de la array

Dada una array arr[] de N elementos enteros, la tarea es elegir un elemento X y aplicar la operación XOR en cada elemento de la array con X de modo que la suma de la array se minimice.
 

Entrada: arr[] = {3, 5, 7, 11, 15} 
Salida: 26 
Representación binaria de los elementos de la array son {0011, 0101, 0111, 1011, 1111} 
Tomamos xor de cada elemento con 7 para minimizar la suma. 
3 XOR 7 = 0100 (4) 
5 XOR 7 = 0010 (2) 
7 XOR 7 = 0000 (0) 
11 XOR 7 = 1100 (12) 
15 XOR 7 = 1000 (8) 
Suma = 4 + 2 + 0 + 12 + 8 = 26
Entrada: arr[] = {1, 2, 3, 4, 5} 
Salida: 14 
 

Planteamiento: La tarea es encontrar el elemento X con el que tenemos que tomar xor de cada elemento. 
 

• Convierta cada número en forma binaria y actualice la frecuencia del bit (0 o 1) en una array correspondiente a la posición de cada bit en el elemento de la array.
• Ahora, recorra la array y verifique si el elemento en el índice es más de n/2 (para ‘n’ elementos, verificamos si el bit establecido aparece más de n/2 en el índice), y posteriormente, obtenemos el elemento ‘X’
• Ahora, toma xor de ‘X’ con todos los elementos y devuelve la suma.

A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
const int MAX = 25;
 
// Function to return the minimized sum
int getMinSum(int arr[], int n)
{
    int bits_count[MAX], max_bit = 0, sum = 0, ans = 0;
 
    memset(bits_count, 0, sizeof(bits_count));
 
    // To store the frequency
    // of bit in every element
    for (int d = 0; d < n; d++) {
        int e = arr[d], f = 0;
        while (e > 0) {
            int rem = e % 2;
            e = e / 2;
            if (rem == 1) {
                bits_count[f] += rem;
            }
            f++;
        }
        max_bit = max(max_bit, f);
    }
 
    // Finding element X
    for (int d = 0; d < max_bit; d++) {
        int temp = pow(2, d);
        if (bits_count[d] > n / 2)
            ans = ans + temp;
    }
 
    // Taking XOR of elements and finding sum
    for (int d = 0; d < n; d++) {
        arr[d] = arr[d] ^ ans;
        sum = sum + arr[d];
    }
    return sum;
}
 
// Driver code
int main()
{
    int arr[] = { 3, 5, 7, 11, 15 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << getMinSum(arr, n);
 
    return 0;
}

// Java implementation of the approach
class GFG {
 
    static int MAX = 25;
 
    // Function to return the minimized sum
    static int getMinSum(int arr[], int n)
    {
        int bits_count[] = new int[MAX],
            max_bit = 0, sum = 0, ans = 0;
 
        // To store the frequency
        // of bit in every element
        for (int d = 0; d < n; d++) {
            int e = arr[d], f = 0;
            while (e > 0) {
                int rem = e % 2;
                e = e / 2;
                if (rem == 1) {
                    bits_count[f] += rem;
                }
                f++;
            }
            max_bit = Math.max(max_bit, f);
        }
 
        // Finding element X
        for (int d = 0; d < max_bit; d++) {
            int temp = (int)Math.pow(2, d);
            if (bits_count[d] > n / 2)
                ans = ans + temp;
        }
 
        // Taking XOR of elements and finding sum
        for (int d = 0; d < n; d++) {
            arr[d] = arr[d] ^ ans;
            sum = sum + arr[d];
        }
        return sum;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 3, 5, 7, 11, 15 };
        int n = arr.length;
        System.out.println(getMinSum(arr, n));
    }
}
 
// This code has been contributed by 29AjayKumar

# Python3 implementation of the approach
 
MAX = 25;
 
# Function to return the minimized sum
def getMinSum(arr, n) :
    bits_count = [0]* MAX
    max_bit = 0; sum = 0; ans = 0;
 
    # To store the frequency
    # of bit in every element
    for d in range(n) :
        e = arr[d]; f = 0;
        while (e > 0) :
            rem = e % 2;
            e = e // 2;
            if (rem == 1) :
                bits_count[f] += rem;
                 
            f += 1
             
        max_bit = max(max_bit, f);
     
 
    # Finding element X
    for d in range(max_bit) :
        temp = pow(2, d);
         
        if (bits_count[d] > n // 2) :
            ans = ans + temp;
 
 
    # Taking XOR of elements and finding sum
    for d in range(n) :
        arr[d] = arr[d] ^ ans;
        sum = sum + arr[d];
     
    return sum
     
 
# Driver code
if __name__ == "__main__" :
 
    arr = [ 3, 5, 7, 11, 15 ];
    n = len(arr);
 
    print(getMinSum(arr, n))
 
# This code is contributed by Ryuga

// C# implementation of the approach
using System;
 
class GFG {
 
    static int MAX = 25;
 
    // Function to return the minimized sum
    static int getMinSum(int[] arr, int n)
    {
        int[] bits_count = new int[MAX];
        int max_bit = 0, sum = 0, ans = 0;
 
        // To store the frequency
        // of bit in every element
        for (int d = 0; d < n; d++) {
            int e = arr[d], f = 0;
            while (e > 0) {
                int rem = e % 2;
                e = e / 2;
                if (rem == 1) {
                    bits_count[f] += rem;
                }
                f++;
            }
            max_bit = Math.Max(max_bit, f);
        }
 
        // Finding element X
        for (int d = 0; d < max_bit; d++) {
            int temp = (int)Math.Pow(2, d);
            if (bits_count[d] > n / 2)
                ans = ans + temp;
        }
 
        // Taking XOR of elements and finding sum
        for (int d = 0; d < n; d++) {
            arr[d] = arr[d] ^ ans;
            sum = sum + arr[d];
        }
        return sum;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int[] arr = { 3, 5, 7, 11, 15 };
        int n = arr.Length;
        Console.WriteLine(getMinSum(arr, n));
    }
}
 
/* This code contributed by PrinciRaj1992 */

<script>
// Javascript implementation of the approach
 
const MAX = 25;
 
// Function to return the minimized sum
function getMinSum(arr, n)
{
    let bits_count = new Array(MAX).fill(0),
    max_bit = 0, sum = 0, ans = 0;
 
    // To store the frequency
    // of bit in every element
    for (let d = 0; d < n; d++) {
        let e = arr[d], f = 0;
        while (e > 0) {
            let rem = e % 2;
            e = parseInt(e / 2);
            if (rem == 1) {
                bits_count[f] += rem;
            }
            f++;
        }
        max_bit = Math.max(max_bit, f);
    }
 
    // Finding element X
    for (let d = 0; d < max_bit; d++) {
        let temp = Math.pow(2, d);
        if (bits_count[d] > parseInt(n / 2))
            ans = ans + temp;
    }
 
    // Taking XOR of elements and finding sum
    for (let d = 0; d < n; d++) {
        arr[d] = arr[d] ^ ans;
        sum = sum + arr[d];
    }
    return sum;
}
 
// Driver code
    let arr = [ 3, 5, 7, 11, 15 ];
    let n = arr.length;
 
    document.write(getMinSum(arr, n));
 
</script>

26

Complejidad de tiempo: O(N*log(max_element)), ya que estamos usando bucles anidados, el bucle exterior atraviesa N veces y el bucle interior atraviesa log(max_element) veces. En el recorrido del bucle interno, estamos disminuyendo por división de piso de 2 en cada recorrido equivalente a 1+1/2+1/4+….1/2 ^max_element . Donde N es el número de elementos en la array y max_element es el elemento máximo presente en la array.

Espacio auxiliar: O(1), ya que no estamos utilizando ningún espacio adicional.