在数组的所有对中找到给定表达式的最小值

给定大小为N的数组A，找到表达式的最小值： $(A_i \& A_j) \oplus (A_i | A_j)$在所有对(i，j)上(其中i！= j)。这里$\&$ ， $|$和$\oplus$分别代表按位AND，按位OR和按位XOR。
例子：

Input:  A = [1, 2, 3, 4, 5]
Output:  1
Explanation: 
(A[1] and A[2]) xor (A[1] or A[2]) = 1,
which is minimum possible value.

Input : A = [12, 3, 14, 5, 9, 8]
Output : 1

天真的方法：
遍历具有不同索引的数组的所有对，并在其上找到给定表达式的最小可能值。
下面执行上述方法。

C++

// C++ program to find the minimum
// value of the given expression
// over all pairs of the array
#include 
using namespace std;
 
// Function to find the minimum
// value of the expression
int MinimumValue(int a[], int n)
{
 
    int answer = INT_MAX;
     
    // Iterate over all the pairs
    // and find the minimum value
    for (int i = 0; i < n; i++)
    {
        for (int j = i + 1; j < n; j++)
        {
            answer = min(answer,
                         ((a[i] & a[j])
                          ^ (a[i] | a[j])));
        }
    }
    return answer;
}
// Driver code
int main()
{
    int N = 6;
    int A[N] = { 12, 3, 14, 5, 9, 8 };
 
    cout << MinimumValue(A, N);
 
    return 0;
}

Java

// Java program to find the minimum
// value of the given expression
// over all pairs of the array
import java.io.*;
import java.util.Arrays;
 
class GFG
{
 
// Function to find the minimum
// value of the expression
static int MinimumValue(int a[], int n)
{
 
    int answer = Integer.MAX_VALUE;
     
    // Iterate over all the pairs
    // and find the minimum value
    for (int i = 0; i < n; i++)
    {
        for (int j = i + 1; j < n; j++)
        {
            answer = Math.min(answer, ((a[i] & a[j])
                                  ^ (a[i] | a[j])));
        }
    }
    return answer;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 6;
    int[] A = new int[]{ 12, 3, 14, 5, 9, 8 };
    System.out.print(MinimumValue(A, N));
 
}
}
 
// This code is contributed by shivanisinghss2110

Python3

# Python3 program to find the minimum
# value of the given expression
# over all pairs of the array
import sys
 
# Function to find the minimum
# value of the expression
def MinimumValue(a,n):
    answer = sys.maxsize
     
    # Iterate over all the pairs
    # and find the minimum value
    for i in range(n):
        for j in range(i + 1, n, 1):
            answer = min(answer,((a[i] & a[j])^(a[i] | a[j])))
    return answer
 
# Driver code
if __name__ == '__main__':
    N = 6
    A = [12, 3, 14, 5, 9, 8]
 
    print(MinimumValue(A, N))
 
# This code is contributed by Bhupendra_Singh

C#

// C# program to find the minimum
// value of the given expression
// over all pairs of the array
using System;
 
class GFG{
 
// Function to find the minimum
// value of the expression
static int MinimumValue(int []a, int n)
{
    int answer = Int32.MaxValue;
     
    // Iterate over all the pairs
    // and find the minimum value
    for(int i = 0; i < n; i++)
    {
       for(int j = i + 1; j < n; j++)
       {
           answer = Math.Min(answer,
                           ((a[i] & a[j]) ^
                            (a[i] | a[j])));
       }
    }
    return answer;
}
 
// Driver Code
public static void Main()
{
    int N = 6;
    int[] A = new int[]{ 12, 3, 14, 5, 9, 8 };
    Console.Write(MinimumValue(A, N));
}
}
 
// This code is contributed by shivanisinghss2110

Javascript

// JavaScript program to find the minimum
// value of the given expression
// over all pairs of the array
 
// Function to find the minimum
// value of the expression
function MinimumValue(a, n)
{
 
    let answer = Number.MAX_SAFE_INTEGER;
     
    // Iterate over all the pairs
    // and find the minimum value
    for (let i = 0; i < n; i++)
    {
        for (let j = i + 1; j < n; j++)
        {
            answer = Math.min(answer,
                        ((a[i] & a[j])
                        ^ (a[i] | a[j])));
        }
    }
    return answer;
}
 
// Driver code
    let N = 6;
    let A = [ 12, 3, 14, 5, 9, 8 ];
    document.write(MinimumValue(A, N));
 
// This code is contributed by Surbhi Tyagi.

输出：

1

时间复杂度： O(N ² )
高效方法

• 让我们简化给定的表达式：

+ represents bitwise OR 
. represents bitwise AND 
^ represents bitwise XOR 
‘ represents 1’s complement
(x . y) ^ (x + y) = (x . y) . (x + y)’ + (x . y)’ . (x + y) (using definition of XOR) 
= (x . y) . (x’ . y’) + (x’+ y’) . (x + y) (De morgan’s law) 
= x.x’.y.y’ + x’.x + x’.y + y’.x + y.y 
= 0 + 0 + x’.y + y’.x + 0 
= x ^ y 
 

• 由于表达式简化为最小xor值对，因此我们可以简单地使用本文中提到的算法来高效地找到相同的值。

下面执行上述方法。

C++

// C++ program to find the minimum
// value of the given expression
// over all pairs of the array
#include 
using namespace std;
 
// Function to find the minimum
// value of the expression
int MinimumValue(int arr[], int n)
{
 
    // The expression simplifies to
    // finding the minimum xor
    // value pair
 
    // Sort given array
    sort(arr, arr + n);
 
    int minXor = INT_MAX;
    int val = 0;
 
    // Calculate min xor of
    // consecutive pairs
    for (int i = 0; i < n - 1; i++)
    {
        val = arr[i] ^ arr[i + 1];
        minXor = min(minXor, val);
    }
 
    return minXor;
}
 
// Driver code
int main()
{
    int N = 6;
    int A[N] = { 12, 3, 14, 5, 9, 8 };
 
    cout << MinimumValue(A, N);
 
    return 0;
}

Java

// Java program to find the minimum
// value of the given expression
// over all pairs of the array
import java.io.*;
import java.util.Arrays;
 
class GFG {
 
// Function to find the minimum
// value of the expression
static int MinimumValue(int arr[], int n)
{
     
    // The expression simplifies to
    // finding the minimum xor
    // value pair
    // Sort given array
    Arrays.sort(arr);
 
    int minXor = Integer.MAX_VALUE;
    int val = 0;
 
    // Calculate min xor of
    // consecutive pairs
    for(int i = 0; i < n - 1; i++)
    {
       val = arr[i] ^ arr[i + 1];
       minXor = Math.min(minXor, val);
    }
 
    return minXor;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 6;
    int[] A = new int[]{ 12, 3, 14, 5, 9, 8 };
 
    System.out.print(MinimumValue(A, N));
}
}
 
// This code is contributed by shivanisinghss2110

Python3

# Python3 program to find the minimum
# value of the given expression
# over all pairs of the array
import sys
 
# Function to find the minimum
# value of the expression
def MinimumValue(arr, n):
 
    # The expression simplifies
    # to finding the minimum
    # xor value pair
    # Sort given array
    arr.sort();
 
    minXor = sys.maxsize;
    val = 0;
 
    # Calculate min xor of
    # consecutive pairs
    for i in range(0, n - 1):
        val = arr[i] ^ arr[i + 1];
        minXor = min(minXor, val);
     
    return minXor;
 
# Driver code
if __name__ == '__main__':
     
    N = 6;
    A = [ 12, 3, 14, 5, 9, 8 ];
 
    print(MinimumValue(A, N));
     
# This code is contributed by sapnasingh4991

C＃

// C# program to find the minimum
// value of the given expression
// over all pairs of the array
using System;
 
class GFG{
 
// Function to find the minimum
// value of the expression
static int MinimumValue(int []arr, int n)
{
     
    // The expression simplifies to
    // finding the minimum xor
    // value pair
    // Sort given array
    Array.Sort(arr);
 
    int minXor = Int32.MaxValue;
    int val = 0;
 
    // Calculate min xor of
    // consecutive pairs
    for(int i = 0; i < n - 1; i++)
    {
       val = arr[i] ^ arr[i + 1];
       minXor = Math.Min(minXor, val);
    }
     
    return minXor;
}
 
// Driver Code
public static void Main()
{
    int N = 6;
    int[] A = new int[]{ 12, 3, 14, 5, 9, 8 };
 
    Console.Write(MinimumValue(A, N));
}
}
 
// This code is contributed by shivanisinghss2110

Java脚本

    // Javascript program to find the minimum
    // value of the given expression
    // over all pairs of the array
     
    // Function to find the minimum
    // value of the expression
    function MinimumValue(arr, n)
    {
 
        // The expression simplifies to
        // finding the minimum xor
        // value pair
 
        // Sort given array
        arr.sort();
 
        let minXor = Number.MAX_VALUE;
        let val = 0;
 
        // Calculate min xor of
        // consecutive pairs
        for (let i = 0; i < n - 1; i++)
        {
            val = arr[i] ^ arr[i + 1];
            minXor = Math.min(minXor, val);
        }
 
        return minXor;
    }
     
    let N = 6;
    let A = [ 12, 3, 14, 5, 9, 8 ];
  
    document.write(MinimumValue(A, N));
     
    //This code is contributed by divyeshrabadiya07

输出：

1

时间复杂度： O(N * log(N))

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