Find the minimum value of the given expression over all pairs of the array

Given an array A of size N, find the minimum value of the expression :  (A_i \& A_j) \oplus (A_i | A_j) over all pairs (i, j) (where i != j). Here  \& ,  | and  \oplus represent bitwise AND, bitwise OR and bitwise XOR resprectively.

Examples:

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

Naive Approach:
Iterate through all the pairs of the array with different index and find the minimum possible value of given expression over them.

Below the implementation of the above approach.

C++

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// C++ program to find the minimum
// value of the given expression
// over all pairs of the array
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:



1

Time Complexity : O(N2)

Efficient Approach

  • Let’s simplify the given expression :

    + 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

  • Since the expression simplifies to minimum xor value pair, we can simply use the algorithm mentioned in this article to find the same efficiently.

Below the implementation of the above approach.

C++

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// C++ program to find the minimum
// value of the given expression
// over all pairs of the array
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

1

Time Complexity : O(N * log(N))

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