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 : over all pairs (i, j) (where i != j). Here , and 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++

// 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;
}

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# 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

<script>
// 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.
</script>

Output:

Time Complexity : O(N²)

Auxiliary Space: O(1)
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++

// 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;
}

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# 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

Javascript

<script>

    // 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
</script>

Output:

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

Auxiliary Space: O(1)

Article Tags :

Algorithms

Bit Magic

DSA

Greedy

Mathematical