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

• Difficulty Level : Hard
• Last Updated : 22 Mar, 2021

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 and A) xor (A or A) = 1,
which is minimum possible value.

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

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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 ``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

 ``
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++

 `// 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`

## Javascript

 ``
Output:
`1`

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

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