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 expressionint 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 codeint 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 arrayimport java.io.*; import java.util.Arrays; class GFG {// Function to find the minimum // value of the expressionstatic 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 codepublic 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 arrayimport sys# Function to find the minimum # value of the expressiondef 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 codeif __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 arrayusing System; class GFG{ // Function to find the minimum // value of the expressionstatic 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 Codepublic 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:
1
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 expressionint 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 codeint 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 arrayimport java.io.*; import java.util.Arrays; class GFG {// Function to find the minimum // value of the expressionstatic 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 codepublic 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 arrayimport sys# Function to find the minimum# value of the expressiondef 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 codeif __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 arrayusing System; class GFG{ // Function to find the minimum // value of the expressionstatic 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 Codepublic 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:
1
Time Complexity : O(N * log(N))
Auxiliary Space: O(1)
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