给定一个大小为 N 的数组 A,找到表达式的最小值: 在所有对 (i, j) (其中 i != j)。这里 , 和分别表示按位与、按位或、按位异或。
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,并在它们上找到给定表达式的最小可能值。
下面执行上面的方法。
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
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
输出:
1
时间复杂度: O(N 2 )
有效的方法
- 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异或值对,我们可以简单地使用本文中提到的算法来有效地找到相同的值。
下面执行上面的方法。
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
蟒蛇3
# 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
输出:
1
时间复杂度: O(N * log(N))
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