给定N个元素的数组，任务是找到K的最小值，以使K与所有数组元素的按位XOR运算产生相同的元素集。如果找不到任何K值，则打印“ -1” 。
例子：
Input: arr[] = { 1, 0, 2, 3}
Output: 1
Explanation:
For K = 1,
1 xor 1 = 0
1 xor 0 = 1
1 xor 2 = 3
1 xor 3 = 2
Thus, K = 1 is the least possible positive value which leaves the array unaltered.
Input: arr[] = { 7, 1, 2, 3, 8}
Output: -1
幼稚的方法：幼稚的方法是对K在[1，1024]范围内的所有可能值进行迭代，并检查K与数组中所有元素的按位XOR是否提供相同的数组元素。如果对于K的任何最小值，按位XOR生成相同的数组，则打印该K值，否则打印“ -1” 。
时间复杂度： O(K * N 2 )
辅助空间： O(1)
高效的方法：可以通过使用额外的空间来优化上述方法。步骤如下：
- 将所有元素插入集合。
- 对[0，1024]范围内的K的所有可能值进行迭代。
- 对于集合中的每个元素，找到与K的按位XOR。
- 按位XOR之后与集合中的元素生成的所有元素的K的第一个值与给定集合的值相同，然后打印K的值。
- 如果没有获得这样的K ，则打印“ -1” 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the minimum
// value of K in given range
int min_value(int arr[], int N)
{
int x, X, K;
// Declare a set
set S;
for (int i = 0; i < N; i++) {
S.insert(arr[i]);
}
// Initialize count variable
int count = 0;
// Iterate in range [1, 1024]
for (int i = 1; i <= 1024; i++) {
// counter set as 0
count = 0;
// Iterating through the Set
for (auto it = S.begin(); it != S.end(); it++)
// Check if the XOR
// calculated is present
// in the Set
{
X = ((i | *it) - (i & *it));
// If the value of Bitwise XOR
// inside the given set then
// increment count
if (S.find(X) != S.end()) {
count++;
}
}
// Check if the value of count is
// equal to the size of set
if (count == S.size()) {
K = i;
// Return minimum value of K
return K;
}
}
// If no such K is found
return -1;
}
// Driver Code
int main()
{
// Given array
int arr[] = { 1, 0, 3, 3, 0, 2 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
cout << min_value(arr, N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG {
// Function to find the minimum
// value of K in given range
static int min_value(int arr[], int N)
{
int x, X, K;
// Declare a set
HashSet S = new HashSet();
for (int i = 0; i < N; i++) {
S.add(arr[i]);
}
// Initialize count variable
int count = 0;
// Iterate in range [1, 1024]
for (int i = 1; i <= 1024; i++) {
// counter set as 0
count = 0;
// Iterating through the Set
for (int it : S) {
// Check if the XOR
// calculated is present
// in the Set
X = ((i | it) - (i & it));
// If the value of Bitwise XOR
// inside the given set then
// increment count
if (S.contains(X)) {
count++;
}
}
// Check if the value of count is
// equal to the size of set
if (count == S.size()) {
K = i;
// Return minimum value of K
return K;
}
}
// If no such K is found
return -1;
}
// Driver Code
public static void main(String[] args)
{
// Given array
int arr[] = { 1, 0, 3, 3, 0, 2 };
int N = arr.length;
// Function Call
System.out.print(min_value(arr, N));
}
}
// This code is contributed by Rajput-Ji Python3
# Python3 program for the above approach
# Function to find the minimum
# value of K in given range
def min_value(arr, N):
x, X, K = 0, 0, 0
# Declare a set
S = set()
for i in range(N):
S.add(arr[i])
# Initialize count variable
count = 0
# Iterate in range [1, 1024]
for i in range(1, 1024):
# counter set as 0
count = 0
# Iterating through the Set
for it in S:
# Check if the XOR
# calculated is present
# in the Set
X = ((i | it) - (i & it))
# If the value of Bitwise XOR
# inside the given set then
# increment count
if X in S:
count += 1
# Check if the value of count is
# equal to the size of set
if (count == len(S)):
K = i
# Return minimum value of K
return K
# If no such K is found
return -1
# Driver Code
# Given array
arr = [1, 0, 3, 3, 0, 2]
N = len(arr)
# Function Call
print(min_value(arr, N))
# This code is contributed by shubhamsingh10C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
// Function to find the minimum
// value of K in given range
static int min_value(int[] arr, int N)
{
// int x;
int X, K;
// Declare a set
HashSet S = new HashSet();
for (int i = 0; i < N; i++) {
S.Add(arr[i]);
}
// Initialize count variable
int count = 0;
// Iterate in range [1, 1024]
for (int i = 1; i <= 1024; i++) {
// counter set as 0
count = 0;
// Iterating through the Set
foreach(int it in S)
{
// Check if the XOR
// calculated is present
// in the Set
X = ((i | it) - (i & it));
// If the value of Bitwise XOR
// inside the given set then
// increment count
if (S.Contains(X)) {
count++;
}
}
// Check if the value of count is
// equal to the size of set
if (count == S.Count) {
K = i;
// Return minimum value of K
return K;
}
}
// If no such K is found
return -1;
}
// Driver code
static void Main()
{
// Given array
int[] arr = { 1, 0, 3, 3, 0, 2 };
int N = arr.Length;
// Function call
Console.Write(min_value(arr, N));
}
}
// This code is contributed by divyeshrabadiya07 输出：
1
时间复杂度： O(K * N * log 2 N)
辅助空间： O(1)