求一个数组中 k 个元素的最大异或
原文:https://www . geeksforgeeks . org/find-max-xor-of-k-elements in-a-array/
给定一个由 N 个整数和一个整数 K 组成的数组 arr[] 。任务是找到给定数组的大小 K 的最大异或子集。 举例:
输入: arr[] = {2，5，4，1，3，7，6，8}，K = 3 输出: 15 我们通过选择 4，5，6，8 输入: arr[] = {3，4，7，7，9}，K = 3 输出: 14
天真方法:迭代数组的所有大小子集 K ，并在它们之间找到最大异或。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the maximum xor for a
// subset of size k from the given array
int Max_Xor(int arr[], int n, int k)
{
// Initialize result
int maxXor = INT_MIN;
// Traverse all subsets of the array
for (int i = 0; i < (1 << n); i++) {
// __builtin_popcount() returns the number
// of sets bits in an integer
if (__builtin_popcount(i) == k) {
// Initialize current xor as 0
int cur_xor = 0;
for (int j = 0; j < n; j++) {
// If jth bit is set in i then include
// jth element in the current xor
if (i & (1 << j))
cur_xor = cur_xor ^ arr[j];
}
// Update maximum xor so far
maxXor = max(maxXor, cur_xor);
}
}
return maxXor;
}
// Driver code
int main()
{
int arr[] = { 2, 5, 4, 1, 3, 7, 6, 8 };
int n = sizeof(arr) / sizeof(int);
int k = 3;
cout << Max_Xor(arr, n, k);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return the maximum xor for a
// subset of size k from the given array
static int Max_Xor(int arr[], int n, int k)
{
// Initialize result
int maxXor = Integer.MIN_VALUE;
// Traverse all subsets of the array
for (int i = 0; i < (1 << n); i++)
{
// __builtin_popcount() returns the number
// of sets bits in an integer
if (Integer.bitCount(i) == k)
{
// Initialize current xor as 0
int cur_xor = 0;
for (int j = 0; j < n; j++)
{
// If jth bit is set in i then include
// jth element in the current xor
if ((i & (1 << j)) == 0)
cur_xor = cur_xor ^ arr[j];
}
// Update maximum xor so far
maxXor = Math.max(maxXor, cur_xor);
}
}
return maxXor;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 2, 5, 4, 1, 3, 7, 6, 8 };
int n = arr.length;
int k = 3;
System.out.println(Max_Xor(arr, n, k));
}
}
// This code is contributed by PrinciRaj1992
Python 3
# Python implementation of the approach
MAX = 10000
MAX_ELEMENT = 50
dp =[[[-1 for i in range(MAX)]
for j in range(MAX_ELEMENT)]
for k in range(MAX_ELEMENT)]
# Function to return the maximum xor for a
# subset of size j from the given array
def Max_Xor(arr, i, j, mask, n):
if (i >= n):
# If the subset is complete then return
# the xor value of the selected elements
if (j == 0):
return mask
else:
return 0
# Return if already calculated for some
# mask and j at the i'th index
if (dp[i][j][mask] != -1):
return dp[i][j][mask]
# Initialize answer to 0
ans = 0
# If we can still include elements in our subset
# include the i'th element
if (j > 0):
ans = Max_Xor(arr, i + 1, j - 1, mask ^ arr[i], n)
# Exclude the i'th element
# ans store the max of both operations
ans = max(ans, Max_Xor(arr, i + 1, j, mask, n))
dp[i][j][mask] = ans
return ans
# Driver code
arr = [2, 5, 4, 1, 3, 7, 6, 8]
n = len(arr)
k = 3
print(Max_Xor(arr, 0, k, 0, n))
# This code is contributed by shubhamsingh10
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the maximum xor for a
// subset of size k from the given array
static int Max_Xor(int []arr, int n, int k)
{
// Initialize result
int maxXor = int.MinValue;
// Traverse all subsets of the array
for (int i = 0; i < (1 << n); i++)
{
// __builtin_popcount() returns the number
// of sets bits in an integer
if (bitCount(i) == k)
{
// Initialize current xor as 0
int cur_xor = 0;
for (int j = 0; j < n; j++)
{
// If jth bit is set in i then include
// jth element in the current xor
if ((i & (1 << j)) == 0)
cur_xor = cur_xor ^ arr[j];
}
// Update maximum xor so far
maxXor = Math.Max(maxXor, cur_xor);
}
}
return maxXor;
}
static int bitCount(long x)
{
int setBits = 0;
while (x != 0)
{
x = x & (x - 1);
setBits++;
}
return setBits;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 2, 5, 4, 1, 3, 7, 6, 8 };
int n = arr.Length;
int k = 3;
Console.WriteLine(Max_Xor(arr, n, k));
}
}
// This code is contributed by Princi Singh
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the maximum xor for a
// subset of size k from the given array
function Max_Xor(arr, n, k)
{
// Initialize result
let maxXor = Number.MIN_VALUE;
// Traverse all subsets of the array
for (let i = 0; i < (1 << n); i++) {
// bitCount() returns the number
// of sets bits in an integer
if (bitCount(i) == k) {
// Initialize current xor as 0
let cur_xor = 0;
for (let j = 0; j < n; j++) {
// If jth bit is set in i then include
// jth element in the current xor
if (i & (1 << j))
cur_xor = cur_xor ^ arr[j];
}
// Update maximum xor so far
maxXor = Math.max(maxXor, cur_xor);
}
}
return maxXor;
}
function bitCount(x)
{
let setBits = 0;
while (x != 0)
{
x = x & (x - 1);
setBits++;
}
return setBits;
}
// Driver code
let arr = [ 2, 5, 4, 1, 3, 7, 6, 8 ];
let n = arr.length;
let k = 3;
document.write(Max_Xor(arr, n, k));
</script>
Output:
15
高效途径:问题可以使用动态规划解决。创建一个 dp 表 dp[i][j][mask] ，该表在第I索引处(包括或不包括)存储可能的最大异或，并且 j 表示我们可以包含在 K 元素的子集中的剩余元素的数量。掩码是直到第 i 第T17】个索引为止所选所有元素的异或。 注意:由于 dp 阵列的空间要求，这种方法仅适用于较小的阵列。 以下是上述办法的实施:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define MAX 10000
#define MAX_ELEMENT 50
int dp[MAX_ELEMENT][MAX_ELEMENT][MAX];
// Function to return the maximum xor for a
// subset of size j from the given array
int Max_Xor(int arr[], int i, int j, int mask, int n)
{
if (i >= n) {
// If the subset is complete then return
// the xor value of the selected elements
if (j == 0)
return mask;
else
return 0;
}
// Return if already calculated for some
// mask and j at the i'th index
if (dp[i][j][mask] != -1)
return dp[i][j][mask];
// Initialize answer to 0
int ans = 0;
// If we can still include elements in our subset
// include the i'th element
if (j > 0)
ans = Max_Xor(arr, i + 1, j - 1, mask ^ arr[i], n);
// Exclude the i'th element
// ans store the max of both operations
ans = max(ans, Max_Xor(arr, i + 1, j, mask, n));
return dp[i][j][mask] = ans;
}
// Driver code
int main()
{
int arr[] = { 2, 5, 4, 1, 3, 7, 6, 8 };
int n = sizeof(arr) / sizeof(int);
int k = 3;
memset(dp, -1, sizeof(dp));
cout << Max_Xor(arr, 0, k, 0, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.util.*;
class GFG
{
static int MAX = 10000;
static int MAX_ELEMENT = 50;
static int [][][] dp = new int[MAX_ELEMENT][MAX_ELEMENT][MAX];
// Function to return the maximum xor for a
// subset of size j from the given array
static int Max_Xor(int arr[], int i,
int j, int mask, int n)
{
if (i >= n)
{
// If the subset is complete then return
// the xor value of the selected elements
if (j == 0)
return mask;
else
return 0;
}
// Return if already calculated for some
// mask and j at the i'th index
if (dp[i][j][mask] != -1)
return dp[i][j][mask];
// Initialize answer to 0
int ans = 0;
// If we can still include elements in our subset
// include the i'th element
if (j > 0)
ans = Max_Xor(arr, i + 1, j - 1,
mask ^ arr[i], n);
// Exclude the i'th element
// ans store the max of both operations
ans = Math.max(ans, Max_Xor(arr, i + 1, j,
mask, n));
return dp[i][j][mask] = ans;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 2, 5, 4, 1, 3, 7, 6, 8 };
int n = arr.length;
int k = 3;
for(int i = 0; i < MAX_ELEMENT; i++)
{
for(int j = 0; j < MAX_ELEMENT; j++)
{
for(int l = 0; l < MAX; l++)
dp[i][j][l] = -1;
}
}
System.out.println(Max_Xor(arr, 0, k, 0, n));
}
}
// This code is contributed by Princi Singh
Python 3
# Python implementation of the approach
MAX = 10000
MAX_ELEMENT = 50
dp =[[[-1 for i in range(MAX)] for j in range(MAX_ELEMENT)] for k in range(MAX_ELEMENT)]
# Function to return the maximum xor for a
# subset of size j from the given array
def Max_Xor(arr, i, j, mask, n):
if (i >= n):
# If the subset is complete then return
# the xor value of the selected elements
if (j == 0):
return mask
else:
return 0
# Return if already calculated for some
# mask and j at the i'th index
if (dp[i][j][mask] != -1):
return dp[i][j][mask]
# Initialize answer to 0
ans = 0
# If we can still include elements in our subset
# include the i'th element
if (j > 0):
ans = Max_Xor(arr, i + 1, j - 1, mask ^ arr[i], n)
# Exclude the i'th element
# ans store the max of both operations
ans = max(ans, Max_Xor(arr, i + 1, j, mask, n))
dp[i][j][mask] = ans
return ans
# Driver code
arr = [2, 5, 4, 1, 3, 7, 6, 8]
n = len(arr)
k = 3
print(Max_Xor(arr, 0, k, 0, n))
# This code is contributed by shubhamsingh10
C
// C# implementation of the approach
using System;
class GFG
{
static int MAX = 10000;
static int MAX_ELEMENT = 50;
static int [,,] dp = new int[MAX_ELEMENT, MAX_ELEMENT, MAX];
// Function to return the maximum xor for a
// subset of size j from the given array
static int Max_Xor(int[] arr, int i,
int j, int mask, int n)
{
if (i >= n)
{
// If the subset is complete then return
// the xor value of the selected elements
if (j == 0)
return mask;
else
return 0;
}
// Return if already calculated for some
// mask and j at the i'th index
if (dp[i,j,mask] != -1)
return dp[i,j,mask];
// Initialize answer to 0
int ans = 0;
// If we can still include elements in our subset
// include the i'th element
if (j > 0)
ans = Max_Xor(arr, i + 1, j - 1,
mask ^ arr[i], n);
// Exclude the i'th element
// ans store the max of both operations
ans = Math.Max(ans, Max_Xor(arr, i + 1, j,
mask, n));
return dp[i,j,mask] = ans;
}
// Driver code
public static void Main ()
{
int[] arr = { 2, 5, 4, 1, 3, 7, 6, 8 };
int n = arr.Length;
int k = 3;
for(int i = 0; i < MAX_ELEMENT; i++)
{
for(int j = 0; j < MAX_ELEMENT; j++)
{
for(int l = 0; l < MAX; l++)
dp[i,j,l] = -1;
}
}
Console.WriteLine(Max_Xor(arr, 0, k, 0, n));
}
}
// This code is contributed by target_2.
Output:
15
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