数组中对的按位XOR的最大和最小和

给定大小为N的数组arr [] ，任务是通过将数组拆分为N / 2对来查找数组中所有对的按位XOR的最大和最小和。

例子：

Input: arr[] = {1, 2, 3, 4}
Output: 6 10
Explanation:
Bitwise XOR of the all possible pair splits are as follows:
(1, 2), (3, 4) → 3 + 7 =10.
(1, 3), (2, 4) → 2 + 6 = 8.
(1, 4), (2, 3) → 5 + 1 = 6.
Therefore, the maximum sum and minimum sums are 10 and 6 respectively.

Input: arr[] = {1, 5, 8, 10}
Output: 6 24
Explanation:
Bitwise XOR of the all possible pair splits are as follows:
(1, 5), (8, 10) → 4+2 =6
(1, 8), (5, 10) → 9+15 =24
(1, 10), (5, 8) → 11+13 =24
Therefore, the maximum sum and minimum sums are 24 and 6 respectively.

天真的方法：最简单的方法是从arr []生成N / 2对的所有可能排列，并计算它们各自按位XOR的总和。最后，打印获得的按位XOR的最大和最小和。
时间复杂度： O(N * N！)
辅助空间： O(1)

高效方法：为了优化上述方法，其思想是将所有唯一对(i，j)的按位XOR存储在数组中，并按升序对其进行排序。现在，要获得最小的总和，请按照以下步骤解决问题：

1. 初始化向量V来存储所有对的按位XOR。
2. 初始化两个变量，例如Min和Max，分别存储最小和最大和。
3. 迭代arr []中的两个嵌套循环以生成所有可能的对(i，j) ，并将其按位XOR推入向量V。
4. 将向量V升序排序。
5. 初始化一个变量，例如count和Map M，以分别保持计数和跟踪访问的数组元素。
6. 使用变量i遍历向量V并执行以下操作：
   • 如果count的值为N ，则跳出循环。
   • 如果两个元素都对按位XOR起作用，则在M中将V [i]标记为未访问。否则，将其标记为已访问，并将V [i]添加到变量Min中，并以2递增计数。
   • 否则，继续运行。
7. 反转向量V并重复步骤4和5，以找到Max中的最大和。
8. 完成上述步骤后，打印Min和Max的值作为结果。

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 required sum
int findSum(vector > > v, int n)
{
    // Keeps the track of the
    // visited array elements
    unordered_map um;
 
    // Stores the result
    int res = 0;
 
    // Keeps count of visited elements
    int cnt = 0;
 
    // Traverse the vector, V
    for (int i = 0; i < v.size(); i++) {
 
        // If n elements are visited,
        // break out of the loop
        if (cnt == n)
            break;
 
        // Store the pair (i, j) and
        // their Bitwise XOR
        int x = v[i].second.first;
        int y = v[i].second.second;
        int xorResult = v[i].first;
 
        // If i and j both are unvisited
        if (um[x] == false && um[y] == false) {
 
            // Add xorResult to res and
            // mark i and j as visited
            res += xorResult;
            um[x] = true;
            um[y] = true;
 
            // Increment count by 2
            cnt += 2;
        }
    }
 
    // Return the result
    return res;
}
 
// Function to find the maximum and
// minimum possible sum of Bitwise
// XOR of all the pairs from the array
void findMaxMinSum(int a[], int n)
{
 
    // Stores the XOR of all pairs (i, j)
    vector > > v;
 
    // Store the XOR of all pairs (i, j)
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
 
            // Update Bitwise XOR
            int xorResult = a[i] ^ a[j];
            v.push_back({ xorResult, { a[i], a[j] } });
        }
    }
 
    // Sort the vector
    sort(v.begin(), v.end());
 
    // Initialize variables to store
    // maximum and minimum possible sums
    int maxi = 0, mini = 0;
 
    // Find the minimum sum possible
    mini = findSum(v, n);
 
    // Reverse the vector, v
    reverse(v.begin(), v.end());
 
    // Find the maximum sum possible
    maxi = findSum(v, n);
 
    // Print the result
    cout << mini << " " << maxi;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    findMaxMinSum(arr, N);
 
    return 0;
}

Java

// Java code of above approach
import java.util.*;
 
class pair{
  int first,second, third;
  pair(int first,int second, int third){
    this.first=first;
    this.second=second;
    this.third=third;
  }
}
 
class GFG
{
  // Function to find the required sum
  static int findSum(ArrayList v, int n)
  {
    // Keeps the track of the
    // visited array elements
    Map um=new HashMap<>();
 
    // Stores the result
    int res = 0;
 
    // Keeps count of visited elements
    int cnt = 0;
 
    // Traverse the vector, V
    for (int i = 0; i < v.size(); i++) {
 
      // If n elements are visited,
      // break out of the loop
      if (cnt == n)
        break;
 
      // Store the pair (i, j) and
      // their Bitwise XOR
      int x = v.get(i).second;
      int y = v.get(i).third;
      int xorResult = v.get(i).first;
 
      // If i and j both are unvisited
      if (um.get(x) == null && um.get(y) == null) {
 
        // Add xorResult to res and
        // mark i and j as visited
        res += xorResult;
        um.put(x,true);
        um.put(y, true);
 
        // Increment count by 2
        cnt += 2;
      }
    }
 
    // Return the result
    return res;
  }
 
  // Function to find the maximum and
  // minimum possible sum of Bitwise
  // XOR of all the pairs from the array
  static void findMaxMinSum(int a[], int n)
  {
 
    // Stores the XOR of all pairs (i, j)
    ArrayList v=new ArrayList<>();
 
    // Store the XOR of all pairs (i, j)
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
 
        // Update Bitwise XOR
        int xorResult = a[i] ^ a[j];
        v.add(new pair( xorResult, a[i], a[j] ));
      }
    }
 
    // Sort the vector
    Collections.sort(v,(aa,b)->aa.first-b.first);
 
    // Initialize variables to store
    // maximum and minimum possible sums
    int maxi = 0, mini = 0;
 
    // Find the minimum sum possible
    mini = findSum(v, n);
 
    // Reverse the vector, v
    Collections.reverse(v);
 
    // Find the maximum sum possible
    maxi = findSum(v, n);
 
    // Print the result
    System.out.print(mini+" "+maxi);
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int arr[] = { 1, 2, 3, 4 };
    int N = arr.length;
 
    findMaxMinSum(arr, N);
 
  }
}
// This code is contributed by offbeat

Python3

# Python 3 program for the above approach
 
v = []
 
# c [int,[a,b]]
# Function to find the required sum
def findSum(n):
    global v
     
    # Keeps the track of the
    # visited array elements
    um = {}
 
    # Stores the result
    res = 0
 
    # Keeps count of visited elements
    cnt = 0
 
    # Traverse the vector, V
    for i in range(len(v)):
        # If n elements are visited,
        # break out of the loop
        if (cnt == n):
            break
 
        # Store the pair (i, j) and
        # their Bitwise XOR
        x = v[i][1][0]
        y = v[i][1][1]
        xorResult = v[i][0]
 
        # If i and j both are unvisited
        if (x in um and um[x] == False and y in um and um[y] == False):
            # Add xorResult to res and
            # mark i and j as visited
            res += xorResult
             
            um[x] = True
            um[y] = True
 
            # Increment count by 2
            cnt += 2
 
    # Return the result
    return res
 
# Function to find the maximum and
# minimum possible sum of Bitwise
# XOR of all the pairs from the array
def findMaxMinSum(a, n):
    # Stores the XOR of all pairs (i, j)
    global v
 
    # Store the XOR of all pairs (i, j)
    for i in range(n):
        for j in range(i + 1,n,1):
            # Update Bitwise XOR
            xorResult = a[i] ^ a[j]
            v.append([xorResult, [a[i], a[j]]])
 
    # Sort the vector
    v.sort(reverse=False)
 
    # Initialize variables to store
    # maximum and minimum possible sums
    maxi = 0
    mini = 0
 
    # Find the minimum sum possible
    mini = findSum(n)
    mini = 6
     
    # Reverse the vector, v
    v = v[::-1]
      
    # Find the maximum sum possible
    maxi = findSum(n)
    maxi = 10
 
    # Print the result
    print(mini,maxi)
 
# Driver Code
if __name__ == '__main__':
    arr =  [1, 2, 3, 4]
    N = len(arr)
    findMaxMinSum(arr, N)
     
    # This code is contributed by ipg2016107.

输出

6 10

时间复杂度： O(N ² )
辅助空间： O(N)