📜  计算给定XOR的所有对

📅  最后修改于: 2021-05-25 09:24:44             🧑  作者: Mango

给定一个由不同的正整数组成的数组和一个数字x,请找到XOR等于x的数组中整数对的数量。
例子:

Input : arr[] = {5, 4, 10, 15, 7, 6}, x = 5
Output : 1
Explanation :  (10 ^ 15) = 5

Input : arr[] = {3, 6, 8, 10, 15, 50}, x = 5
Output : 2
Explanation : (3 ^ 6) = 5 and (10 ^ 15) = 5 

一种简单的解决方案是遍历每个元素,并检查是否存在另一个与XOR等于x的数字。该解决方案花费O(n 2 )时间。

有效解决此问题需要O(n)时间。该想法基于以下事实:当且仅当arr [i] ^ x等于arr [j]时,arr [i] ^ arr [j]等于x。

1) Initialize result as 0.
2) Create an empty hash set "s".
3) Do following for each element arr[i] in arr[]
   (a)    If x ^ arr[i] is in "s", then increment result by 1.
   (b)    Insert arr[i] into the hash set "s".
3) return result.
C++
// C++ program to Count all pair with given XOR
// value x
#include
  
using namespace std;
  
// Returns count of pairs in arr[0..n-1] with XOR
// value equals to x.
int xorPairCount(int arr[], int n, int x)
{
    int result = 0; // Initialize result
  
    // create empty set that stores the visiting 
    // element of array. 
    // Refer below post for details of unordered_set
    // https://www.geeksforgeeks.org/unorderd_set-stl-uses/
    unordered_set s;
  
    for (int i=0; iJava
// Java program to Count all pair with 
// given XOR value x 
import java.util.*;
  
class GFG 
{
  
    // Returns count of pairs in arr[0..n-1] with XOR 
    // value equals to x. 
    static int xorPairCount(int arr[], int n, int x) 
    {
        int result = 0; // Initialize result 
  
        // create empty set that stores the visiting 
        // element of array. 
        // Refer below post for details of unordered_set 
        // https://www.geeksforgeeks.org/unorderd_set-stl-uses/ 
        HashSet s = new HashSet();
  
        for (int i = 0; i < n; i++) 
        {
            // If there exist an element in set s 
            // with XOR equals to x^arr[i], that means 
            // there exist an element such that the 
            // XOR of element with arr[i] is equal to 
            // x, then increment count. 
            if (s.contains(x ^ arr[i]) && 
                    (x ^ arr[i]) == (int) s.toArray()[s.size() - 1]) 
            {
                result++;
            }
  
            // Make element visited 
            s.add(arr[i]);
        }
  
        // return total count of 
        // pairs with XOR equal to x 
        return result;
    }
  
    // Driver code 
    public static void main(String[] args) 
    {
        int arr[] = {5, 4, 10, 15, 7, 6};
        int n = arr.length;
        int x = 5;
        System.out.print("Count of pairs with given XOR = "
                + xorPairCount(arr, n, x));
    }
}
  
// This code contributed by Rajput-Ji


Python3
# Python3 program to count all the pair 
# with given xor
  
# Returns count of pairs in arr[0..n-1] 
# with XOR value equals to x.
def xorPairCount(arr, n, x):
    result = 0 # Initialize result
      
    # create empty set that stores the 
    # visiting element of array. 
    s = set()
    for i in range(0, n):
          
        # If there exist an element in set s
        # with XOR equals to x^arr[i], that 
        # means there exist an element such 
        # that the XOR of element with arr[i]  
        # is equal to x, then increment count.
        if(x ^ arr[i] in s):
            result = result + 1
              
        # Make element visited
        s.add(arr[i])
    return result
      
# Driver Code
if __name__ == "__main__":
    arr = [5, 4, 10, 15, 7, 6]
    n = len(arr)
    x = 5
    print("Count of pair with given XOR = " +
                str(xorPairCount(arr, n, x)))
  
# This code is contributed by Anubhav Natani


C#
// C# program to Count all pair with 
// given XOR value x 
using System;
using System.Collections.Generic;
  
class GFG 
{
  
    // Returns count of pairs in arr[0..n-1] with XOR 
    // value equals to x. 
    static int xorPairCount(int []arr, int n, int x) 
    {
        int result = 0; // Initialize result 
  
        // create empty set that stores the visiting 
        // element of array. 
        // Refer below post for details of unordered_set 
        // https://www.geeksforgeeks.org/unorderd_set-stl-uses/ 
        HashSet s = new HashSet();
  
        for (int i = 0; i < n; i++) 
        {
            // If there exist an element in set s 
            // with XOR equals to x^arr[i], that means 
            // there exist an element such that the 
            // XOR of element with arr[i] is equal to 
            // x, then increment count. 
            if (s.Contains(x ^ arr[i])) 
            {
                result++;
            }
  
            // Make element visited 
            s.Add(arr[i]);
        }
  
        // return total count of 
        // pairs with XOR equal to x 
        return result;
    }
  
    // Driver code 
    public static void Main() 
    {
        int []arr = {5, 4, 10, 15, 7, 6};
        int n = arr.Length;
        int x = 5;
        Console.WriteLine("Count of pairs with given XOR = "
                + xorPairCount(arr, n, x));
    }
}
  
/* This code contributed by PrinciRaj1992 */


C++
// C++ program to Count all pair with given XOR
// value x
#include
  
using namespace std;
  
// Returns count of pairs in arr[0..n-1] with XOR
// value equals to x.
int xorPairCount(int arr[], int n, int x)
{
    int result = 0; // Initialize result
  
    // create empty map that stores counts of
    // individual elements of array.
    unordered_map m;
  
    for (int i=0; iJava
// Java program to Count all pair with given XOR
// value x
import java.util.*;
  
class GFG
{
  
// Returns count of pairs in arr[0..n-1] with XOR
// value equals to x.
static int xorPairCount(int arr[], int n, int x)
{
    int result = 0; // Initialize result
  
    // create empty map that stores counts of
    // individual elements of array.
    Map m = new HashMap<>();
  
    for (int i = 0;  i < n ; i++)
    {
        int curr_xor = x^arr[i];
  
        // If there exist an element in map m
        // with XOR equals to x^arr[i], that means
        // there exist an element such that the
        // XOR of element with arr[i] is equal to
        // x, then increment count.
        if (m.containsKey(curr_xor))
            result += m.get(curr_xor);
  
        // Increment count of current element
        if(m.containsKey(arr[i]))
        {
            m.put(arr[i], m.get(arr[i]) + 1);
        }
        else{
            m.put(arr[i], 1);
        }
    }
    // return total count of pairs with XOR equal to x
    return result;
}
  
// Driver code
public static void main(String[] args) 
{
    int arr[] = {2, 5, 2};
    int n = arr.length;
    int x = 0;
    System.out.println("Count of pairs with given XOR = "
        + xorPairCount(arr, n, x));
}
}
  
// This code has been contributed by 29AjayKumar


Python3
# Python3 program to Count all pair with 
# given XOR value x
  
# Returns count of pairs in arr[0..n-1] 
# with XOR value equals to x.
def xorPairCount(arr, n, x):
  
    result = 0 # Initialize result
  
    # create empty map that stores counts 
    # of individual elements of array.
    m = dict()
  
    for i in range(n):
      
        curr_xor = x ^ arr[i]
  
        # If there exist an element in map m
        # with XOR equals to x^arr[i], that
        # means there exist an element such that 
        # the XOR of element with arr[i] is equal 
        # to x, then increment count.
        if (curr_xor in m.keys()):
            result += m[curr_xor]
  
        # Increment count of current element
        if arr[i] in m.keys():
            m[arr[i]] += 1
        else:
            m[arr[i]] = 1
      
    # return total count of pairs
    # with XOR equal to x
    return result
  
# Driver Code
arr = [2, 5, 2]
n = len(arr)
x = 0
print("Count of pairs with given XOR = ",
                 xorPairCount(arr, n, x))
  
# This code is contributed by Mohit Kumar


C#
// C# program to Count all pair with given XOR
// value x
using System; 
using System.Collections.Generic;
  
class GFG
{
  
// Returns count of pairs in arr[0..n-1] with XOR
// value equals to x.
static int xorPairCount(int []arr, int n, int x)
{
    int result = 0; // Initialize result
  
    // create empty map that stores counts of
    // individual elements of array.
    Dictionary m = new Dictionary();
  
    for (int i = 0; i < n ; i++)
    {
        int curr_xor = x^arr[i];
  
        // If there exist an element in map m
        // with XOR equals to x^arr[i], that means
        // there exist an element such that the
        // XOR of element with arr[i] is equal to
        // x, then increment count.
        if (m.ContainsKey(curr_xor))
            result += m[curr_xor];
  
        // Increment count of current element
        if(m.ContainsKey(arr[i]))
        {
            var val = m[arr[i]];
            m.Remove(arr[i]);
            m.Add(arr[i], val + 1);
        }
        else
        {
            m.Add(arr[i], 1);
        }
    }
      
    // return total count of pairs with XOR equal to x
    return result;
}
  
// Driver code
public static void Main(String[] args) 
{
    int []arr = {2, 5, 2};
    int n = arr.Length;
    int x = 0;
    Console.WriteLine("Count of pairs with given XOR = "
        + xorPairCount(arr, n, x));
}
}
  
// This code has been contributed by 29AjayKumar


输出:

Count of pairs with given XOR = 1

时间复杂度: O(n)

如何处理重复项?
如果输入数组中有重复项,则上述有效解决方案将不起作用。例如,上述解决方案对于{2,2,5}和{5,2,2}产生不同的结果。为了处理重复项,我们存储所有元素的出现次数。我们使用unordered_map而不是unordered_set。

C++

// C++ program to Count all pair with given XOR
// value x
#include
  
using namespace std;
  
// Returns count of pairs in arr[0..n-1] with XOR
// value equals to x.
int xorPairCount(int arr[], int n, int x)
{
    int result = 0; // Initialize result
  
    // create empty map that stores counts of
    // individual elements of array.
    unordered_map m;
  
    for (int i=0; iJava
// Java program to Count all pair with given XOR
// value x
import java.util.*;
  
class GFG
{
  
// Returns count of pairs in arr[0..n-1] with XOR
// value equals to x.
static int xorPairCount(int arr[], int n, int x)
{
    int result = 0; // Initialize result
  
    // create empty map that stores counts of
    // individual elements of array.
    Map m = new HashMap<>();
  
    for (int i = 0;  i < n ; i++)
    {
        int curr_xor = x^arr[i];
  
        // If there exist an element in map m
        // with XOR equals to x^arr[i], that means
        // there exist an element such that the
        // XOR of element with arr[i] is equal to
        // x, then increment count.
        if (m.containsKey(curr_xor))
            result += m.get(curr_xor);
  
        // Increment count of current element
        if(m.containsKey(arr[i]))
        {
            m.put(arr[i], m.get(arr[i]) + 1);
        }
        else{
            m.put(arr[i], 1);
        }
    }
    // return total count of pairs with XOR equal to x
    return result;
}
  
// Driver code
public static void main(String[] args) 
{
    int arr[] = {2, 5, 2};
    int n = arr.length;
    int x = 0;
    System.out.println("Count of pairs with given XOR = "
        + xorPairCount(arr, n, x));
}
}
  
// This code has been contributed by 29AjayKumar

Python3

# Python3 program to Count all pair with 
# given XOR value x
  
# Returns count of pairs in arr[0..n-1] 
# with XOR value equals to x.
def xorPairCount(arr, n, x):
  
    result = 0 # Initialize result
  
    # create empty map that stores counts 
    # of individual elements of array.
    m = dict()
  
    for i in range(n):
      
        curr_xor = x ^ arr[i]
  
        # If there exist an element in map m
        # with XOR equals to x^arr[i], that
        # means there exist an element such that 
        # the XOR of element with arr[i] is equal 
        # to x, then increment count.
        if (curr_xor in m.keys()):
            result += m[curr_xor]
  
        # Increment count of current element
        if arr[i] in m.keys():
            m[arr[i]] += 1
        else:
            m[arr[i]] = 1
      
    # return total count of pairs
    # with XOR equal to x
    return result
  
# Driver Code
arr = [2, 5, 2]
n = len(arr)
x = 0
print("Count of pairs with given XOR = ",
                 xorPairCount(arr, n, x))
  
# This code is contributed by Mohit Kumar

C#

// C# program to Count all pair with given XOR
// value x
using System; 
using System.Collections.Generic;
  
class GFG
{
  
// Returns count of pairs in arr[0..n-1] with XOR
// value equals to x.
static int xorPairCount(int []arr, int n, int x)
{
    int result = 0; // Initialize result
  
    // create empty map that stores counts of
    // individual elements of array.
    Dictionary m = new Dictionary();
  
    for (int i = 0; i < n ; i++)
    {
        int curr_xor = x^arr[i];
  
        // If there exist an element in map m
        // with XOR equals to x^arr[i], that means
        // there exist an element such that the
        // XOR of element with arr[i] is equal to
        // x, then increment count.
        if (m.ContainsKey(curr_xor))
            result += m[curr_xor];
  
        // Increment count of current element
        if(m.ContainsKey(arr[i]))
        {
            var val = m[arr[i]];
            m.Remove(arr[i]);
            m.Add(arr[i], val + 1);
        }
        else
        {
            m.Add(arr[i], 1);
        }
    }
      
    // return total count of pairs with XOR equal to x
    return result;
}
  
// Driver code
public static void Main(String[] args) 
{
    int []arr = {2, 5, 2};
    int n = arr.Length;
    int x = 0;
    Console.WriteLine("Count of pairs with given XOR = "
        + xorPairCount(arr, n, x));
}
}
  
// This code has been contributed by 29AjayKumar

输出:

Count of pairs with given XOR = 1

时间复杂度: O(n)