计算具有给定 XOR 的子数组的数量

给定一个整数数组 arr[] 和一个数字 m，计算其元素异或为 m 的子数组的数量。
例子：

Input : arr[] = {4, 2, 2, 6, 4}, m = 6
Output : 4
Explanation : The subarrays having XOR of 
              their elements as 6 are {4, 2}, 
              {4, 2, 2, 6, 4}, {2, 2, 6},
               and {6}

Input : arr[] = {5, 6, 7, 8, 9}, m = 5
Output : 2
Explanation : The subarrays having XOR of
              their elements as 5 are {5}
              and {5, 6, 7, 8, 9}

一个简单的解决方案是使用两个循环遍历 arr[] 的所有可能的子数组，并计算其元素的 XOR 为 m 的子数组的数量。

C++

// A simple C++ Program to count all subarrays having
// XOR of elements as given value m
#include 
using namespace std;
 
// Simple function that returns count of subarrays
// of arr with XOR value equals to m
long long subarrayXor(int arr[], int n, int m)
{
    long long ans = 0; // Initialize ans
 
    // Pick starting point i of subarrays
    for (int i = 0; i < n; i++) {
        int xorSum = 0; // Store XOR of current subarray
 
        // Pick ending point j of subarray for each i
        for (int j = i; j < n; j++) {
            // calculate xorSum
            xorSum = xorSum ^ arr[j];
 
            // If xorSum is equal to given value,
            // increase ans by 1.
            if (xorSum == m)
                ans++;
        }
    }
    return ans;
}
 
// Driver program to test above function
int main()
{
    int arr[] = { 4, 2, 2, 6, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int m = 6;
 
    cout << "Number of subarrays having given XOR is "
         << subarrayXor(arr, n, m);
    return 0;
}

Java

// A simple Java Program to count all
// subarrays having XOR of elements
// as given value m
public class GFG {
 
    // Simple function that returns
    // count of subarrays of arr with
    // XOR value equals to m
    static long subarrayXor(int arr[],
                             int n, int m)
    {
         
        // Initialize ans
        long ans = 0;
 
        // Pick starting point i of
        // subarrays
        for (int i = 0; i < n; i++)
        {
             
            // Store XOR of current
            // subarray
            int xorSum = 0;
 
            // Pick ending point j of
            // subarray for each i
            for (int j = i; j < n; j++)
            {
                 
                // calculate xorSum
                xorSum = xorSum ^ arr[j];
 
                // If xorSum is equal to
                // given value, increase
                // ans by 1.
                if (xorSum == m)
                    ans++;
            }
        }
         
        return ans;
    }
 
    // Driver code
    public static void main(String args[])
    {
 
        int[] arr = { 4, 2, 2, 6, 4 };
        int n = arr.length;
        int m = 6;
 
        System.out.println("Number of subarrays"
                       + " having given XOR is "
                       + subarrayXor(arr, n, m));
    }
}
 
// This code is contributed by Sam007.

Python3

# A simple Python3 Program to count all subarrays having
# XOR of elements as given value m
  
# Simple function that returns count of subarrays
# of arr with XOR value equals to m
def subarrayXor(arr, n, m):
    ans = 0 # Initialize ans
  
    # Pick starting po i of subarrays
    for i in range(0,n):
         
        xorSum = 0 # Store XOR of current subarray
  
        # Pick ending po j of subarray for each i
        for j  in range(i,n):
            # calculate xorSum
            xorSum = xorSum ^ arr[j]
  
            # If xorSum is equal to given value,
            # increase ans by 1.
            if (xorSum == m):
                ans+=1
    return ans
  
# Driver program to test above function
def main():
    arr = [ 4, 2, 2, 6, 4 ]
    n = len(arr)
    m = 6
  
    print("Number of subarrays having given XOR is "
         , subarrayXor(arr, n, m))
 
if __name__ == '__main__':
    main()
     
#this code contributed by 29AjayKumar

C#

// A simple C# Program to count all
// subarrays having XOR of elements
// as given value m
using System;
 
class GFG {
     
    // Simple function that returns
    // count of subarrays of arr
    // with XOR value equals to m
    static long subarrayXor(int[] arr,
                            int n, int m)
    {
         
        // Initialize ans
        long ans = 0;
 
        // Pick starting point i of
        // subarrays
        for (int i = 0; i < n; i++)
        {
             
            // Store XOR of current
            // subarray
            int xorSum = 0;
 
            // Pick ending point j of
            // subarray for each i
            for (int j = i; j < n; j++)
            {
                 
                // calculate xorSum
                xorSum = xorSum ^ arr[j];
 
                // If xorSum is equal to
                // given value, increase
                // ans by 1.
                if (xorSum == m)
                    ans++;
            }
        }
         
        return ans;
    }
 
    // Driver Program
    public static void Main()
    {
        int[] arr = { 4, 2, 2, 6, 4 };
        int n = arr.Length;
        int m = 6;
 
        Console.Write("Number of subarrays"
                  + " having given XOR is "
                  + subarrayXor(arr, n, m));
    }
}
 
// This code is contributed by Sam007.

Javascript

 
// A simple Javascript Program to
// count all subarrays having
// XOR of elements as given value m
 
// Simple function that
// returns count of subarrays
// of arr with XOR value equals to m
function subarrayXor(arr, n, m)
{
    let ans = 0; // Initialize ans
 
    // Pick starting point i of subarrays
    for (let i = 0; i < n; i++)
    {
    // Store XOR of current subarray
        let xorSum = 0;
 
        // Pick ending point j of
        // subarray for each i
        for (let j = i; j < n; j++) {
            // calculate xorSum
            xorSum = xorSum ^ arr[j];
 
            // If xorSum is equal to given value,
            // increase ans by 1.
            if (xorSum == m)
                ans++;
        }
    }
    return ans;
}
 
// Driver program to test above function
    let arr = [ 4, 2, 2, 6, 4 ];
    let n = arr.length;
    let m = 6;
 
    document.write(
    "Number of subarrays having given XOR is "
         + subarrayXor(arr, n, m)
         );
 

输出：

Number of subarrays having given XOR is 4

上述解决方案的时间复杂度为 O(n ² )。

有效的方法：
一个有效的解决方案在 O(n) 时间内解决了上述问题。让我们把 [i+1, j] 范围内所有元素的 XOR 称为 A，将 [0, i] 范围内的所有元素称为 B，将 [0, j] 范围内的所有元素称为 C。如果我们对 B 进行 XOR与C，[0, i]中的重叠元素从B和C归零，我们得到[i+1, j]范围内所有元素的异或，即A。由于A = B XOR C，我们有B = A XOR C。现在，如果我们知道 C 的值并且我们将 A 的值作为 m，我们将得到 A 的计数作为满足该关系的所有 B 的计数。本质上，我们得到了每个 C 具有 XOR-sum m 的所有子阵列的计数。当我们计算整个 C 的计数总和时，我们得到了答案。

1) Initialize ans as 0.
2) Compute xorArr, the prefix xor-sum array.
3) Create a map mp in which we store count of 
   all prefixes with XOR as a particular value. 
4) Traverse xorArr and for each element in xorArr
   (A) If m^xorArr[i] XOR exists in map, then 
       there is another previous prefix with 
       same XOR, i.e., there is a subarray ending
       at i with XOR equal to m. We add count of
       all such subarrays to result. 
   (B) If xorArr[i] is equal to m, increment ans by 1.
   (C) Increment count of elements having XOR-sum 
       xorArr[i] in map by 1.
5) Return ans.

C++

// C++ Program to count all subarrays having
// XOR of elements as given value m with
// O(n) time complexity.
#include 
using namespace std;
 
// Returns count of subarrays of arr with XOR
// value equals to m
long long subarrayXor(int arr[], int n, int m)
{
    long long ans = 0; // Initialize answer to be returned
 
    // Create a prefix xor-sum array such that
    // xorArr[i] has value equal to XOR
    // of all elements in arr[0 ..... i]
    int* xorArr = new int[n];
 
    // Create map that stores number of prefix array
    // elements corresponding to a XOR value
    unordered_map mp;
 
    // Initialize first element of prefix array
    xorArr[0] = arr[0];
 
    // Computing the prefix array.
    for (int i = 1; i < n; i++)
        xorArr[i] = xorArr[i - 1] ^ arr[i];
 
    // Calculate the answer
    for (int i = 0; i < n; i++) {
       
        // Find XOR of current prefix with m.
        int tmp = m ^ xorArr[i];
 
        // If above XOR exists in map, then there
        // is another previous prefix with same
        // XOR, i.e., there is a subarray ending
        // at i with XOR equal to m.
        ans = ans + ((long long)mp[tmp]);
 
        // If this subarray has XOR equal to m itself.
        if (xorArr[i] == m)
            ans++;
 
        // Add the XOR of this subarray to the map
        mp[xorArr[i]]++;
    }
 
    // Return total count of subarrays having XOR of
    // elements as given value m
    return ans;
}
 
// Driver program to test above function
int main()
{
    int arr[] = { 4, 2, 2, 6, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int m = 6;
 
    cout << "Number of subarrays having given XOR is "
         << subarrayXor(arr, n, m);
    return 0;
}

Java

// Java Program to count all subarrays having
// XOR of elements as given value m with
// O(n) time complexity.
import java.util.*;
 
class GFG {
 
    // Returns count of subarrays of arr with XOR
    // value equals to m
    static long subarrayXor(int arr[], int n, int m)
    {
        long ans = 0; // Initialize answer to be returned
 
        // Create a prefix xor-sum array such that
        // xorArr[i] has value equal to XOR
        // of all elements in arr[0 ..... i]
        int[] xorArr = new int[n];
 
        // Create map that stores number of prefix array
        // elements corresponding to a XOR value
        HashMap mp
            = new HashMap();
 
        // Initialize first element of prefix array
        xorArr[0] = arr[0];
 
        // Computing the prefix array.
        for (int i = 1; i < n; i++)
            xorArr[i] = xorArr[i - 1] ^ arr[i];
 
        // Calculate the answer
        for (int i = 0; i < n; i++) {
            // Find XOR of current prefix with m.
            int tmp = m ^ xorArr[i];
 
            // If above XOR exists in map, then there
            // is another previous prefix with same
            // XOR, i.e., there is a subarray ending
            // at i with XOR equal to m.
            ans = ans
                  + (mp.containsKey(tmp) == false
                         ? 0
                         : ((long)mp.get(tmp)));
 
            // If this subarray has XOR equal to m itself.
            if (xorArr[i] == m)
                ans++;
 
            // Add the XOR of this subarray to the map
            if (mp.containsKey(xorArr[i]))
                mp.put(xorArr[i], mp.get(xorArr[i]) + 1);
            else
                mp.put(xorArr[i], 1);
        }
 
        // Return total count of subarrays having XOR of
        // elements as given value m
        return ans;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 4, 2, 2, 6, 4 };
        int n = arr.length;
        int m = 6;
 
        System.out.print(
            "Number of subarrays having given XOR is "
            + subarrayXor(arr, n, m));
    }
}
 
// This code is contributed by PrinciRaj1992

蟒蛇3

# Python3 Program to count all subarrays
# having XOR of elements as given value m
# with O(n) time complexity.
 
# Returns count of subarrays of arr
# with XOR value equals to m
def subarrayXor(arr, n, m):
 
    ans = 0 # Initialize answer to be returned
 
    # Create a prefix xor-sum array such that
    # xorArr[i] has value equal to XOR
    # of all elements in arr[0 ..... i]
    xorArr =[0 for _ in range(n)]
 
    # Create map that stores number of prefix array
    # elements corresponding to a XOR value
    mp = dict()
 
    # Initialize first element
    # of prefix array
    xorArr[0] = arr[0]
 
    # Computing the prefix array.
    for i in range(1, n):
        xorArr[i] = xorArr[i - 1] ^ arr[i]
 
    # Calculate the answer
    for i in range(n):
         
        # Find XOR of current prefix with m.
        tmp = m ^ xorArr[i]
 
        # If above XOR exists in map, then there
        # is another previous prefix with same
        # XOR, i.e., there is a subarray ending
        # at i with XOR equal to m.
        if tmp in mp.keys():
            ans = ans + (mp[tmp])
 
        # If this subarray has XOR
        # equal to m itself.
        if (xorArr[i] == m):
            ans += 1
 
        # Add the XOR of this subarray to the map
        mp[xorArr[i]] = mp.get(xorArr[i], 0) + 1
 
    # Return total count of subarrays having
    # XOR of elements as given value m
    return ans
 
# Driver Code
arr = [4, 2, 2, 6, 4]
n = len(arr)
m = 6
 
print("Number of subarrays having given XOR is",
                        subarrayXor(arr, n, m))
 
# This code is contributed by mohit kumar

C＃

// C# Program to count all subarrays having
// XOR of elements as given value m with
// O(n) time complexity.
using System;
using System.Collections.Generic;
 
class GFG {
 
    // Returns count of subarrays of arr with XOR
    // value equals to m
    static long subarrayXor(int[] arr, int n, int m)
    {
        long ans = 0; // Initialize answer to be returned
 
        // Create a prefix xor-sum array such that
        // xorArr[i] has value equal to XOR
        // of all elements in arr[0 ..... i]
        int[] xorArr = new int[n];
 
        // Create map that stores number of prefix array
        // elements corresponding to a XOR value
        Dictionary mp
            = new Dictionary();
 
        // Initialize first element of prefix array
        xorArr[0] = arr[0];
 
        // Computing the prefix array.
        for (int i = 1; i < n; i++)
            xorArr[i] = xorArr[i - 1] ^ arr[i];
 
        // Calculate the answer
        for (int i = 0; i < n; i++) {
            // Find XOR of current prefix with m.
            int tmp = m ^ xorArr[i];
 
            // If above XOR exists in map, then there
            // is another previous prefix with same
            // XOR, i.e., there is a subarray ending
            // at i with XOR equal to m.
            ans = ans
                  + (mp.ContainsKey(tmp) == false
                         ? 0
                         : ((long)mp[tmp]));
 
            // If this subarray has XOR equal to m itself.
            if (xorArr[i] == m)
                ans++;
 
            // Add the XOR of this subarray to the map
            if (mp.ContainsKey(xorArr[i]))
                mp[xorArr[i]] = mp[xorArr[i]] + 1;
            else
                mp.Add(xorArr[i], 1);
        }
 
        // Return total count of subarrays having XOR of
        // elements as given value m
        return ans;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int[] arr = { 4, 2, 2, 6, 4 };
        int n = arr.Length;
        int m = 6;
 
        Console.Write(
            "Number of subarrays having given XOR is "
            + subarrayXor(arr, n, m));
    }
}
 
// This code is contributed by Rajput-Ji

Javascript

// Javascript Program to count all subarrays having
// XOR of elements as given value m with
// O(n) time complexity.
     
    // Returns count of subarrays of arr with XOR
    // value equals to m
    function subarrayXor(aee,n,m)
    {
        let ans = 0; // Initialize answer to be returned
  
        // Create a prefix xor-sum array such that
        // xorArr[i] has value equal to XOR
        // of all elements in arr[0 ..... i]
        let xorArr = new Array(n);
  
        // Create map that stores number of prefix array
        // elements corresponding to a XOR value
        let mp = new Map();
  
        // Initialize first element of prefix array
        xorArr[0] = arr[0];
  
        // Computing the prefix array.
        for (let i = 1; i < n; i++)
            xorArr[i] = xorArr[i - 1] ^ arr[i];
  
        // Calculate the answer
        for (let i = 0; i < n; i++) {
            // Find XOR of current prefix with m.
            let tmp = m ^ xorArr[i];
  
            // If above XOR exists in map, then there
            // is another previous prefix with same
            // XOR, i.e., there is a subarray ending
            // at i with XOR equal to m.
            ans = ans
                  + (mp.has(tmp) == false
                         ? 0
                         : (mp.get(tmp)));
  
            // If this subarray has XOR equal to m itself.
            if (xorArr[i] == m)
                ans++;
  
            // Add the XOR of this subarray to the map
            if (mp.has(xorArr[i]))
                mp.set(xorArr[i], mp.get(xorArr[i]) + 1);
            else
                mp.set(xorArr[i], 1);
        }
  
        // Return total count of subarrays having XOR of
        // elements as given value m
        return ans;
    }
     
     // Driver code
    let arr=[4, 2, 2, 6, 4];
     
    let n = arr.length;
    let m = 6;
    document.write("Number of subarrays having given XOR is "
            + subarrayXor(arr, n, m));
     
// This code is contributed by unknown2108

输出：

Number of subarrays having given XOR is 4

时间复杂度： O(n)
替代方法：使用Python字典存储前缀异或

蟒蛇3

from collections import defaultdict
def subarrayXor(arr, n, m):
    HashTable=defaultdict(bool)
    HashTable[0]=1
    count=0
    curSum=0
    for i in arr:
        curSum^=i
        if HashTable[curSum^m]:
            count+=HashTable[curSum^m]
        HashTable[curSum]+=1
    return(count)
         
 
# Driver program to test above function
def main():
    arr = [ 5, 6, 7, 8, 9 ]
    n = len(arr)
    m = 5
 
    print("Number of subarrays having given XOR is "
        , subarrayXor(arr, n, m))
 
if __name__ == '__main__':
    main()
     
 # This code is contributed by mrmechanical26052000

输出：

Number of subarrays having given XOR is 4

时间复杂度： O(n)

空间复杂度： O(n)

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