计算索引 (i, j, k) 的三元组，使得 [i, j) 之间元素的 XOR 等于 [j, k]

给定一个整数数组Arr 。任务是计算三元组(i, j, k) 的数量，使得A _i ^ A _i+1 ^ A _i+2 ^ …。 ^ A _j-1 = A _j ^ A _j+1 ^ A _j+2 ^ ….. ^ A _k ，并且0 < (i, j, k) < N ，其中N是数组的大小。
其中^是两个数字的按位异或。
例子：

Input: Arr = [5, 2, 7] 
Output: 2 
Explanation: 
The triplets are (0, 2, 2) since 5 ^ 2 = 7 and (0, 1, 2) since 2 ^ 7 = 5. 
Input: Arr = [3, 6, 12, 8, 6, 2, 1, 5] 
Output: 6 
 

方法

• 让我们简化给定的表达式：

Ai ^ Ai + 1 ^ ...Aj - 1 = Aj ^ Aj + 1 ^ ...Ak

Taking XOR with Aj ^ Aj + 1 ^ ...Ak on both sides

Ai ^ Ai + 1 ^ ...Aj - 1 ^ Aj ^ Aj + 1 ^ ...Ak = 0

• 因此，具有 XOR 0 的子数组 [i, k] 将具有k – i 个三元组，因为可以从 [i+1, k] 中选择任意 j，并且该三元组将满足条件。
• 现在的问题简化为找出 XOR 为 0 的所有子数组的长度，并且对于每个这样的长度 L，将 L – 1 添加到答案中。
• 在前缀 XOR 数组中，如果在 2 个索引处，例如 L 和 R，前缀 XOR 值相同 – 那么这意味着子数组 [L + 1, R] = 0 的 XOR。
• 为了找到计数，我们将存储以下内容：

Say we are at index i, and the prefix XOR at i = x.

count[x] = Frequency of x in prefix XOR array before i

ways[x] -> For each all j < i, if prefixXor[j] = x,
then ways[x] += (j+1)

• 这些可用于使用以下公式计算三胞胎：

Triplets += count[x] * i  - ways[x]

下面是上述方法的实现：

C++

// C++ program  to count the Number of
// triplets in array having subarray
// XOR equal
#include 
using namespace std;
 
// Function return the count of
// triplets having subarray XOR equal
int CountOfTriplets(int a[], int n)
{
    int answer = 0;
 
    // XOR value till i
    int x = 0;
 
    // Count and ways array as defined
    // above
    int count[100005] = { 0 };
    int ways[100005] = { 0 };
 
    for (int i = 0; i < n; i++)
    {
        x ^= a[i];
 
        // Using the formula stated
        answer += count[x] * i - ways[x];
 
        // Increase the frequency of x
        count[x]++;
 
        // Add i+1 to ways[x] for upcoming
        // indices
        ways[x] += (i + 1);
    }
    return answer;
}
 
// Driver code
int main()
{
 
    int Arr[] = { 3, 6, 12, 8, 6, 2, 1, 5 };
 
    int N = sizeof(Arr) / sizeof(Arr[0]);
 
    cout << CountOfTriplets(Arr, N);
 
    return 0;
}

Java

// Java program to count the Number of
// triplets in array having subarray
// XOR equal
import java.io.*;
import java.util.Arrays;
import java.util.ArrayList;
import java.lang.*;
import java.util.Collections;
 
class GFG
{
 
// Function return the count of
// triplets having subarray XOR equal
static int CountOfTriplets(int a[], int n)
{
    int answer = 0;
 
    // XOR value till i
    int x = 0;
 
    // Count and ways array as defined
    // above
    int count[] = new int[100005];
    int ways[] = new int[100005];
     
 
    for (int i = 0; i < n; i++)
    {
        x ^= a[i];
 
        // Using the formula stated
        answer += count[x] * i - ways[x];
 
        // Increase the frequency of x
        count[x]++;
 
        // Add i+1 to ways[x] for upcoming
        // indices
        ways[x] += (i + 1);
    }
    return answer;
}
 
// Driver code
public static void main(String[] args)
{
 
    int Arr[] = { 3, 6, 12, 8, 6, 2, 1, 5 };
 
    int N = Arr.length;
 
    System.out.print(CountOfTriplets(Arr, N));
 
}
}
// This code is contributed by shivanisinghss2110

Python3

# Python3 program  to count the Number of
# triplets in array having subarray
# XOR equal
 
# Function return the count of
# triplets having subarray XOR equal
def CountOfTriplets(a,n):
    answer = 0
 
    # XOR value till i
    x = 0
 
    # Count and ways array as defined
    # above
    count = [0 for i in range(100005)]
    ways = [0 for i in range(100005)]
 
    for i in range(n):
        x ^= a[i]
 
        # Using the formula stated
        answer += count[x] * i - ways[x]
 
        # Increase the frequency of x
        count[x] += 1
 
        # Add i+1 to ways[x] for upcoming
        # indices
        ways[x] += (i + 1)
    return answer
 
# Driver code
if __name__ == '__main__':
    Arr =  [3, 6, 12, 8, 6, 2, 1, 5]
 
    N = len(Arr)
 
    print(CountOfTriplets(Arr, N))
 
# This code is contributed by Bhupendra_Singh

C#

// C# program to count the Number of
// triplets in array having subarray
// XOR equal
using System;
 
class GFG {
 
// Function return the count of
// triplets having subarray XOR equal
static int CountOfTriplets(int []a, int n)
{
    int answer = 0;
 
    // XOR value till i
    int x = 0;
 
    // Count and ways array as defined
    // above
    int []count = new int[100005];
    int []ways = new int[100005];
     
 
    for(int i = 0; i < n; i++)
    {
       x ^= a[i];
        
       // Using the formula stated
       answer += count[x] * i - ways[x];
        
       // Increase the frequency of x
       count[x]++;
        
       // Add i+1 to ways[x] for upcoming
       // indices
       ways[x] += (i + 1);
    }
     
    return answer;
}
 
// Driver code
public static void Main(String[] args)
{
 
    int []Arr = { 3, 6, 12, 8, 6, 2, 1, 5 };
    int N = Arr.Length;
 
    Console.Write(CountOfTriplets(Arr, N));
 
}
}
 
// This code is contributed by Rohit_ranjan

Javascript

 
    // Javascript program  to count the Number of
    // triplets in array having subarray
    // XOR equal
     
    // Function return the count of
    // triplets having subarray XOR equal
    function CountOfTriplets(a, n)
    {
        let answer = 0;
 
        // XOR value till i
        let x = 0;
 
        // Count and ways array as defined
        // above
        let count = new Array(100005);
        let ways = new Array(100005);
        count.fill(0);
        ways.fill(0);
 
        for (let i = 0; i < n; i++)
        {
            x ^= a[i];
 
            // Using the formula stated
            answer += count[x] * i - ways[x];
 
            // Increase the frequency of x
            count[x]++;
 
            // Add i+1 to ways[x] for upcoming
            // indices
            ways[x] = ways[x] + i + 1;
        }
        return answer;
    }
     
    let Arr = [ 3, 6, 12, 8, 6, 2, 1, 5 ];
   
    let N = Arr.length;
   
    document.write(CountOfTriplets(Arr, N));
 

输出：

6

时间复杂度： O(N)

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