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📜  计算索引(i,j,k)的三元组,以使[i,j)之间的元素的XOR等于[j,k]

📅  最后修改于: 2021-04-29 05:50:01             🧑  作者: Mango

给定一个整数数组Arr 。任务是计算三元组(i,j,k)的数量,以使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) ,其中N是数组的大小。
其中^是两个数字的按位异或。
例子:

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]中选择任何j,并且三元组将满足条件。
  • 现在,问题减少到查找XOR为0的所有子数组的长度,并且对于每个这样的长度L,在答案中加上L – 1。
  • 在前缀XOR数组中,如果在2个索引处(例如L和R),则前缀XOR值相同–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)
  • 这些可以使用以下公式来计算三胞胎:
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


输出:
6

时间复杂度: O(N)