子数组的XOR(元素范围)|套装2

给定一个大小为N和Q个查询的整数arr [] 。每个查询的格式为(L，R) ，其中L和R是数组的索引。任务是找到子数组arr [L…R]的XOR值。
例子：

Input: arr[] = {2, 5, 1, 6, 1, 2, 5} query[] = {{1, 4}}
Output: 3
The XOR value of arr[1…4] is 3.

Input: arr[] = {2, 5, 1, 6, 1, 2, 5} query[] = {{0, 6}}
Output: 6
The XOR value of arr[0…6] is 6.

O(N)辅助空间方法：请参考本文以了解O(N)辅助空间方法。

方法：在这里我们将讨论一个恒定的空间解决方案，但是我们将修改输入数组。

1. 将输入数组从索引1更新为N – 1 ，以便arr [i]存储从arr [0]到arr [i]的XOR。

   arr[i] = XOR(arr[0], arr[1], .., arr[i])

2. 要处理从L到R的查询，只需返回arr [L-1] ^ arr [R]即可。

   arr[L-1] = arr[0] ^ arr[1] ^ arr[2]…. ^ arr[L-1]
   arr[R] = arr[0] ^ arr[1]……arr[L-1] ^ arr[L] ^ arr[L+1]…… ^ arr[R]
   arr[L-1] ^ arr[R] = arr[L] ^ arr[L+1]…. ^ arr[R]

   where ^ denotes XOR

例如：

Consider the example: arr[] = { 3, 2, 4, 5, 1, 1, 5, 3 }, query[] = {{1, 4 }, { 3, 7}}
After taking XOR of consecutive elements, arr[] = {3, 1, 5, 0, 1, 0, 5, 6}
For first query {1, 4} ans = arr[0] ^ arr[4] = 3 ^ 1 = 2
For second query {3, 7} ans = arr[2] ^ arr[7] = 5 ^ 6 = 3

下面是上述方法的实现：

C++

// C++ program to find XOR
// in a range from L to R
  
#include 
using namespace std;
  
// Function to find XOR
// in a range from L to R
void find_Xor(int arr[],
              pair querry[],
              int N, int Q)
{
    // Compute xor from arr[0] to arr[i]
    for (int i = 1; i < N; i++) {
        arr[i] = arr[i] ^ arr[i - 1];
    }
  
    int ans = 0;
  
    // process every query
    // in constant time
    for (int i = 0; i < Q; i++) {
  
        // if L==0
        if (querry[i].first == 0)
            ans = arr[querry[i].second];
        else
            ans = arr[querry[i].first - 1]
                  ^ arr[querry[i].second];
  
        cout << ans << endl;
    }
}
  
// Driver Code
int main()
{
    int arr[] = { 3, 2, 4, 5,
                  1, 1, 5, 3 };
    int N = 8;
    int Q = 2;
    pair query[Q]
        = { { 1, 4 },
            { 3, 7 } };
  
    // query[]
    find_Xor(arr, query, N, Q);
    return 0;
}

Java

// Java program to find XOR
// in a range from L to R
class GFG{
      
static class pair
{ 
    int first, second; 
    public pair(int first, int second) 
    { 
        this.first = first; 
        this.second = second; 
    } 
} 
  
// Function to find XOR
// in a range from L to R
static void find_Xor(int arr[],
                     pair querry[],
                     int N, int Q)
{
      
    // Compute xor from arr[0] to arr[i]
    for(int i = 1; i < N; i++)
    {
       arr[i] = arr[i] ^ arr[i - 1];
    }
      
    int ans = 0;
  
    // Process every query
    // in constant time
    for(int i = 0; i < Q; i++)
    {
          
       // If L==0
       if (querry[i].first == 0)
           ans = arr[querry[i].second];
       else
           ans = arr[querry[i].first - 1] ^
                 arr[querry[i].second];
  
       System.out.print(ans + "\n");
    }
}
  
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 3, 2, 4, 5,
                  1, 1, 5, 3 };
    int N = 8;
    int Q = 2;
      
    pair query[] = { new pair(1, 4),
                     new pair(3, 7) };
  
    // query[]
    find_Xor(arr, query, N, Q);
}
}
  
// This code is contributed by gauravrajput1

Python3

# Python3 program to find XOR 
# in a range from L to R
  
# Function to find XOR 
# in a range from L to R 
def find_Xor(arr, query, N, Q):
      
    # Compute xor from arr[0] to arr[i]
    for i in range(1, N):
        arr[i] = arr[i] ^ arr[i - 1]
          
    ans = 0
      
    # Process every query 
    # in constant time 
    for i in range(Q):
          
        # If L == 0
        if query[i][0] == 0:
            ans = arr[query[i][1]]
        else:
            ans = (arr[query[i][0] - 1] ^ 
                   arr[query[i][1]])
        print(ans)
      
# Driver code
def main():
      
    arr = [ 3, 2, 4, 5, 1, 1, 5, 3 ]
    N = 8
    Q = 2
      
    # query[]
    query = [ [ 1, 4 ],
              [ 3, 7 ] ]
                
    find_Xor(arr, query, N, Q)
      
main()
  
# This code is contributed by Stuti Pathak

C#

// C# program to find XOR
// in a range from L to R
using System;
  
class GFG{    
class pair
{ 
    public int first, second; 
    public pair(int first, int second) 
    { 
        this.first = first; 
        this.second = second; 
    } 
} 
   
// Function to find XOR
// in a range from L to R
static void find_Xor(int []arr,
                     pair []querry,
                     int N, int Q)
{
       
    // Compute xor from arr[0] to arr[i]
    for(int i = 1; i < N; i++)
    {
       arr[i] = arr[i] ^ arr[i - 1];
    }
       
    int ans = 0;
   
    // Process every query
    // in constant time
    for(int i = 0; i < Q; i++)
    {
           
       // If L==0
       if (querry[i].first == 0)
           ans = arr[querry[i].second];
       else
           ans = arr[querry[i].first - 1] ^
                 arr[querry[i].second];
   
       Console.Write(ans + "\n");
    }
}
   
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 3, 2, 4, 5,
                  1, 1, 5, 3 };
    int N = 8;
    int Q = 2;
       
    pair []query = { new pair(1, 4),
                     new pair(3, 7) };
   
    // query[]
    find_Xor(arr, query, N, Q);
}
}
  
// This code is contributed by 29AjayKumar

输出：

2
3

时间复杂度： O(N + Q)
辅助空间： O(1)