子数组的异或（元素范围）| 2套

给定一个大小为N和Q的数组整数arr[]查询。每个查询的形式为(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：O(N) 辅助空间方法请参考这篇文章。

方法：这里我们将讨论一个常数空间解决方案，但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] 。

例如：

Consider the example: arr[] = { 3, 2, 4, 5, 1, 1, 5, 3 }, query[] = {{1, 4 }, { 3, 7}} 
After taking the 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 the 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

Javascript

 
// Javascript program to find XOR
// in a range from L to R
 
// Function to find XOR
// in a range from L to R
function find_Xor(arr, querry, N, Q)
{
    // Compute xor from arr[0] to arr[i]
    for (var i = 1; i < N; i++) {
        arr[i] = arr[i] ^ arr[i - 1];
    }
 
    var ans = 0;
 
    // process every query
    // in constant time
    for (var i = 0; i < Q; i++) {
 
        // if L==0
        if (querry[i][0] == 0)
            ans = arr[querry[i][1]];
        else
            ans = arr[querry[i][0] - 1]
                  ^ arr[querry[i][1]];
 
        document.write( ans + "<br>");
    }
}
 
// Driver Code
var arr = [3, 2, 4, 5,
              1, 1, 5, 3];
var N = 8;
var Q = 2;
var query
    = [[1, 4],
        [3, 7]];
// query[]
find_Xor(arr, query, N, Q);
 
 

输出：

2
3

时间复杂度： O (N + Q)
辅助空间： O (1)

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