求一个数 X，使得给定数组的每个元素加上 X 后的 XOR 为 0

给定一个包含正整数的奇数长度 N数组arr[] 。任务是找到一个正整数X ，这样，将X添加到arr[]的所有元素，然后对所有元素进行XOR得到0 。如果不存在这样的X ，则返回-1 。

例子：

Input: arr[] = {2, 4, 5}
Output: 1
Explanation: Following are the operations performed in arr[] to get the desired result.
Adding 1 to each element in arr[] updates arr[] to arr[] = {3, 5, 6}
Now XOR of all the elements in arr[] is 3^5^6 = 0. 
Therefore, 1 is the required answer.  

Input: arr[] = {4, 5, 13}
Output: -1
Explanation: No such x exists for fulfilling the desired conditions. 

方法：奇数个1的XOR = 1，而偶数个1 = 0。这个想法可以用来解决给定的问题。请按照以下步骤操作：

• 初始化变量X = 0 。
• 数组元素的二进制表示将用于遍历元素并确定X 。
• 从第 0 位开始遍历到第 63 位。
• 如果在任何位位置，数组元素的设置位 (1) 总数为奇数，则将 2 的幂与 X 相加。
• 如果在迭代完成后在任何位位置有奇数个 1，则不存在这样的 X。否则，打印 X 作为答案。

请参见下面的插图：

Illustration:

Case-1 (X possible): Take arr[] = { 2, 4, 5} 

	5th	4th	3rd	2nd	1st	0th
arr[0]	0	0	0	0	1	0
arr[1]	0	0	0	1	0	0
arr[2]	0	0	0	1	0	1
X	0	0	0	0	0	0

• Initially, X = 0 0 0 0 0 0, at 0th position set(1) bits are odd, so in order to make the set bits even,  flip the bits at 0th position. So, for flipping the bits just add (0 0 0 0 0 1) to all the elements of arr[] and in X.
• Now, the table will look like the following:

	5th	4th	3rd	2nd	1st	0th
arr[0]	0	0	0	0	1	1
arr[1]	0	0	0	1	0	1
arr[2]	0	0	0	1	1	0
XOR	0	0	0	0	0	0
X	0	0	0	0	0	1

• Now, the XOR of (arr[0]+x) ^  ( arr[1]+x) ^ arr[2]+x) = 0, result will be 0. So, print  res = X.

在不可能的情况下采取以下 X

Case-2: Take example: arr[] = { 4, 5, 13 } 

	5th	4th	3rd	2nd	1st	0th
arr[0]	0	0	0	1	0	0
arr[1]	0	0	0	1	0	1
arr[2]	0	0	1	1	0	1
XOR	0	0	1	1	0	0
X	0	0	0	0	0	0

          XOR = Arr[0] ^ Arr[1] ^ Arr[2] = 1 1 0 0,   Here there are odd number of 1‘s at 2nd and 3rd bits.

• So, add 2pow(2nd) to all elements of arr and in X, then again will take the XOR, after this the elements become:

	5th	4th	3rd	2nd	1st	0th
arr[0]	0	0	1	0	0	0
arr[1]	0	0	1	0	0	1
arr[2]	0	1	0	0	0	1
XOR	0	1	0	0	0	0
X	0	0	0	1	0	0

If this keeps on going the left most 1 in XOR keeps on moving left.

下面是上述方法的实现：

C++

// C++ program for above approach
#include 
using namespace std;
 
// Function to find required result
long long solve(vector& a,
                int n)
{
    long long res = 0, j = 0, one = 1;
 
    // For 64 Bit
    while (j < 64) {
        // j is traversing each bit
        long long Xor = 0;
        long long powerOf2 = one << j;
 
        for (auto x : a)
            Xor ^= x;
 
        if (j == 63 && (Xor & powerOf2))
            return -1;
 
        if (Xor & powerOf2) {
            res += powerOf2;
            for (int i = 0; i < n; i++)
                a[i] += powerOf2;
        }
        j++;
    }
    return res;
}
 
// Driver Code
int main()
{
 
    // Size of arr[]
    int N = 3;
    vector arr = { 2, 4, 5 };
 
    cout << solve(arr, N) << '\n';
 
    return 0;
}

Java

// Java program for above approach
class GFG{
 
// Function to find required result
static long solve(int[] a,
                int n)
{
    long res = 0, j = 0, one = 1;
 
    // For 64 Bit
    while (j < 64)
    {
       
        // j is traversing each bit
        long Xor = 0;
        long powerOf2 = one << j;
 
        for (int x : a)
            Xor ^= x;
 
        if (j == 63 && (Xor & powerOf2)!=0)
            return -1;
 
        if ((Xor & powerOf2)!=0) {
            res += powerOf2;
            for (int i = 0; i < n; i++)
                a[i] += powerOf2;
        }
        j++;
    }
    return res;
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Size of arr[]
    int N = 3;
    int[] arr = { 2, 4, 5 };
 
    System.out.print(solve(arr, N));
}
}
 
// This code is contributed by shikhasingrajput

Python3

# python program for above approach
 
# Function to find required result
def solve(a, n):
 
    res = 0
    j = 0
    one = 1
 
    # For 64 Bit
    while (j < 64):
                # j is traversing each bit
        Xor = 0
        powerOf2 = one << j
 
        for x in a:
            Xor ^= x
 
        if (j == 63 and (Xor & powerOf2)):
            return -1
 
        if (Xor & powerOf2):
            res += powerOf2
            for i in range(0, n):
                a[i] += powerOf2
 
        j += 1
 
    return res
 
# Driver Code
if __name__ == "__main__":
 
        # Size of arr[]
    N = 3
    arr = [2, 4, 5]
 
    print(solve(arr, N))
 
    # This code is contributed by rakeshsahni

C#

// C# program for above approach
using System;
class GFG
{
 
    // Function to find required result
    static long solve(int[] a, int n)
    {
        int res = 0, j = 0, one = 1;
 
        // For 64 Bit
        while (j < 64)
        {
 
            // j is traversing each bit
            long Xor = 0;
            long powerOf2 = one << j;
 
            foreach (int x in a)
                Xor ^= x;
 
            if (j == 63 && (Xor & powerOf2) != 0)
                return -1;
 
            if ((Xor & powerOf2) != 0)
            {
                res += (int)powerOf2;
                for (int i = 0; i < n; i++)
                    a[i] += (int)powerOf2;
            }
            j++;
        }
        return res;
    }
 
    // Driver Code
    public static void Main()
    {
 
        // Size of arr[]
        int N = 3;
        int[] arr = { 2, 4, 5 };
 
        Console.Write(solve(arr, N));
    }
}
 
// This code is contributed by gfgking

Javascript

 
// JavaScript program for above approach
 
// Function to find required result
function solve(a, n)
{
    let res = 0, j = 0, one = 1;
     
    // For 64 Bit
    while (j < 64)
    {
         
        // j is traversing each bit
        let Xor = 0;
        let powerOf2 = one << j;
     
        for(let x of a)
            Xor ^= x;
     
        if (j == 63 && (Xor & powerOf2))
            return -1;
     
        if (Xor & powerOf2)
        {
            res += powerOf2;
            for(let i = 0; i < n; i++)
                a[i] += powerOf2;
        }
        j++;
    }
    return res;
}
 
// Driver Code
 
// Size of arr[]
let N = 3;
let arr = [ 2, 4, 5 ];
 
document.write(solve(arr, N) + '<br>');
 
// This code is contributed by Potta Lokesh
 

输出

1

时间复杂度： O(N*logN)
辅助空间： O(1)