在混洗后的另一个数组中查找缺失的数字

给定一个n个正整数的数组“ arr1”。 arr1 []的内容被复制到另一个数组’arr2’，但是数字被重新排列，一个元素被删除。查找缺少的元素(不使用任何额外的空间，并且时间复杂度为O(n))。

例子 ：

Input : arr1[] = {4, 8, 1, 3, 7}, 
        arr2[] = {7, 4, 3, 1}
Output : 8

Input : arr1[] = {12, 10, 15, 23, 11, 30}, 
        arr2[] = {15, 12, 23, 11, 30}
Output : 10

一个简单的解决方案是逐个考虑第一个数组的每个元素，然后在第二个数组中搜索。一旦找到缺少的元素，我们将返回。该解决方案的时间复杂度为O(n ² )

一个有效的解决方案基于异或。每个元素的组合出现次数是两次，一个在“ arr1”中，另一个在“ arr2”中，除了一个元素在“ arr1”中仅出现一次。我们知道(a Xor a)=0。因此，只需对两个数组的元素进行XOR。结果将是缺少的数字。

C++

// C++ implementation to find the
// missing number in shuffled array
// C++ implementation to find the
// missing number in shuffled array
#include 
using namespace std;
  
// Returns the missing number
// Size of arr2[] is n-1
int missingNumber(int arr1[], int arr2[], 
                                   int n)
{
    // Missing number 'mnum'
    int mnum = 0;
  
    // 1st array is of size 'n'
    for (int i = 0; i < n; i++)
        mnum = mnum ^ arr1[i];
  
    // 2nd array is of size 'n - 1'
    for (int i = 0; i < n - 1; i++)
        mnum = mnum ^ arr2[i];
  
    // Required missing number
    return mnum;
}
  
// Driver Code
int main()
{
    int arr1[] = {4, 8, 1, 3, 7};
    int arr2[] = {7, 4, 3, 1};
    int n = sizeof(arr1) / sizeof(arr1[0]);
    cout << "Missing number = "
        << missingNumber(arr1, arr2, n);
    return 0;
}

Java

// Java implementation to find the
// missing number in shuffled array
  
class GFG 
{
    // Returns the missing number
    // Size of arr2[] is n-1
    static int missingNumber(int arr1[], 
                             int arr2[],
                                  int n)
    {
        // Missing number 'mnum'
        int mnum = 0;
      
        // 1st array is of size 'n'
        for (int i = 0; i < n; i++)
            mnum = mnum ^ arr1[i];
      
        // 2nd array is of size 'n - 1'
        for (int i = 0; i < n - 1; i++)
            mnum = mnum ^ arr2[i];
      
        // Required missing number
        return mnum;
    }
      
    // Driver Code
    public static void main (String[] args) 
    {
        int arr1[] = {4, 8, 1, 3, 7};
        int arr2[] = {7, 4, 3, 1};
        int n = arr1.length;
          
        System.out.println("Missing number = "
            + missingNumber(arr1, arr2, n));
    }
}

Python3

# Python3 implementation to find the
# missing number in shuffled array
  
# Returns the missing number
# Size of arr2[] is n - 1
def missingNumber(arr1, arr2, n):
  
    # missing number 'mnum'
    mnum = 0
  
    # 1st array is of size 'n'
    for i in range(n):
        mnum = mnum ^ arr1[i]
  
    # 2nd array is of size 'n - 1'
    for i in range(n - 1):
        mnum = mnum ^ arr2[i]
  
    # Required missing number
    return mnum
  
# Driver Code
arr1 = [4, 8, 1, 3, 7]
arr2= [7, 4, 3, 1]
n = len(arr1)
print("Missing number = ", 
    missingNumber(arr1, arr2, n))
          
# This code is contributed by Anant Agarwal.

C#

// C# implementation to find the
// missing number in shuffled array
using System;
  
class GFG
{
    // Returns the missing number
    // Size of arr2[] is n-1
    static int missingNumber(int []arr1, 
                             int []arr2, 
                                  int n)
    {
        // Missing number 'mnum'
        int mnum = 0;
      
        // 1st array is of size 'n'
        for (int i = 0; i < n; i++)
            mnum = mnum ^ arr1[i];
      
        // 2nd array is of size 'n - 1'
        for (int i = 0; i < n - 1; i++)
            mnum = mnum ^ arr2[i];
      
        // Required missing number
        return mnum;
    }
      
    // Driver Code
    public static void Main () 
    {
        int []arr1 = {4, 8, 1, 3, 7};
        int []arr2 = {7, 4, 3, 1};
        int n = arr1.Length;
          
    Console.Write("Missing number = "
            + missingNumber(arr1, arr2, n));
    }
}
  
// This code is contributed by nitin mittal.

输出：

Missing number = 8

复杂度：O(n)时间复杂度和O(1)额外空间。