通过将两个 0 或两个 1 对替换为 0 并将 10 或 01 对替换为 1 来减少二进制数组

给定一个大小为N的二进制数组arr[] ，任务是在执行一组操作后找到数组中剩余的最后一个数。在每个操作中，选择任意两个数字并执行以下操作：

如果两个数字相同，则将它们从数组中删除并插入0 。
如果两个数字不同，则删除它们并插入1 。

例子：

Input: arr[]={0, 0, 1}
Output: 1
Explanation: There are two possible sequence of operations as follows:

arr[] = {0, 0, 1}, delete (0, 1) and insert 0 => arr[] = {0, 0}, delete (0, 0) and insert 1=> arr[] = {1}.
arr[] = {0, 0, 1}, delete (0, 0) and insert 0 => arr[] = {0, 1}, delete (0, 1) and insert 1=> arr[] = {1}.

Hence the remaining element is 1.

Input: arr[]={1, 0, 0, 0, 1}
Output: 0

编程需要懂一点英语

方法：可以根据以下观察解决给定的问题：

2 个相同的数字被 0 取代。
2 个不同的数字被 1 取代。

现在，为每个结果创建一个表：

仔细观察上表，可以注意到该表表示的是按位异或运算。因此，剩余的整数将等于给定数组元素的按位异或，可以进一步简化为 1 的频率为偶数，结果为0 ，否则为1 。

下面是上述方法的实现。

C++

// C++ program of the above approach
#include 
using namespace std;
 
// Function to find last remaining
// integer in the given array
int lastNumber(vector& arr)
{
 
    // Variable to store the
    // frequency of 1
    int one = 0;
 
    // Loop to iterate the
    // given array
    for (int x : arr) {
        if (x == 1) {
            one += 1;
        }
    }
 
    // If frequency of 1 is even
    if (one % 2 == 0)
        return 0;
 
    // If frequency of 1 is odd
    return 1;
}
 
// Driver Code
int main()
{
    vector arr = { 1, 0, 0, 0, 1 };
    cout << lastNumber(arr);
}

Java

// Java program of the above approach
import java.util.ArrayList;
class GFG {
 
  // Function to find last remaining
  // integer in the given array
  static Integer lastNumber(ArrayList arr)
  {
 
    // Variable to store the
    // frequency of 1
    int one = 0;
 
    // Loop to iterate the
    // given array
    for (int x : arr) {
      if (x == 1) {
        one += 1;
      }
    }
 
    // If frequency of 1 is even
    if (one % 2 == 0)
      return 0;
 
    // If frequency of 1 is odd
    return 1;
  }
 
  // Driver Code
  public static void main(String args[]) {
    ArrayList arr = new ArrayList();
    arr.add(1);
    arr.add(0);
    arr.add(0);
    arr.add(0);
    arr.add(1);
 
    System.out.println(lastNumber(arr));
  }
}
 
// This code is contributed by gfgking

Python3

# python program of the above approach
 
# Function to find last remaining
# integer in the given array
def lastNumber(arr):
 
    # Variable to store the
    # frequency of 1
    one = 0
 
    # Loop to iterate the
    # given array
    for x in arr:
        if (x == 1):
            one += 1
 
    # If frequency of 1 is even
    if (one % 2 == 0):
        return 0
 
    # If frequency of 1 is odd
    return 1
 
# Driver Code
if __name__ == "__main__":
 
    arr = [1, 0, 0, 0, 1]
    print(lastNumber(arr))
 
# This code is contributed by rakeshsahni

C#

// C# program of the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find last remaining
// integer in the given array
static int lastNumber(List arr)
{
     
    // Variable to store the
    // frequency of 1
    int one = 0;
 
    // Loop to iterate the
    // given array
    foreach(int x in arr)
    {
        if (x == 1)
        {
            one += 1;
        }
    }
 
    // If frequency of 1 is even
    if (one % 2 == 0)
        return 0;
 
    // If frequency of 1 is odd
    return 1;
}
 
// Driver Code
public static void Main()
{
    List arr = new List(){ 1, 0, 0, 0, 1 };
     
    Console.WriteLine(lastNumber(arr));
}
}
 
// This code is contributed by ukasp

Javascript

输出

时间复杂度： O(N)
辅助空间： O(1)