缩小数组，使每个元素最多出现2次

给定大小为N的排序数组arr ，任务是缩小数组，以使每个元素最多可以出现两次。
例子：

Input: arr[] = {1, 2, 2, 2, 3} 
Output: {1, 2, 2, 3} 
Explanation: 
Remove 2 once, as it occurs more than 2 times.

Input: arr[] = {3, 3, 3} 
Output: {3, 3} 
Explanation: 
Remove 3 once, as it occurs more than 2 times.  

方法：这可以借助两个指针算法来解决。

1. 从左侧开始travsering数组，并保持两个指针。
2. 一个指针(让我说i)用于迭代数组。
3. 第二个指针(让我们说st)向前移动以查找下一个唯一元素，第i个元素出现了两次以上。

下面是上述方法的实现：

CPP

// C++ program to reduce the array
// such that each element appears
// at most 2 times
 
#include 
using namespace std;
 
// Function to remove duplicates
void removeDuplicates(int arr[], int n)
{
    // Initalise 2nd pointer
    int st = 0;
 
    // Itereate over the array
    for (int i = 0; i < n; i++) {
 
        if (i < n - 2
            && arr[i] == arr[i + 1]
            && arr[i] == arr[i + 2])
            continue;
 
        // Updating the 2nd pointer
        else {
            arr[st] = arr[i];
            st++;
        }
    }
 
    cout << "{";
    for (int i = 0; i < st; i++) {
        cout << arr[i];
 
        if (i != st - 1)
            cout << ", ";
    }
    cout << "}";
}
 
// Driver code
int main()
{
    int arr[]
        = { 1, 1, 1, 2,
            2, 2, 3, 3,
            3, 3, 3, 3,
            4, 5 };
 
    int n = sizeof(arr)
            / sizeof(arr[0]);
 
    // Function call
    removeDuplicates(arr, n);
 
    return 0;
}

Java

// Java program to reduce the array
// such that each element appears
// at most 2 times
class GFG
{
 
// Function to remove duplicates
static void removeDuplicates(int arr[], int n)
{
    // Initalise 2nd pointer
    int st = 0;
 
    // Itereate over the array
    for (int i = 0; i < n; i++) {
 
        if (i < n - 2
            && arr[i] == arr[i + 1]
            && arr[i] == arr[i + 2])
            continue;
 
        // Updating the 2nd pointer
        else {
            arr[st] = arr[i];
            st++;
        }
    }
 
    System.out.print("{");
    for (int i = 0; i < st; i++) {
        System.out.print(arr[i]);
 
        if (i != st - 1)
            System.out.print(", ");
    }
    System.out.print("}");
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 1, 1, 2,
                  2, 2, 3, 3,
                  3, 3, 3, 3,
                  4, 5 };
 
    int n = arr.length;
 
    // Function call
    removeDuplicates(arr, n);
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 program to reduce the array
# such that each element appears
# at most 2 times
 
# Function to remove duplicates
def removeDuplicates(arr, n) :
 
    # Initalise 2nd pointer
    st = 0;
 
    # Itereate over the array
    for i in range(n) :
 
        if (i < n - 2 and arr[i] == arr[i + 1]
            and arr[i] == arr[i + 2]) :
            continue;
 
        # Updating the 2nd pointer
        else :
            arr[st] = arr[i];
            st += 1;
 
    print("{",end="")
    for i in range(st) :
        print(arr[i],end="");
         
        if (i != st - 1) :
            print(", ",end="");
     
    print("}",end="");
 
# Driver code
if __name__ == "__main__" :
 
    arr = [ 1, 1, 1, 2,
            2, 2, 3, 3,
            3, 3, 3, 3,
            4, 5 ];
 
    n = len(arr);
     
    # Function call
    removeDuplicates(arr, n);
 
# This code is contributed by Yash_R

C#

// C# program to reduce the array
// such that each element appears
// at most 2 times
using System;
 
class GFG
{
  
// Function to remove duplicates
static void removeDuplicates(int []arr, int n)
{
    // Initalise 2nd pointer
    int st = 0;
  
    // Itereate over the array
    for (int i = 0; i < n; i++) {
  
        if (i < n - 2
            && arr[i] == arr[i + 1]
            && arr[i] == arr[i + 2])
            continue;
  
        // Updating the 2nd pointer
        else {
            arr[st] = arr[i];
            st++;
        }
    }
  
    Console.Write("{");
    for (int i = 0; i < st; i++) {
        Console.Write(arr[i]);
  
        if (i != st - 1)
            Console.Write(", ");
    }
    Console.Write("}");
}
  
// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 1, 1, 2,
                  2, 2, 3, 3,
                  3, 3, 3, 3,
                  4, 5 };
  
    int n = arr.Length;
  
    // Function call
    removeDuplicates(arr, n);
}
}
 
// This code is contributed by sapnasingh4991

输出

{1, 1, 2, 2, 3, 3, 4, 5}

时间复杂度： O(N)
空间复杂度： O(1)

另一种方法：使用Counter()函数

• 使用计数器函数计算所有元素的频率。
• 拿一个空清单。
• 遍历数组。
• 如果任何元素的频率大于或等于2，则将其频率设为1并将其附加到列表中。
• 如果任何元素的频率等于1，则将其频率设为0，并将其附加到列表中。
• 打印列表。

下面是上述方法的实现：

Python3

# Python3 program to reduce the array
# such that each element appears
# at most 2 times
from collections import Counter
 
# Function to remove duplicates
def removeDuplicates(arr, n):
    freq = Counter(arr)
     
    # Taking empty list
    l = []
    for i in range(n):
       
        if(freq[arr[i]] >= 2):
           
            # Making frequency to 1
            freq[arr[i]] = 1
            l.append(arr[i])
             
        elif(freq[arr[i]] == 1):
             
            # Making frequency to 0
            # and appending to list
            l.append(arr[i])
            freq[arr[i]] = 0
             
    # Printing the list
    for i in l:
        print(i, end=" ")
 
 
# Driver code
if __name__ == "__main__":
 
    arr = [1, 1, 1, 2,
           2, 2, 3, 3,
           3, 3, 3, 3,
           4, 5]
 
    n = len(arr)
 
    # Function call
    removeDuplicates(arr, n)
 
# This code is contributed by vikkycirus

输出

1 1 2 2 3 3 4 5 

时间复杂度： O(N)

空间复杂度： O(N)