通过将元素更改为负数来最大化 Array 中唯一元素的数量

给定包含N个整数的数组arr 。任务是找到数组中唯一元素的最大数量，如果数组中的每个元素都可以更改为负数，即可以将数组中的X更改为-X 。

例子：

Input: arr[] = {-1, 3, 2, 3, 2}
Output: 5
Explanation: Change one 2 to -2 and another 3 to -3 to get the arr[]={-1, 3, 2, -3, -2}, having 5 unique values.

Input: arr[] = {1, 2, 2, 2, 3, 3, 3}
Output: 5

方法：可以在这个问题中使用一个集合。请按照以下步骤解决：

• 从i=0到i遍历数组，并且对于每个元素arr[i] ：
  • 检查集合中是否存在abs(arr[i]) 。
  • 如果它不存在，则将其插入集合中。
  • 否则在集合中插入原始值。
• 由于集合仅包含原始值，因此答案将是集合的大小。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the number of unique elements
int CountUniqueElements(int arr[], int N)
{
    // Set to hold unique values
    unordered_set s;
    int i = 0;
 
    // Loop to determine and
    // store all unique values
    while (i < N) {
        // Positive value of arr[i]
        int val1 = abs(arr[i]);
 
        // Negative value of arr[i]
        int val2 = -val1;
 
        // Checking if val1 is present or not
        // If not then insert val1 in the set
        // Insert val2 in the set
        if (s.count(val1)) {
            s.insert(val2);
        }
 
        // Else inserting the original value
        else {
            s.insert(val1);
        }
        i++;
    }
 
    // Return the count of unique
    // values in the Array
    return s.size();
}
 
// Driver Code
int main()
{
    // Declaring Array of size 7
    int arr[] = { 1, 2, 2, 2, 3, 3, 3 };
    int N = sizeof(arr) / sizeof(int);
 
    cout << CountUniqueElements(arr, N);
    return 0;
}

---

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
  // Function to find the number of unique elements
  static int CountUniqueElements(int arr[], int N)
  {
     
    // Set to hold unique values
    Set s = new HashSet();
    int i = 0;
 
    // Loop to determine and
    // store all unique values
    while (i < N)
    {
       
      // Positive value of arr[i]
      int val1 = Math.abs(arr[i]);
 
      // Negative value of arr[i]
      int val2 = -val1;
 
      // Checking if val1 is present or not
      // If not then insert val1 in the set
      // Insert val2 in the set
      if (s.contains(val1)) {
        s.add(val2);
      }
 
      // Else inserting the original value
      else {
        s.add(val1);
      }
      i++;
    }
 
    // Return the count of unique
    // values in the Array
    return s.size();
  }
 
  // Driver Code
  public static void main (String[] args)
  {
     
    // Declaring Array of size 7
    int arr[] = { 1, 2, 2, 2, 3, 3, 3 };
    int N =arr.length;
    System.out.print(CountUniqueElements(arr, N));
  }
}
 
// This code is contributed by hrithikgarg03188.

Python3

# Python 3 program for the above approach
 
# Function to find the number of unique elements
def CountUniqueElements(arr, N):
 
    # Set to hold unique values
    s = set([])
    i = 0
 
    # Loop to determine and
    # store all unique values
    while (i < N):
        # Positive value of arr[i]
        val1 = abs(arr[i])
 
        # Negative value of arr[i]
        val2 = -val1
 
        # Checking if val1 is present or not
        # If not then insert val1 in the set
        # Insert val2 in the set
        if (list(s).count(val1)):
            s.add(val2)
 
        # Else inserting the original value
        else:
            s.add(val1)
 
        i += 1
 
    # Return the count of unique
    # values in the Array
    return len(s)
 
# Driver Code
if __name__ == "__main__":
 
    # Declaring Array of size 7
    arr = [1, 2, 2, 2, 3, 3, 3]
    N = len(arr)
 
    print(CountUniqueElements(arr, N))
 
    # This code is contributed by ukasp.

C#

// C# program for the above approach
 
using System;
using System.Collections.Generic;
 
class GFG
{
 
  // Function to find the number of unique elements
  static int CountUniqueElements(int[] arr, int N)
  {
 
    // Set to hold unique values
    HashSet s = new HashSet();
    int i = 0;
 
    // Loop to determine and
    // store all unique values
    while (i < N)
    {
 
      // Positive value of arr[i]
      int val1 = Math.Abs(arr[i]);
 
      // Negative value of arr[i]
      int val2 = -val1;
 
      // Checking if val1 is present or not
      // If not then insert val1 in the set
      // Insert val2 in the set
      if (s.Contains(val1))
      {
        s.Add(val2);
      }
 
      // Else inserting the original value
      else
      {
        s.Add(val1);
      }
      i++;
    }
 
    // Return the count of unique
    // values in the Array
    return s.Count;
  }
 
  // Driver Code
  public static void Main()
  {
 
    // Declaring Array of size 7
    int[] arr = { 1, 2, 2, 2, 3, 3, 3 };
    int N = arr.Length;
    Console.Write(CountUniqueElements(arr, N));
  }
}
 
// This code is contributed by gfgking

Javascript

      // JavaScript code for the above approach
 
      // Function to find the number of unique elements
      function CountUniqueElements(arr, N)
      {
       
          // Set to hold unique values
          let s = new Set();
          let i = 0;
 
          // Loop to determine and
          // store all unique values
          while (i < N)
          {
           
              // Positive value of arr[i]
              let val1 = Math.abs(arr[i]);
 
              // Negative value of arr[i]
              let val2 = -val1;
 
              // Checking if val1 is present or not
              // If not then insert val1 in the set
              // Insert val2 in the set
              if (s.has(val1)) {
                  s.add(val2);
              }
 
              // Else inserting the original value
              else {
                  s.add(val1);
              }
              i++;
          }
 
          // Return the count of unique
          // values in the Array
          return s.size;
      }
 
      // Driver Code
 
      // Declaring Array of size 7
      let arr = [1, 2, 2, 2, 3, 3, 3];
      let N = arr.length;
 
      document.write(CountUniqueElements(arr, N));
 
       // This code is contributed by Potta Lokesh
  

输出

5

时间复杂度： O(N)
辅助空间： O(N)