计数每次删除后剩余的成对相等数组元素

给定大小为N的数组arr [] ，每个数组元素arr [i]的任务是计算可以通过从数组中删除arr [i]而获得的相等元素对的数量。

例子：

Input: arr[] = { 1, 1, 1, 2 } 
Output: 1 1 1 3 
Explanation: 
Removing arr[0] from the array modifies arr[] to { 1, 1, 2 } and count of pairs of equal elements = 1 
Removing arr[1] from the array modifies arr[] to { 1, 1, 2 } and count of pairs of equal elements = 1 
Removing arr[2] from the array modifies arr[] to { 1, 1, 2 } and count of pairs of equal elements = 1 
Removing arr[3] from the array modifies arr[] to { 1, 1, 1 } and count of pairs of equal elements = 3 
Therefore, the required output is 1 1 1 3.

Input: arr[] = { 2, 3, 4, 3, 2 } 
Output: 1 1 2 1 1

天真的方法：解决此问题的最简单方法是遍历数组，并对于每个^第i^个元素从数组中删除arr [i]并打印保留在数组中的相等数组元素对的计数。

时间复杂度： O(N ² )
辅助空间： O(N)

高效方法：请按照以下步骤解决问题：

• 初始化一个映射，例如mp ，以存储数组中每个不同元素的频率。
• 初始化一个变量，例如cntPairs ，以存储相等数组元素对的总数。
• 遍历地图，并通过将cntPairs的值增加(mp [i] *(mp [i] – 1))/ 2来存储相等元素对的总数。
• 最后，遍历数组。对于每个^第i^个元素，输出(cntPairs – mp [i] + 1)的值，该值表示通过从数组中删除arr [i]来获得相等数组元素对的数量。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to count pairs of equal elements
// by removing arr[i] from the array
void pairs_after_removing(int arr[], int N)
{
    // Stores total count of
    // pairs of equal elements
    int cntPairs = 0;
 
    // Store frequency of each
    // distinct array element
    unordered_map mp;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Update frequency of arr[i]
        mp[arr[i]]++;
    }
 
    // Traverse the map
    for (auto element : mp) {
 
        // Stores key of an element
        int i = element.first;
        cntPairs += mp[i] * (mp[i] - 1) / 2;
    }
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Stores count of pairs of equal
        // element by removing arr[i]
        int pairs_after_arr_i_removed
            = cntPairs + 1 - mp[arr[i]];
 
        cout << pairs_after_arr_i_removed << ' ';
    }
    return;
}
 
// Driver Code
int main()
{
    // Given Array
    int arr[] = { 2, 3, 4, 3, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    pairs_after_removing(arr, N);
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
  
class GFG{
  
// Function to count pairs of equal elements
// by removing arr[i] from the array
static void pairs_after_removing(int arr[], int N)
{
     
    // Stores total count of
    // pairs of equal elements
    int cntPairs = 0;
  
    // Store frequency of each
    // distinct array element
    Map mp = new HashMap();
  
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
         
        // Update frequency of arr[i]
        mp.put(arr[i], mp.getOrDefault(arr[i], 0) + 1);
    }
  
    // Traverse the map
    for(Map.Entry element : mp.entrySet())
    {
         
        // Stores key of an element
        int i = element.getKey();
        cntPairs += mp.get(i) * (mp.get(i) - 1) / 2;
    }
  
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
         
        // Stores count of pairs of equal
        // element by removing arr[i]
        int pairs_after_arr_i_removed = cntPairs +
                           1 - mp.get(arr[i]);
  
        System.out.print(pairs_after_arr_i_removed + " ");
    }
    return;
}
  
// Driver code
public static void main(String[] args)
{
     
    // Given Array
    int arr[] = { 2, 3, 4, 3, 2 };
    int N = arr.length;
  
    pairs_after_removing(arr, N);
}
}
 
// This code is contributed by susmitakundugoaldanga

Python3

# python program to implement
# the above approach
 
# Function to count pairs of equal elements
# by removing arr[i] from the array
def pairs_after_removing(arr, N):
     
    # Stores total count of
    # pairs of equal elements
    cntPairs = 0
 
    # Store frequency of each
    # distinct array element
    mp = {}
 
    # Traverse the array
    for i in arr:
 
        # Update frequency of arr[i]
        mp[i] = mp.get(i, 0) + 1
 
    # Traverse the map
    for element in mp:
 
        # Stores key of an element
        i = element
        cntPairs += mp[i] * (mp[i] - 1) // 2
 
    # Traverse the array
    for i in range(N):
 
        # Stores count of pairs of equal
        # element by removing arr[i]
        pairs_after_arr_i_removed = cntPairs + 1 - mp[arr[i]]
 
        print(pairs_after_arr_i_removed, end = ' ')
    return
 
# Driver Code
if __name__ == '__main__':
   
    # Given Array
    arr = [2, 3, 4, 3, 2]
    N = len(arr)
    pairs_after_removing(arr, N)
 
# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
   
class GFG
{
   
// Function to count pairs of equal elements
// by removing arr[i] from the array
static void pairs_after_removing(int[] arr, int N)
{
      
    // Stores total count of
    // pairs of equal elements
    int cntPairs = 0;
   
    // Store frequency of each
    // distinct array element
     Dictionary mp = new Dictionary();
   
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
          
        // Update frequency of arr[i]
        if(mp.ContainsKey(arr[i]))
            {
                mp[arr[i]]++;
            }
            else
            {
                mp[arr[i]] = 1;
            }
    }
   
    // Traverse the map
    foreach(KeyValuePair element in mp)
    {
          
        // Stores key of an element
        int i = element.Key;
        cntPairs += mp[i] * (mp[i] - 1) / 2;
    }
   
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
          
        // Stores count of pairs of equal
        // element by removing arr[i]
        int pairs_after_arr_i_removed = cntPairs +
                           1 - mp[arr[i]];
   
        Console.Write(pairs_after_arr_i_removed + " ");
    }
    return;
}
   
// Driver code
public static void Main()
{
      
    // Given Array
    int[] arr = { 2, 3, 4, 3, 2 };
    int N = arr.Length;
   
    pairs_after_removing(arr, N);
}
}
 
// This code is contributed by sanjoy_62

输出：

1 1 2 1 1

时间复杂度： O(N)
辅助空间： O(N)