在数组中查找频率等于其他元素的频率之和的元素

给定一个整数数组arr [] ，任务是在数组中找到一个元素，该元素的频率等于该数组其他元素的频率之和。

例子：

Input: arr[] = {1, 2, 2, 3, 3, 3}
Output: 3
Explanation:
Frequencies of elements of the array –
Frequency(3) = 3
Frequency(2) = 2
Frequency(1) = 1
Here, Frequency of the element 3 is equal to the
sum of frequencies of other elements of the array.

Input: arr[] = {1, 2, 3}
Output: -1
Explanation:
In the above-given array, there is no such
element whose frequency is equal to the sum of
frequencies of other elements of the array.

方法：问题中的关键观察因素是：如果数组的长度为奇数，则将不存在此类元素；而对于偶数长度的数组，请计算数组中每个元素的频率，然后最后检查数组中的任何元素阵列的频率等于阵列长度的一半。

下面是上述方法的实现：

C++

// C++ implementation to find the
// element whose frequency is equal 
// to the sum of frequencies of 
// other elements of the array
  
#include 
using namespace std;
  
// Function to check that any element 
// have frequency equal to the sum of 
// frequency of other elements of the array
bool isFrequencyEqual(int arr[], int len)
{
    // Check that if the array length
    // is odd, Then no solution possible
    if (len % 2 == 1){
        cout << "No Such Element";
        return false;
    }
  
    // Hash-map to store the frequency
    // of elements of array
    map freq;
      
    // Loop to find the frequency 
    // of elements of array
    for (int i = 0; i < len; i++)
        freq[arr[i]]++;
          
    // Loop to check if any element 
    // of the array have frequency 
    // equal to the half length
    for (int i = 0; i < len; i++){
        if (freq[arr[i]] == len / 2){
            cout << arr[i] << endl;
            return true;
        }
    }
          
    cout << "No such element";
    return false;
}
  
// Driver Code
int main()
{
    int arr[6] = { 1, 2, 2, 3, 3, 3 };
    int n = 6;
      
    // Function Call
    isFrequencyEqual(arr, n);
    return 0;
}

Java

// Java implementation to find the
// element whose frequency is equal 
// to the sum of frequencies of 
// other elements of the array
import java.util.*;
  
class GFG{
   
// Function to check that any element 
// have frequency equal to the sum of 
// frequency of other elements of the array
static boolean isFrequencyEqual(int arr[], int len)
{
    // Check that if the array length
    // is odd, Then no solution possible
    if (len % 2 == 1){
        System.out.print("No Such Element");
        return false;
    }
   
    // Hash-map to store the frequency
    // of elements of array
    HashMap freq = new HashMap();
       
    // Loop to find the frequency 
    // of elements of array
    for (int i = 0; i < len; i++)
        if(freq.containsKey(arr[i])){
            freq.put(arr[i], freq.get(arr[i])+1);
        }
        else{
            freq.put(arr[i], 1);
        }
           
    // Loop to check if any element 
    // of the array have frequency 
    // equal to the half length
    for (int i = 0; i < len; i++){
        if (freq.containsKey(arr[i]) && freq.get(arr[i]) == len / 2){
            System.out.print(arr[i] +"\n");
            return true;
        }
    }
           
    System.out.print("No such element");
    return false;
}
   
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 2, 3, 3, 3 };
    int n = 6;
       
    // Function Call
    isFrequencyEqual(arr, n);
}
}
  
// This code is contributed by Rajput-Ji

C#

// C# implementation to find the
// element whose frequency is equal 
// to the sum of frequencies of 
// other elements of the array
using System;
using System.Collections.Generic;
  
class GFG{
    
// Function to check that any element 
// have frequency equal to the sum of 
// frequency of other elements of the array
static bool isFrequencyEqual(int []arr, int len)
{
    // Check that if the array length
    // is odd, Then no solution possible
    if (len % 2 == 1){
        Console.Write("No Such Element");
        return false;
    }
    
    // Hash-map to store the frequency
    // of elements of array
    Dictionary freq = new Dictionary();
        
    // Loop to find the frequency 
    // of elements of array
    for (int i = 0; i < len; i++)
        if(freq.ContainsKey(arr[i])){
            freq[arr[i]] = freq[arr[i]]+1;
        }
        else{
            freq.Add(arr[i], 1);
        }
            
    // Loop to check if any element 
    // of the array have frequency 
    // equal to the half length
    for (int i = 0; i < len; i++){
        if (freq.ContainsKey(arr[i]) && freq[arr[i]] == len / 2){
            Console.Write(arr[i] +"\n");
            return true;
        }
    }
            
    Console.Write("No such element");
    return false;
}
    
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 2, 3, 3, 3 };
    int n = 6;
        
    // Function Call
    isFrequencyEqual(arr, n);
}
}
  
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation to find the
# element whose frequency is equal 
# to the sum of frequencies of 
# other elements of the array
  
# Function to check that any element 
# have frequency equal to the sum of 
# frequency of other elements of the array
def isFrequencyEqual(arr, length) :
  
    # Check that if the array length
    # is odd, Then no solution possible
    if (length % 2 == 1) :
        print("No Such Element");
        return False;
  
    # Hash-map to store the frequency
    # of elements of array
    freq = dict.fromkeys(arr, 0);
      
    # Loop to find the frequency 
    # of elements of array
    for i in range(length) :
        freq[arr[i]] += 1;
          
    # Loop to check if any element 
    # of the array have frequency 
    # equal to the half length
    for i in range(length) :
        if (freq[arr[i]] == length / 2) :
            print(arr[i]);
            return True;
          
    print("No such element",end="");
    return False;
  
# Driver Code
if __name__ == "__main__" :
  
    arr = [ 1, 2, 2, 3, 3, 3 ];
    n = 6;
      
    # Function Call
    isFrequencyEqual(arr, n);
      
# This code is contributed by Yash_R

输出：

3

时间复杂度： O(N)
空间复杂度： O(N)