查找数组中频率等于其他元素频率总和的元素

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, the frequency of element 3 is equal to the 
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：问题中的关键观察是如果数组的长度是奇数，那么就不会有这样的元素，而在偶数长度数组的情况下，计算数组中每个元素的频率，然后最后，检查数组中频率等于数组半长的任何元素。

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

Javascript

// Javascript program
 
// Function to check that any element
// have frequency equal to the sum of
// frequency of other elements of the array
function isFrequencyEqual(arr, len)
{
    // Check that if the array length
    // is odd, Then no solution possible
    if (len % 2 == 1){
        document.write("No Such Element");
        return false;
    }
   
    // Hash-map to store the frequency
    // of elements of array
    var freq = {};
    for (var i = 0; i < len; i++)
        freq[arr[i]] = 0;
    // Loop to find the frequency
    // of elements of array
    for (var 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 (var i = 0; i < len; i++){
        if (freq[arr[i]] == len / 2){
            document.write(arr[i]);
            return true;
        }
    }
           
    document.write("No such element");
    return false;
}
var arr = [ 1, 2, 2, 3, 3, 3];
var n = 6;
 
isFrequencyEqual(arr, n);

输出：

3

时间复杂度： O(N)
空间复杂度： O(N)