Encuentre el elemento que tiene una frecuencia diferente a la de otros elementos de la array

Dada una array de N enteros. Cada elemento de la array aparece el mismo número de veces excepto un elemento. La tarea es encontrar este elemento.

Ejemplos:

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

Input : arr[] = {0, 1, 2, 4, 4}
Output : 4

La idea es usar una frecuencia de tabla hash para almacenar las frecuencias de elementos dados. Una vez que tenemos frecuencias en la tabla hash, podemos recorrer la tabla para encontrar el único valor que es diferente de los demás.

A continuación se muestra la implementación de la idea anterior:

// C++ program to find the element having
// different frequency than other array
// elements having same frequency
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the element having
// different frequency from other array
// elements with same frequency
int findElement(int arr[], int n)
{
    // Store frequencies of elements
    unordered_map<int, int> freq;
    for (int i = 0; i < n; i++) {
  
        // increase the value by 1 for every
        // time the element occurs in an array
        freq[arr[i]]++;
    }
  
    // Below code is used find the only different
    // value in freq. 
    auto it = freq.begin();
    int fst_fre = it->second, fst_ele = it->first;
    if (freq.size() <= 2)
        return fst_fre;
    it++;
    int sec_fre = it->second, sec_ele = it->first;
    it++;
    int trd_fre = it->second, trd_ele = it->first;
    if (sec_fre == fst_fre && sec_fre != trd_fre)
        return trd_ele;
    if (sec_fre == trd_fre && sec_fre != fst_fre)
        return fst_ele;
    if (fst_fre == trd_fre && sec_fre != fst_fre)
        return sec_ele;
  
    // We reach here when first three frequencies are same
    it++;
    for (; it != freq.end(); it++) {
        if (it->second != fst_fre)
            return it->first;
    }
  
    return -1;
}
  
// Driver code
int main()
{
    int arr[] = { 0, 1, 2, 4, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findElement(arr, n) << endl;
    return 0;
}

# Python program to find the element having 
# different frequency than other array 
# elements having same frequency 
  
# Function for above implementation
def findElement(arr, n) :
      
    # Empty dictionary to hold the values
    freq = {}
      
    # Initialization of frequencies of each 
    # element to 0
    for i in range(0, n) :
          
        freq[arr[i]] = 0
          
    # Count of frequencies of elements
    for i in range(0, n) :
          
        freq[arr[i]] = freq[arr[i]] + 1
       
    # Storing the first value of dictionary
    trd_ele = freq[0]
      
    # Variable to hold the final result
    position = -1
      
    # Following loop iterates through the dictionary
    # and checks if frequencies are different 
    # from the frequency of the first element
    for i in freq :
          
        flag = freq[i]
          
        if trd_ele != flag :
              
            # Difference has been detected
            position = i
            break
              
    # Following lines of code checks if the first
    # element is itself the required anomaly by 
    # comparing the frequencies of first 3 elements
    fst_ele = freq[1]
    sec_ele = freq[2]
      
    if trd_ele != fst_ele :
          
        if trd_ele != sec_ele :
              
            for i in freq :
                  
                # First element is the desired result
                position = i
                break
      
    # Final result is returned
    return position
      
  
# Driver code
arr = [ 0, 1, 2, 4, 4 ]
# Variable to store length of array
n = len(arr)
print (findElement(arr, n))
  
# This code is contributed by Pratik Basu 

4

Tiempo Complejidad : O(n)
Espacio Auxiliar : O(n)