Find the element having different frequency than other array elements

Given an array of N integers. Each element in the array occurs same number of times except one element. The task is to find this element.

Examples:

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

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

The idea is to use a hash table freq to store the frequencies of given elements. Once we have frequencies in the hash table, we can traverse the table to find the only value which is different from others.

Below is the implementation of the above idea :

C++

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// 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;
}

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Python3

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# 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 

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Output:

4

Time Complexity : O(n)
Auxiliary Space : O(n)

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Improved By : PratikBasu