Find the element having different frequency than other array elements

Given an array of N integers. Each element in the array occurs the 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 the others. 

Implementation:

C++

// 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;
}

Java

import java.util.*;

public class DifferentFrequencyElement {
    // Function to find the element having different frequency
    // from other array elements with the same frequency
    static int findElement(int[] arr) {
        // Store frequencies of elements
        HashMap<Integer, Integer> freq = new HashMap<>();
        for (int i = 0; i < arr.length; i++) {
            // Increase the value by 1 for every time the element occurs in the array
            freq.put(arr[i], freq.getOrDefault(arr[i], 0) + 1);
        }

        // Initialize variables to store frequencies and elements
        int firstFreq = -1, secondFreq = -1, thirdFreq = -1;
        int firstElement = -1, secondElement = -1, thirdElement = -1;

        for (Map.Entry<Integer, Integer> entry : freq.entrySet()) {
            int element = entry.getKey();
            int frequency = entry.getValue();

            // Update the variables based on frequency
            if (firstFreq == -1) {
                firstFreq = frequency;
                firstElement = element;
            } else if (frequency == firstFreq) {
                // If the current frequency is the same as the first frequency, update secondFreq and secondElement
                secondFreq = frequency;
                secondElement = element;
            } else if (frequency == secondFreq) {
                // If the current frequency is the same as the second frequency, update thirdFreq and thirdElement
                thirdFreq = frequency;
                thirdElement = element;
            } else {
                // If the frequency is different from both firstFreq and secondFreq, return the element
                return element;
            }
        }

        // If the third frequency is different, return the third element
        if (secondFreq == firstFreq && secondFreq != thirdFreq)
            return thirdElement;
        if (secondFreq == thirdFreq && secondFreq != firstFreq)
            return firstElement;
        if (firstFreq == thirdFreq && secondFreq != firstFreq)
            return secondElement;

        // We reach here when the first three frequencies are the same
        for (Map.Entry<Integer, Integer> entry : freq.entrySet()) {
            if (entry.getValue() != firstFreq)
                return entry.getKey();
        }

        return -1;
    }

    public static void main(String[] args) {
        int[] arr = { 0, 1, 2, 4, 4 };
        int result = findElement(arr);
        System.out.println(result);
    }
}

Python3

# 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

C#

using System;
using System.Collections.Generic;

class DifferentFrequencyElement
{
    // Function to find the element having different frequency
    // from other array elements with the same frequency
    static int FindElement(int[] arr)
    {
        // Store frequencies of elements
        Dictionary<int, int> freq = new Dictionary<int, int>();
        foreach (int num in arr)
        {
            // Increase the value by 1 for every time the element occurs in the array
            if (freq.ContainsKey(num))
            {
                freq[num]++;
            }
            else
            {
                freq[num] = 1;
            }
        }

        // Initialize variables to store frequencies and elements
        int firstFreq = -1, secondFreq = -1, thirdFreq = -1;
        int firstElement = -1, secondElement = -1, thirdElement = -1;

        foreach (KeyValuePair<int, int> entry in freq)
        {
            int element = entry.Key;
            int frequency = entry.Value;

            // Update the variables based on frequency
            if (firstFreq == -1)
            {
                firstFreq = frequency;
                firstElement = element;
            }
            else if (frequency == firstFreq)
            {
                // If the current frequency is the same as the first frequency, update secondFreq and secondElement
                secondFreq = frequency;
                secondElement = element;
            }
            else if (frequency == secondFreq)
            {
                // If the current frequency is the same as the second frequency, update thirdFreq and thirdElement
                thirdFreq = frequency;
                thirdElement = element;
            }
            else
            {
                // If the frequency is different from both firstFreq and secondFreq, return the element
                return element;
            }
        }

        // If the third frequency is different, return the third element
        if (secondFreq == firstFreq && secondFreq != thirdFreq)
            return thirdElement;
        if (secondFreq == thirdFreq && secondFreq != firstFreq)
            return firstElement;
        if (firstFreq == thirdFreq && secondFreq != firstFreq)
            return secondElement;

        // We reach here when the first three frequencies are the same
        foreach (KeyValuePair<int, int> entry in freq)
        {
            if (entry.Value != firstFreq)
                return entry.Key;
        }

        return -1;
    }

    static void Main()
    {
        int[] arr = { 0, 1, 2, 4, 4 };
        int result = FindElement(arr);
        Console.WriteLine(result);
    }
}

Javascript

// JavaScript implementation
function findElement(arr, n) {

    // Empty object to hold the values
    let freq = {};

    // Initialization of frequencies of each
    // element to 0
    for(let i = 0; i < n; i++) {
        freq[arr[i]] = 0;
    }

    // Count of frequencies of elements
    for(let i = 0; i < n; i++) {
        freq[arr[i]] = freq[arr[i]] + 1;
    }

    // Storing the first value of object
    let trd_ele = freq[0];

    // Variable to hold the final result
    let position = -1;

    // Following loop iterates through the object
    // and checks if frequencies are different
    // from the frequency of the first element
    for(let i in freq) {
        let 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
    let fst_ele = freq[1];
    let sec_ele = freq[2];

    if (trd_ele !== fst_ele) {
        if (trd_ele !== sec_ele) {
            for(let i in freq) {
                // First element is the desired result
                position = i;
                break;
            }
        }
    }

    // Final result is returned
    return position;

}

// Driver code
let arr = [ 0, 1, 2, 4, 4 ];
// Variable to store length of array
let n = arr.length;
console.log(findElement(arr, n));

// This code is contributed by  phasing17

Output

4

Complexity Analysis:

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