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# Find majority element using Hashing

Given an array of size N, find the majority element. The majority element is the element that appears more than times in the given array.

Examples:

Input: [3, 2, 3]
Output: 3

Input: [2, 2, 1, 1, 1, 2, 2]
Output: 2

The problem has been solved using 4 different methods in the previous post. In this post hashing based solution is implemented. We count occurrences of all elements. And if count of any element becomes more than n/2, we return it.

Hence if there is a majority-element, it will be the value of the key.

Below is the implementation of the above approach:

## C++

 #includeusing namespace std;  #define ll long long int  // function to print the majority Numberint majorityNumber(int arr[], int n){    int ans = -1;    unordered_map<int, int>freq;    for (int i = 0; i < n; i++)    {        freq[arr[i]]++;        if (freq[arr[i]] > n / 2)            ans = arr[i];    }    return ans;}   // Driver codeint main(){    int a[] = {2, 2, 1, 1, 1, 2, 2};    int n = sizeof(a) / sizeof(int);    cout << majorityNumber(a, n);     return 0;}  // This code is contributed // by sahishelangia

## Java

 import java.util.*;  class GFG {  // function to print the majority Numberstatic int majorityNumber(int arr[], int n){    int ans = -1;    HashMap freq = new HashMap();                                              for (int i = 0; i < n; i++)    {        if(freq.containsKey(arr[i]))        {            freq.put(arr[i], freq.get(arr[i]) + 1);        }        else        {            freq.put(arr[i], 1);        }        if (freq.get(arr[i]) > n / 2)            ans = arr[i];    }    return ans;}   // Driver codepublic static void main(String[] args) {    int a[] = {2, 2, 1, 1, 1, 2, 2};    int n = a.length;    System.out.println(majorityNumber(a, n));}}   // This code is contributed by Princi Singh

## Python3

 # function to print the # majorityNumberdef majorityNumber(nums):          # stores the num count     num_count = {}          # iterate in the array     for num in nums:                  if num in num_count:            num_count[num] += 1        else:            num_count[num] = 1      for num in num_count:        if num_count[num] > len(nums)/2:            return num    return -1  # Driver Codea = [2, 2, 1, 1, 1, 2, 2]print(majorityNumber(a))

## C#

 // C# implementation of the approachusing System;using System.Collections.Generic;      class GFG {  // function to print the majority Numberstatic int majorityNumber(int []arr, int n){    int ans = -1;    Dictionary<int,               int> freq = new Dictionary<int,                                          int>();                                              for (int i = 0; i < n; i++)    {        if(freq.ContainsKey(arr[i]))        {            freq[arr[i]] = freq[arr[i]] + 1;        }        else        {            freq.Add(arr[i], 1);        }        if (freq[arr[i]] > n / 2)            ans = arr[i];    }    return ans;}   // Driver codepublic static void Main(String[] args) {    int []a = {2, 2, 1, 1, 1, 2, 2};    int n = a.Length;    Console.WriteLine(majorityNumber(a, n));}}  // This code is contributed by Rajput-Ji

## Javascript

 

Output

2

Complexity Analysis:

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