Majority Element | Set-2 (Hashing)

Given an array of size N, find the majority element. The majority element is the element that appears more than \floor{\frac{n}{2}} 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++

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#include<bits/stdc++.h>
using namespace std;
  
#define ll long long int
  
// function to print the majority Number
int 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 code
int 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

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Java

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import java.util.*;
  
class GFG 
{
  
// function to print the majority Number
static int majorityNumber(int arr[], int n)
{
    int ans = -1;
    HashMap<Integer,
            Integer> freq = new HashMap<Integer,
                                        Integer>();
                                          
    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 code
public 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

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Python

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# function to print the 
# majorityNumber
def majorityNumber(nums):
      
    # stores the key count 
    key, count = None, 0
      
    # iterate in the array 
    for num in nums:
          
        # key is 0
        if key is None:
            key, count = num, 1
        else:
            if key == num:
                count += 1
            else:
                count -= 1
  
        if count == 0:
            key = None
  
    return key
  
# Driver Code
a = [2, 2, 1, 1, 1, 2, 2]
print majorityNumber(a)

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C#

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// C# implementation of the approach
using System;
using System.Collections.Generic;
      
class GFG 
{
  
// function to print the majority Number
static 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 code
public 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

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Output:

2

Below is the time and space complexity of the above algorithm:-

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



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