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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

#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 

Java

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 

Python

# function to print the  
# majorityNumber 
def 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 Code 
a = [2, 2, 1, 1, 1, 2, 2] 
print majorityNumber(a) 

C#

// 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 

Output:

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

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

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python

C#

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