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Majority Element | Set-2 (Hashing)

  • Difficulty Level : Easy
  • Last Updated : 07 Dec, 2021

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++




#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

Python3




# 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

Javascript




<script>
 
// function to print the majority Number
function majorityNumber(arr, n)
{
    let ans = -1;
    let freq = new Map();
    for (let i = 0; i < n; i++)
    {
        freq[arr[i]]++;
 
        if(freq.has(arr[i])){
            freq.set(arr[i], freq.get(arr[i]) + 1)
        }else{
            freq.set(arr[i], 1)
        }
 
        if (freq.get(arr[i]) > n / 2)
            ans = arr[i];
    }
    return ans;
}
 
// Driver code
 
    let a = [2, 2, 1, 1, 1, 2, 2];
    let n = a.length;
    document.write(majorityNumber(a, n));
 
// This code is contributed
// by _saurabh_jaiswal
 
</script>
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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