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++

#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

Complexity Analysis:

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