Majority Element II Leetcode Solution

In this problem, we are given an array of integers. The goal is to find all the elements which occur more than ⌊N / 3⌋ time in the array where N = size of the array and ⌊ ⌋ is the floor operator. We need to return an array of such elements.

 

Range of elements: -10^9 to 10^9

Example

Array = {1 , 2 , 3 , 3 , 2}

2 3

Explanation: ⌊N / 3⌋ = ⌊5 / 3⌋ = 1. Now, the integers 2 and 3 have frequencies equal to 2, which is greater 1. So, we print them.

Array = {1 , 2 , 3 , 4}

No majority elements

Explanation: We did not find any element whose frequency is greater than 1 in this case. So, we print "No Majority elements".

Approach(Storing Frequencies)

As the title suggests, we can store the frequency of every element in the array using a hash table and then check for elements having a frequency greater than ⌊N / 3⌋. In this way, we can find all elements that satisfy the condition.

Algorithm

 

- Initialize a HashMap-frequency to store the frequencies of elements in the array and a list/vector result to store the majority elements

- For every element i in the array:

 

- Increment its frequency: frequency++(or set as 1 if not present already)

 

- For every key in the hashmap:

 

- If frequency > N / 3

 

- Add it to the result

 

- Return the list result

 

Implementation of Majority Element II Leetcode Solution

C++ Program

#include

using namespace std;

 

vector majorityElement(vector& a)

{

int N = a.size();

 

www.tutorialcup.com/leetcode-solutions/majority-element-i...

Done