In all other cases, decrease cnt1 and cnt2 by 1.

After processing all elements, el1 and el2 should be the candidate elements for majority. To confirm:

Use another loop to manually check the counts of el1 and el2 in the array.
If either el1 or el2's count is greater than floor(N/3), it is considered a valid majority element.

Code

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    // Function to find majority elements in an array
    vector<int> majorityElementTwo(vector<int>& nums) {
        
        // Size of the array
        int n = nums.size(); 

        // Counts for elements el1 and el2
        int cnt1 = 0, cnt2 = 0;
        
        /*Initialize Element 1 and 
        Element 2 with INT_MIN value*/
        int el1 = INT_MIN, el2 = INT_MIN;

        /*Find the potential candidates using
        Boyer Moore's Voting Algorithm*/
        for (int i = 0; i < n; i++) {
            
            if (cnt1