CareerCup

This can be broken down to two parts.
1. Get a frequency count for every unique number. This is unavoidable step.
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2. Now, those buckets can be sorted in decreasing order of frequency, and then top n can be selected. But that requires sorting those many buckets, one per unique no.
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2. Now, those buckets can be safely inserted inside a heap with size n.
Your problem is solved.
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Given the total no of unique elements are less than 100, I would suggest sorting is a better solution. As the frequency will be integer, radix sort is in fact a faster solution.
[ geeksforgeeks.org/radix-sort/ ]

So, generic pet answer of *heap* might not be actually a good idea. But then... your call.

The algorithim has already been explained by NoOne. This is the heap solution in a python one liner the cost is nlogk

from heapq import nlargest
from collections import Counter

def getFirstK(array, n):
    return map(lambda x: x[1], 
               nlargest(n, [ (v, k) for k, v in Counter(array).iteritems() ]))

C++ implementation

#include <string>
#include <iostream>
#include <map>
#include <vector>
#include <algorithm>

using namespace std;

int main()
{
	int nArray[] = {
		5, 4, 2, 3, 6,		7, 8, 3, 4, 5,
		6, 7, 9, 3, 2,		3, 5, 6, 10, 5,
		3, 2, 5, 2, 3
	};
	int N = 5;	// Get top 5 frequent numbers
	map<int, int> mCounts;
	for (auto number : nArray)
		mCounts[number] ++;
	vector<pair<int, int>> counts(mCounts.begin(), mCounts.end());	// Get counts into sortable container
	sort(counts.rbegin(), counts.rend(), [](pair<int, int> a, pair<int, int>b) {	// Sort according to counts
		return (a.second < b.second);
	});
	for (int i = 0; (i < N) && i < counts.size(); ++i)	// Display frequent N numbers
		cout << i + 1 << "] " << counts[i].first << ": " << counts[i].second << " times" << endl;
	cout << "Press enter to close.";
	getline(cin, string());
	return 0;
}