Migratory Birds - HackerRank Challenge | C++ Implementation

You have been asked to help study the population of birds migrating across the continent. Each type of bird you are interested in will be identified by an integer value. Each time a particular kind of bird is spotted, its id number will be added to your array of sightings. You would like to be able to find out which type of bird is most common given a list of sightings. Your task is to print the type number of that bird and if two or more types of birds are equally common, choose the type with the smallest ID number.

For example, assume your bird sightings are of types arr = [1, 1, 2, 2, 3]. There are two each of types 1 and 2, and one sighting of type 3. Pick the lower of the two types seen twice: type 1.

Read the full problem here : Migratory Birds

To solve the problem first the input array, types storing types of birds, must be sorted. An array type_count stores count of bird of each type.

std::sort(types.begin(), types.end());
std::vector<unsigned int> type_count(6, 0);

Since there are only 5 types of birds with ids 1, 2, 3, 4 and 5 so type_count[0] = 0.

We will start iteration on types from index 1 and check if id of bird at that index and id of bird at previous index is equal. If it is equal i.e. types[i] == types[i-1], integer variable count is incremented. count counts the number of birds with that id(type[-1]).

If types[i] is not equal to types[i-1] then type_count for id types[i-1] is updated with the value stored in count.

unsigned int count = 1;
for (unsigned int i = 1; i < types.size(); ++i)
{
    if (types[i] == types[i-1])
    {
        count++;
    }
    else if (types[i] != types[i-1])
    {
        type_count[ types[i-1] ] = count;
        count = 1;
    }
}

When the loop is executed and we are out of the for loop then value stored in count is assigned to type_count at index which is equal to largest id number. This index is accessed by types.back() because types is sorted and types.back() returns last element which is the largest id of the birds.

type_count[ types.back() ] = count;

The program should return the id of birds which has the highest number of count and if two ids have same count then lowest id is returned. In other words, index of the highest element of type_count is returned and if it contains two or more occurrence of highest value then lower index of these highest value is returned.

Integer variable max_val stores the maximum number of birds with same type and min_index stores the minimum index of the max_val in type_count.

unsigned int max_val = type_count[0];
unsigned short  min_index = 0;

for (unsigned short i = 1; i < type_count.size(); ++i)
{
    if (type_count[i] > max_val)
    {
        max_val = type_count[i];
        min_index = i;
    }
}

Related: Kangaroo HackerRank Challenge

#include <iostream>
#include <vector>
#include <algorithm>

unsigned short common_type(std::vector<unsigned short>& types)
{
    std::sort(types.begin(), types.end());
    std::vector<unsigned int> type_count(6, 0); //stores count of each bird whose id == index

    unsigned int count = 1;

    for (unsigned int i = 1; i < types.size(); ++i)
    {
        if (types[i] == types[i-1])
        {
            count++;
        }
        else if (types[i] != types[i-1])
        {
            type_count[ types[i-1] ] = count;
            count = 1;
        }
    }
    type_count[ types.back() ] = count;

    unsigned int max_val = type_count[0];
    unsigned short  min_index = 0;

    for (unsigned short i = 1; i < type_count.size(); ++i)
    {
        if (type_count[i] > max_val)
        {
            max_val = type_count[i];
            min_index = i;
        }
    }
    return min_index;
}

int main()
{
    unsigned int num_of_birds;
    std::cin >> num_of_birds;
    std::vector<unsigned short> types(num_of_birds);

    for (unsigned int i = 0; i < num_of_birds; ++i)
    {
        std::cin >> types[i];
    }

    std::cout << common_type(types) << "\n";
}

View this code on Github.