C# solution:

public static int migratoryBirds(List arr)

{
    arr.Sort();
    var types = arr.Distinct().ToList();
    var counts = new List<int>();
    int count = 0;        
    foreach (int type in types)
    {
        for (int i = 0; i < arr.Count; i++)
            {
                if(arr[i] == type)
                {
                    count++;
                }
            }
        counts.Add(count);
        count = 0; 
    }            

    int indexOfType = counts.IndexOf(counts.Max());
    return types[indexOfType];
}

Here is how I solved it using C++. Feel free to leave feedback on my solution:

`cpp

int migratoryBirds(vector arr) {

// Dump each bird type in a hashmap with its frequency
std::unordered_map<int, int> map;
for(int a : arr) {
    ++map[a];
}

// Then, check each type (start from type 5)
// Return type with highest frequency (starting from end will result in highestCt having lowest ID)
int resultType = 0;
int highestCt = -1;
for(int i = 5; i > 0; --i) {
    int count = map[i];
    if(count >= highestCt) {
        highestCt = count;
        resultType = i;
    }
}

return resultType;

}`

Java Solution

using HashMaps

 public static int migratoryBirds(List<Integer> arr) {
    // Write your code here
        int curMax = 0 ;
        HashMap<Integer , Integer> map = new HashMap<>();
        for(int num : arr){
            if(map.containsKey(num)){
                int curFreq = map.get(num);
                map.put(num , curFreq+1);
                curMax = Math.max(curMax , map.get(num));
            }
            else{
                map.put(num ,1);
                curMax = Math.max(curMax,1);
            }
        }
        int res = Integer.MAX_VALUE;
        for(int key : map.keySet()){
            if(map.get(key) == curMax){
                res = Math.min(res,key);
            }
        }
        return res;
    }

public static int migratoryBirds(List<Integer> arr) {

Map<Integer,Integer> m=new HashMap<>();

    for(int i: arr){
        if(m.containsKey(i)){
            m.put(i, m.get(i)+1);
        }
        else{
            m.put(i, 1);
        }
    }
    int key=0;
    int value=0;
    for(int i:m.keySet()){
        if(m.get(i)>value){
            value=m.get(i);
            key=i;
        }
    }
    return key;

}

}

Clean problem — reminds me of how I categorized menu items by popularity on the Popeyes menu site. Here's my Python take using Counter:

from collections import Counter

def migratoryBirds(arr):
    counts = Counter(arr)
    return min([k for k, v in counts.items() if v == max(counts.values())])