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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Migratory Birds
  5. Discussions

Migratory Birds

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3588 Discussions

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  • alban_tyrex
     + 0 comments

    Here is my c++ solution you can find the explanation here : https://youtu.be/poQPrHTMe08

    int migratoryBirds(vector<int> arr) {
    map<int, int> mp;
    int cnt = 0, id;
    for(int i = 0; i < arr.size(); i++)mp[arr[i]]++;
    for(auto it = mp.begin(); it != mp.end(); it++){
        if(it->second > cnt || (it->second == cnt && it->first < id)){
            cnt = it->second;
            id = it->first;
        }
    }
    return id;
}

  • ilovemylittlecr1
     + 0 comments

    Best Solution in O(n) time:

    import java.io.IOException;

    public class MigratoryBirds{ public static void main(String[] args) throws IOException { System.out.println(migratoryBirds()); }

    //Worst Case: O(n)
static int migratoryBirds() throws IOException {
    Reader reader = new Reader();
    int n = reader.readInt();

    final int k = 6;
    int[] counts = new int[k];
    int type = 0;
    int maxCount = 0;

    for (int i = 0; i < n; i++) {
        int birdType = reader.readInt();
        counts[birdType]++;

        if(counts[birdType] > maxCount ||
            (counts[birdType] == maxCount && birdType < type)) {
            type = birdType;
            maxCount = counts[birdType];
        }
    }
    return type;
}

    }

  • memax747
     + 1 comment

    PHP easy solution Migratory Birds

    function migratoryBirds($arr) {
    $counts = array_count_values($arr);
    ksort($counts);
    arsort($counts);
    return array_key_first($counts);
}

  • alfonso_oramas
     + 0 comments

    Python 3: Used a key/value pair to map out each value and increment the times the id has been seen like some sort of counter.

    def migratoryBirds(arr):
    integer_dictionary = {}
    for number in arr:
        ## To prevent an issue where the key doesn't exist in our dictionary
        ## before an addtion assignment "+=" we check whether or not to initialize it.
        if number in integer_dictionary:
            integer_dictionary[number] += 1
        else:
            integer_dictionary[number] = 1
    ## Now we find the upmost maximum value.
    maximum_value = max(integer_dictionary.values())
    
    ## And we use that to find our smallest id if there are multiple
    ## high level values.
    keys_with_maximum_values = []
    for key, value in integer_dictionary.items():
        if value == maximum_value:
            keys_with_maximum_values.append(key)
    
    ## Returning the smallest id found.
    return min(keys_with_maximum_values)

  • a_borraccino86
     + 0 comments

    Java

    public static int migratoryBirds(List<Integer> arr) {
    // Write your code here
    
        if(arr == null || arr.isEmpty()) return 0;
        if(arr.size() == 1) return arr.get(0);

        Map<Integer, Integer> counterMap = new HashMap<>();
        int maxFrequency = 0;
        int result = Integer.MAX_VALUE;
        for(Integer bird : arr){
            counterMap.put(bird, counterMap.getOrDefault(bird, 0) + 1);
            Integer frequency = counterMap.get(bird);
            if(frequency > maxFrequency){
                maxFrequency = frequency;
                result = bird;
            }

            if(frequency == maxFrequency){
                result = Math.min(result, bird);
            }
        }

        return result;

}