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  1. Migratory Birds
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Migratory Birds

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190 Discussions

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  • eypates04
     + 1 comment

    fun migratoryBirds(arr: Array): Int {

    val birdMap = mutableMapOf<Int, Int>()
for (bird in arr) {
    birdMap[bird] = birdMap.getOrDefault(bird, 0) + 1
}

val value = birdMap.map { it.value }.max()
val id = birdMap.filterValues { it == value }.keys.min()

return id

    }

  • pradymm
     + 0 comments
    public static int migratoryBirds(List<Integer> arr) {
// Write your code here
    Map<Integer,Integer> countMap = new HashMap<>();
    List<Integer> maxList = new ArrayList<>();
    for(Integer i : arr){
        countMap.put(i, countMap.getOrDefault(i, 0) + 1);
    }

    Integer max = Collections.max(countMap.values());
    for(Map.Entry<Integer, Integer> maps : countMap.entrySet() ){
        if(maps.getValue() == max){
            maxList.add(maps.getKey());
        }
    }
    return Collections.min(maxList);
}

  • harrydunn333
     + 0 comments
    frequency = {}
    for i in arr:
        frequency[i] = frequency.get(i,0) + 1
    maxCount = 0
    bird = None
    for key, values in frequency.items():
        if values > maxCount:
            maxCount = values
            bird = key  
        if values == maxCount and key < bird:
            bird = key  
    return bird

  • kumarsahudeepak1
     + 0 comments
    def migratoryBirds(arr):
    # Write your code here
    d ={}
    for i in arr:
        if i in d.keys():
            d[i]+=1
        else:
            d[i]=1
    maxval = 0
    minkey = 0
    for key,value in d.items():
        if value>maxval:
            maxval=value
            minkey=key
        elif value==maxval:
            if key<minkey:
                minkey=key
    return minkey

  • victor_reial
     + 0 comments

    Python best solution

    If you’re looking for solutions to the 3-month preparation kit in either Python or Rust, you can find them below: my solutions 

    def migratory_birds(arr):
    #Time complexity: O(n)
    #Space complexity (ignoring input): O(1). You could use a frequency array instead
    #of a dictionary and use even less space
    birds_dict = {}
    for bird in arr:
        if bird in birds_dict:
            birds_dict[bird]+=1
        else:
            birds_dict[bird]=1

    most_frequent = [6,0]
    for (bird, frequency) in birds_dict.items():
        if frequency > most_frequent[1]:
            most_frequent = [bird, frequency]
        if (frequency == most_frequent[1]) and (bird < most_frequent[0]):
            most_frequent[0] = bird

    return most_frequent[0]