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]