Given an array of bird sightings where every element represents a
bird type id, determine the id of the most frequently sighted type. If
more than 1 type has been spotted that maximum amount, return the
smallest of their ids.
Example
There are two each of types and , and one sighting of type . Pick the lower of the two types seen twice: type
.
Function Description
Complete the migratoryBirds function in the editor below.
migratoryBirds has the following parameter(s):
int arr[n]: the types of birds sighted
Returns
int: the lowest type id of the most frequently sighted birds
Input Format
The first line contains an integer,
, the size of .
The second line describes as
space-separated integers, each a type number of the bird sighted.
Constraints
It is guaranteed that each type is , , , , or
.
Sample Input 0
61 4 4 4 5 3
Sample Output 0
4
Explanation 0
The different types of birds occur in the following frequencies:
Type
:
bird
Type
:
birds
Type
:
bird
Type
:
birds
Type
:
bird
The type number that occurs at the highest frequency is type
, so we print
as our answer.
Sample Input 1
111 2 3 4 5 4 3 2 1 3 4
Sample Output 1
3
Explanation 1
The different types of birds occur in the following frequencies:
Type
: Type : Type : Type : Type : Two types have a frequency of , and the lower of those is type .
importjava.io.*;
importjava.math.*;
importjava.security.*;
importjava.text.*;
importjava.util.*;
importjava.util.concurrent.*;
importjava.util.function.*;
importjava.util.regex.*;
importjava.util.stream.*;
importstaticjava.util.stream.Collectors.joining;
importstaticjava.util.stream.Collectors.toList;
classResult{
publicstaticintmigratoryBirds(List<Integer>arr){
List<Integer> countList =newArrayList<>();
Integer[] initialCount ={0,0,0,0,0};
countList.addAll(Arrays.asList(initialCount));
int maxCount =0;
for(int i =0; i <arr.size(); i++){
int currentBirdCount = countList.get(arr.get(i)-1);
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