Migratory Birds HackerRank Solution in Java

Java Solution for Migratory Birds HackerRank Problem | Find Lowest Maximum frequently Numbers in Java List or Array

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 1 :

arr = [1, 1, 2, 2, 3]
output = 1

There are two each of types 1 and 2, and one sighting of type 3.  Pick the lower of the two types seen twice: type 1.

Example 2 :

arr = [1, 2, 3, 4, 5, 4, 3, 2, 1, 3, 4]
output = 3

The different types of birds occur in the following frequencies:

• Type 1 : 2 bird 
• Type 2 : 2 bird 
• Type 3 : 3 bird 
• Type 4 : 3 birds
  
• Type 5 : 1 bird 

In this example we have two type of frequent birds : 3 and 4.
From both 3 is lower than 4 so answer will be 3.

Means we have to simply find out maximum frequent number from given list or array which value is lower than other same numbers.

So lets see solutions with explanation.

Solution 1 : Using HashMap

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;
import java.util.Map.Entry;

class Result {

    public static int migratoryBirds(List<Integer> arr) {
        
        Map<Integer, Integer> map = new HashMap<>();
        for (Integer number : arr) {
            if (!map.containsKey(number)) {
                map.put(number, 1);
            } else {
                map.put(number, map.get(number)+1);
            }
        }
        
        int max = 1;
        int ans = 0;
        for (Entry<Integer, Integer> entry : map.entrySet()) {
            if (entry.getValue() > max) {
                ans = entry.getKey();
                max = entry.getValue();
            }
        }
        return ans;
    }

}

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(
                new InputStreamReader(System.in));
        BufferedWriter bufferedWriter = new BufferedWriter(
                new FileWriter(System.getenv("OUTPUT_PATH")));

        int arrCount = Integer.parseInt(bufferedReader
                .readLine().trim());

        List<Integer> arr = Stream.of(
                bufferedReader.readLine()
                .replaceAll("\\s+$", "").split(" "))
                .map(Integer::parseInt)
                .collect(toList());

        int result = Result.migratoryBirds(arr);

        bufferedWriter.write(String.valueOf(result));
        bufferedWriter.newLine();

        bufferedReader.close();
        bufferedWriter.close();
    }
}

Solution explanation :

• We are using HashMap for storing each list value and its frequency.
• Traverse through given list or array, check if current value (bird type id) is already present in map or not. if current value is not present then, store current value as key and 1 as value, if current value does not present in map then store current value as key and current key value + 1.
• After traverse through map and get max lowest value (bird type id).
• Return ans.

Output explanation :

list = [1, 1, 2, 2, 3]

• index = 0, map = []
  
• !map.containsKey(1) becomes true
• map = [1 = 1]
• index = 1, map = [1 = 1]
• !map.containsKey(1) becomes false
• map = [1 = 2]  
• index = 2, map = []
  
• !map.containsKey(2) becomes true
• map = [1 = 2, 2 = 1]
• index = 3, map = [1 = 2, 2 = 1]
• !map.containsKey(2) becomes false
• map = [1 = 2, 2 = 2] 
  
• index = 4, map = [1 = 2, 2 = 2]
  
• !map.containsKey(3) becomes true
• map = [1 = 2, 2 = 2, 3 = 1]
• First loop completed
  
  
• Loop through map
• index = 0, max = 1, ans = 0
• entry.getValue() > max | 2 > 1 becomes true
• ans = entry.getKey() | ans = 1
• max = entry.getValue() | max = 2
• index = 1, max = 2, ans = 1
• entry.getValue() > max | 2 > 2 becomes false
• index = 2, max = 2, ans = 1
• entry.getValue() > max | 1 > 2 becomes false
• ans = 1

We can also done using JAVA 8 but in HackerRank some of TestCses become Time limit exceeded error. But you can learn how to solve this problem using JAVA 8.

Solution 2 : Using Java 8 Stream API

public static int migratoryBirds(List<Integer> list) {
    
    int ans = list.stream()
                .reduce(BinaryOperator
                    .maxBy((o1, o2) ->
                        Collections.frequency(list, o1)
                        - Collections.frequency(list, o2)))
                .orElse(null);
 
    return ans;
}

Learn more about JAVA 8 Stream Reduce() method :

• Java 8 Reduce() Method with examples  

Happy Coding.

See other HackerRank problem and its solutions in Java with Explanation :

• Subarray Division HackerRank solution in Java