Migratory Birds | HackerRank Solution

Hello coders, today we are going to solve Migratory Birds HackerRank Solution which is a Part of HackerRank Algorithm Series.

Task

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

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

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, n, the size of arr.
The second line describes arr as n space-separated integers, each a type number of the bird sighted.

Constraints

• 5 <= n <= 2 x 10⁵
• It is guaranteed that each type is 1, 2, 3, 4, or 5.

Sample Input 0

6
1 4 4 4 5 3

Sample Output 0

4

Explanation 0

The different types of birds occur in the following frequencies:

• Type 1: 1 bird
• Type 2: 0 birds
• Type 3: 1 bird
• Type 4: 3 birds
• Type 5: 1 bird

The type number that occurs at the highest frequency is type 4, so we print 4 as our answer.

Sample Input 1

11
1 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 1: 2
• Type 2: 2
• Type 3: 3
• Type 4: 3
• Type 5: 1

Two types have a frequency of 3, and the lower of those is type 3.

Solution – Migratory Birds

C++

#include <bits/stdc++.h>

using namespace std;

const int maxN = 1e5+10;
int N,A[10];

int main()
{
    cin >> N;
    for (int i=1,x; i <= N; i++) cin >> x, A[x]++;
    int ans = 1;
    for (int i=2; i <= 5; i++)
        if (A[i] > A[ans]) ans = i;
    cout << ans;
}

Python

#!/bin/python3

import math
import os
import random
import re
import sys


def migratoryBirds(arr):
    # Write your code here
    count = [0] * 6
    for i in arr:
        count[i] += 1
    return count.index(max(count))

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    arr_count = int(input().strip())

    arr = list(map(int, input().rstrip().split()))

    result = migratoryBirds(arr)

    fptr.write(str(result) + '\n')

    fptr.close()

Java

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] birds = new int[5];
        for (int i = 0; i < n; i++) birds[in.nextInt()-1]++;
        int max = 0;
        int id = 0;
        for (int i = 0; i < 5; i++) {
            if (birds[i] > max) {
                max = birds[i];
                id = i+1;
            }
        }
        System.out.println(id);
    }
}

Disclaimer: The above Problem (Migratory Birds) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.