Migratory Birds Hacker Rank Solution Best & Easiest

Posted by admin October 5, 2022
Migratory Birds Problem Solution
Migratory Birds Problem Solution

In this post, we will solve the Migratory Birds HackerRank Solution. This problem (Migratory Birds) is a part of the HackerRank Problem Solving series.

Problem:https://www.hackerrank.com/challenges/three-month-preparation-kit-migratory-birds/problem

Migratory Birds Hacker Rank Solution

Problem solution in Python programming:

def migratoryBirds(arr):
    # Write your code here
    arr = sorted(arr)
    freq = {}
    id = 0
    counter = 0
    for i in arr:
        if i in freq:
            freq[i] += 1
        else:
            freq[i] = 1
        if freq[i] > counter:
            id = i
            counter = freq[i]
    
    return id

Problem solution in JavaScript programming:

function migratoryBirds(arr) {
    if (arr.length == 0) {return 0}
    let count = new Array(5).fill(0);
    for (let i = 0; i < arr.length; i++) {
        count[arr[i]-1]++;
    }
    let max = Math.max(...count);
    return count.indexOf(max)+1;
}

Problem solution in Java programming:

public static int migratoryBirds(List<Integer> arr) {
    // Write your code here
        int[] newInt = new int[6];
        int value = 0;
        for(int i = 0; i < arr.size(); i++){
           value = arr.get(i);
           newInt[value] += 1;
        }
        
        int mostFrequent = 0;
        for(int i = 1; i < newInt.length; i++){
            if(newInt[mostFrequent] < newInt[i]){
                mostFrequent = i;
            }
        }
        
        return mostFrequent;
    }

Problem solution in C++ programming:

int migratoryBirds(vector<int> arr) {
    map<int,int> m;
    int max[2]={0};
    for(auto i : arr){
        if(m.find(i)!=m.end()) {
            m[i]+=1;
            if(m[i]>max[1] || (m[i]==max[1] && i<max[0])) {
                max[0]=i;
                max[1]=m[i];
            }
        } else m[i]=1;   
    }
    return max[0];
}