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];
}

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