# Migratory Birds

### Problem Statement :

```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 * 10^5
It is guaranteed that each type is 1, 2, 3, 4, or 5.```

### Solution :

```                            ```Solution in C :

python3  :

import sys
from collections import Counter

n = int(input().strip())
types = list(map(int, input().strip().split(' ')))
freq = Counter(types)
print(max(freq, key=freq.get))

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);
}
}

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;
}

C  :

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
int n;
int i;
int j;
int k;
int max;
int sth;
max = 0;
j = 1;
scanf("%d",&n);
sth = 0;
int *types = malloc(sizeof(int) * n);
for(int types_i = 0; types_i < n; types_i++){
scanf("%d",&types[types_i]);
}
i = 0;
while (j <= 5 )
{
k = 0;
i = 0;
while (i < n)
{
if (types[i] == j)
k++;
i++;
}
if (k > max)
{
sth = j;
max = k;
}
j++;
}
printf("%d", sth);
return 0;
}```
```

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