# Hackerrank - Migratory Birds Solution

You have been asked to help study the population of birds migrating across the continent. Each type of bird you are interested in will be identified by an integer value. Each time a particular kind of bird is spotted, its id number will be added to your array of sightings. You would like to be able to find out which type of bird is most common given a list of sightings. Your task is to print the type number of that bird and if two or more types of birds are equally common, choose the type with the smallest ID number.

For example, assume your bird sightings are of types . There are two each of types and , and one sighting of type . Pick the lower of the two types seen twice: type .

**Function Description**

Complete the *migratoryBirds* function in the editor below. It should return the lowest type number of the most frequently sighted bird.

migratoryBirds has the following parameter(s):

*arr*: an array of integers representing types of birds sighted

**Input Format**

The first line contains an integer denoting , the number of birds sighted and reported in the array .

The second line describes as space-separated integers representing the type numbers of each bird sighted.

**Constraints**

- It is guaranteed that each type is , , , , or .

**Output Format**

Print the type number of the most common bird; if two or more types of birds are equally common, choose the type with the smallest ID number.

**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 , so we print 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 , and the lower of those is type .

### Solution in python

```
input()
def migratoryBirds(arr):
count = [0]*6
for i in arr:
count[i] += 1
return count.index(max(count))
arr = map(int,input().split())
print(migratoryBirds(arr))
```

**Explanation**

First you initialize an array like this arr = [0,0,0,0,0,0] which can also be written as [0]*6

Now when you see bird 1 you will increment index 1 by 1, if you see bird 2 you will increment index 2 and so and so.

For example, you saw the following birds 2,2,2,3,3,5 then our array will become something like this

[0,0,3,2,0,1]

As you you may have noticed.

Index 1 increased by 0 since bird 1 is seen 0 time

Index 2 increased by 3 since bird 2 is seen 3 times

Index 3 increased by 2 since bird 3 is seen 2 times

Index 4 increased by 0 since bird 4 is seen 2 times

Index 5 increased by 1 since bird 5 is seen 1 time

Now max(arr) will give us 3, that is the count of the most seen bird.

arr.index(max(arr)) or arr.index(3) will give us 2 which means the bird in index 2 is seen the maximum number of times.