My Python 3 O(n) solution, iterating once through array to build dictionary or hash map, then count the keys that have a value equal to maximum and append those keys to an array. If more than one key exists, the minimum does nothing:

import os
def migratoryBirds(arr):
    freq = {}
    for val in arr:
        freq[val] = freq.get(val, 0) + 1
    max_val = max(freq.values())
    keys = []
    for k, v in freq.items():
        if v == max_val:
            keys.append(k)
    return min(keys)
        
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()

Here is my simple C++ solution,

int migratoryBirds(vector arr) {

int cnt = 0;
int prevCnt = 0;
int res  = 0;
for( int i =1; i <=5; i++)  // if you read the constraint it says only 1 - 5 types guranteed, so we find the max of each element and the corresponding min
{
    cnt = std::count (arr.begin(), arr.end(), i);

    if( prevCnt < cnt )
    {
        prevCnt = cnt;
        res = i;
    }
    else if( prevCnt == cnt )
    {
        res = min(res, i); // here we get the min of the max cnt if they are the same
    }
}
return res;   

}

Here is my Python solution

def migratoryBirds(arr):
    # Write your code here
    
    min_number = min(arr)
    max_number = max(arr)

    # Find the lowest type id of the most frequently sighted birds
    lowestId = max_number
    for i in range(max_number-1, min_number, -1):
        x = arr.count(i)
        if x >= (arr.count(lowestId)):
            lowestId = i
    return lowestId

def migratoryBirds(arr) freq = {}

# Step 1: Build frequency hash manually
for i in arr
  if freq.has_key?(i)
    freq[i] += 1
  else
    freq[i] = 1
  end
end

# Step 2: Find the highest frequency
max_freq = 0
for key in freq.keys
  if freq[key] > max_freq
    max_freq = freq[key]
  end
end

# Step 3: Among all bird types with max frequency, find the smallest ID
min_id = nil
for key in freq.keys
  if freq[key] == max_freq
    if min_id.nil? || key < min_id
      min_id = key
    end
  end
end

return min_id

end

Python

def migratoryBirds(arr): # Write your code here res = [] for i in range(5): res.append(arr.count(i+1)) return res.index(max(res))+1