The challenge

Find the number with the most digits.

If two numbers in the argument array have the same number of digits, return the first one in the array.

The solution in Python code

Option 1:

def find_longest(xs):
    return max(xs, key=lambda x: len(str(x)))

Option 2:

def find_longest(arr):
    arr.sort(reverse=True) 
    return arr[0]

Option 3:

def find_longest(arr):
    max_lenght = 0
    max_index = 0
    for cur_num in arr:
        lenght = len(str(cur_num))
        if lenght > max_lenght:
            max_lenght = lenght
            max_index = arr.index(cur_num)
    return arr[max_index]

Test cases to validate our solution

import test
from solution import find_longest

@test.describe("Fixed Tests")
def fixed_tests():
    @test.it('Basic Test Cases')
    def basic_test_cases():
        test.assert_equals(find_longest([1, 10, 100]), 100)
        test.assert_equals(find_longest([9000, 8, 800]), 9000)
        test.assert_equals(find_longest([8, 900, 500]), 900)
        test.assert_equals(find_longest([3, 40000, 100]), 40000)
        test.assert_equals(find_longest([1, 200, 100000]), 100000)

@test.describe("Random tests")
def random_tests():
    
    from random import randint
    from functools import reduce
    
    sol=lambda arr: reduce(lambda a,b: b if len(str(a))<len(str(b)) else a,arr)
    
    for _ in range(40):
        arr=[randint(1,10**randint(1,20)) for q in range(randint(1,50))]
        expected = sol(arr)
        @test.it(f"Testing for find_longest({arr})")
        def _():
            test.assert_equals(find_longest(arr[:]),expected)