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)
