Given an array nums
of integers, return how many of them contain an even number of digits.
Example 1:
Input: nums = [12,345,2,6,7896] Output: 2 Explanation: 12 contains 2 digits (even number of digits). 345 contains 3 digits (odd number of digits). 2 contains 1 digit (odd number of digits). 6 contains 1 digit (odd number of digits). 7896 contains 4 digits (even number of digits). Therefore only 12 and 7896 contain an even number of digits.
Example 2:
Input: nums = [555,901,482,1771] Output: 1 Explanation: Only 1771 contains an even number of digits.
Constraints:
1 <= nums.length <= 500
1 <= nums[i] <= 105
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
We want to find how many numbers in a given list have an even number of digits. The brute force method involves checking each number individually and counting its digits. If the digit count is even, we increment a counter.
Here's how the algorithm would work step-by-step:
def find_numbers_with_even_number_of_digits(numbers):
even_digits_count = 0
for number in numbers:
# Convert the number to a string to easily count digits
number_string = str(number)
number_of_digits = len(number_string)
# Check if the number of digits is even
if number_of_digits % 2 == 0:
# Increment the counter if the digit count is even
even_digits_count += 1
return even_digits_count
The problem requires us to count numbers with an even number of digits. The trick is to quickly figure out how many digits a number has and check if that count is even without doing any complicated calculations.
Here's how the algorithm would work step-by-step:
def find_numbers_with_even_number_of_digits(numbers):
even_digit_count = 0
for number in numbers:
digit_count = 0
temporary_number = number
# Count digits by repeatedly dividing by 10
while temporary_number > 0:
temporary_number //= 10
digit_count += 1
# Check if the number of digits is even
if digit_count % 2 == 0:
# Increment the count if even
even_digit_count += 1
return even_digit_count
Case | How to Handle |
---|---|
Null or empty input array | Return 0 immediately, as there are no numbers to evaluate. |
Array containing only single-digit numbers | Count will be 0 as none of the numbers will have even digits. |
Array with extremely large numbers that could cause integer overflow during digit counting | Consider using string conversion to count digits to avoid integer overflow. |
Array containing only negative numbers | Take the absolute value of each number before counting digits, as the negative sign doesn't affect digit count. |
Array containing zeros | Zero has one digit, therefore should not be counted toward the even number count. |
Array with a large number of elements to assess performance implications | Ensure the algorithm has O(n) time complexity by iterating through the array once. |
Array with a mix of positive and negative numbers | The algorithm should correctly handle both positive and negative integers after taking the absolute value. |
Array where all numbers have an even amount of digits | The count should accurately reflect the number of elements in the array. |