Problem Description

You are given an array nums containing integers. Your task is to count how many numbers in this array have an even number of digits.

For example:

• The number 12 has 2 digits (even count)
• The number 345 has 3 digits (odd count)
• The number 2 has 1 digit (odd count)
• The number 6789 has 4 digits (even count)

The solution works by converting each number to a string to easily count its digits. For each number x in the array, len(str(x)) gives us the digit count. We then check if this count is even using the modulo operator (% 2 == 0). The sum() function counts how many numbers satisfy this condition, giving us the final answer.

Intuition

The key insight is that we need to count the digits in each number. While we could count digits mathematically (by repeatedly dividing by 10), the simplest approach is to convert each number to a string since the length of the string representation directly gives us the digit count.

Once we know how to get the digit count, checking if it's even is straightforward - we just need to verify if the count is divisible by 2. In Python, we can check this using count % 2 == 0.

Since we're counting how many numbers satisfy a condition, we can use Python's generator expression with sum(). This works because True evaluates to 1 and False evaluates to 0 in Python. So sum(condition for x in nums) effectively counts how many times the condition is true.

The one-liner solution sum(len(str(x)) % 2 == 0 for x in nums) elegantly combines all these ideas: convert to string, check digit count parity, and count the matches all in a single expression.

Learn more about Math patterns.

Solution Approach

We traverse each element x in the array nums. For the current element x, we directly convert it to a string and then check if its length is even. If it is, we increment the answer by one.

The implementation follows these steps:

1. Iterate through the array: We go through each number x in nums

2. Convert to string: For each number x, we use str(x) to convert it to its string representation

3. Count digits: We use len(str(x)) to get the number of digits in the number

4. Check if even: We check if the digit count is even using the condition len(str(x)) % 2 == 0

5. Count matches: We use Python's sum() function with a generator expression to count how many numbers have an even digit count

The complete solution becomes a single line: sum(len(str(x)) % 2 == 0 for x in nums)

This approach has a time complexity of O(n × d) where n is the number of elements in the array and d is the average number of digits in the numbers (since string conversion takes O(d) time). The space complexity is O(d) for the temporary string created during conversion.

Example Walkthrough

Let's walk through a concrete example with the array nums = [12, 345, 2, 6789, 100].

We'll process each number and check if it has an even number of digits:

1. First number: 12

   • Convert to string: "12"
   • Count digits: len("12") = 2
   • Check if even: 2 % 2 == 0 → True
   • Running count: 1

2. Second number: 345

   • Convert to string: "345"
   • Count digits: len("345") = 3
   • Check if even: 3 % 2 == 0 → False
   • Running count: 1 (no change)

3. Third number: 2

   • Convert to string: "2"
   • Count digits: len("2") = 1
   • Check if even: 1 % 2 == 0 → False
   • Running count: 1 (no change)

4. Fourth number: 6789

   • Convert to string: "6789"
   • Count digits: len("6789") = 4
   • Check if even: 4 % 2 == 0 → True
   • Running count: 2

5. Fifth number: 100

   • Convert to string: "100"
   • Count digits: len("100") = 3
   • Check if even: 3 % 2 == 0 → False
   • Running count: 2 (no change)

Final answer: 2 (the numbers 12 and 6789 have an even number of digits)

Using the one-liner solution: sum(len(str(x)) % 2 == 0 for x in nums) produces:

• For x=12: len(str(12)) % 2 == 0 → True (contributes 1)
• For x=345: len(str(345)) % 2 == 0 → False (contributes 0)
• For x=2: len(str(2)) % 2 == 0 → False (contributes 0)
• For x=6789: len(str(6789)) % 2 == 0 → True (contributes 1)
• For x=100: len(str(100)) % 2 == 0 → False (contributes 0)
• Sum: 1 + 0 + 0 + 1 + 0 = 2

Solution Implementation

Time and Space Complexity

The time complexity is O(n × log M), where n is the length of the array nums and M is the maximum value of the elements in the array.

• The code iterates through each element in nums once, contributing O(n) to the complexity
• For each element x, converting it to a string using str(x) takes O(log₁₀ x) time, which is proportional to the number of digits in x
• In the worst case, the largest element has value M, requiring O(log M) time for conversion
• The len() operation on the string is O(1) after string conversion
• Therefore, the overall time complexity is O(n × log M)

The space complexity is O(log M).

• The generator expression itself uses O(1) space as it processes elements one at a time
• However, for each element x, the str(x) conversion creates a temporary string whose length is proportional to the number of digits in x
• The largest temporary string created would be for the maximum value M, which requires O(log M) space
• Since only one string is created at a time (not stored permanently), the space complexity is O(log M)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Handling Negative Numbers

The current solution doesn't account for negative numbers correctly. When converting a negative number to a string, the minus sign is included in the length count, which can lead to incorrect results.

Example Issue:

• -12 converts to "-12" with length 3 (odd), but the actual digit count should be 2 (even)
• -999 converts to "-999" with length 4 (even), but the actual digit count should be 3 (odd)

Solution: Use the absolute value before converting to string:

return sum(len(str(abs(num))) % 2 == 0 for num in nums)

2. Mathematical Approach vs String Conversion

While string conversion is intuitive and readable, it involves type conversion overhead. For performance-critical applications with large datasets, a mathematical approach might be more efficient.

Alternative Mathematical Solution:

def findNumbers(self, nums: List[int]) -> int:
    count = 0
    for num in nums:
        num = abs(num)  # Handle negative numbers
        digits = 0
        while num > 0:
            digits += 1
            num //= 10
        if digits % 2 == 0:
            count += 1
    return count

Or using logarithms:

import math

def findNumbers(self, nums: List[int]) -> int:
    count = 0
    for num in nums:
        if num == 0:
            digits = 1
        else:
            digits = math.floor(math.log10(abs(num))) + 1
        if digits % 2 == 0:
            count += 1
    return count

3. Edge Case: Zero

The number 0 has 1 digit, which the current solution handles correctly. However, in the mathematical logarithm approach, log10(0) is undefined, requiring special handling as shown above.

4. Memory Consideration

Although the string conversion approach has O(d) space complexity per conversion, creating many temporary strings in a tight loop could cause unnecessary garbage collection pressure in memory-constrained environments. The mathematical approach avoids this by working directly with integers.