#1295

Easy

LeetCode Problem Workspace

Find Numbers with Even Number of Digits

Count the integers in an array that have an even number of digits using a direct array traversal with digit math.

Problem Statement

Given an array of integers, your task is to determine how many numbers contain an even number of digits. For each element, count its digits and only include it if the count is even.

For example, given nums = [12,345,2,6,7896], 12 has 2 digits, 345 has 3 digits, 2 has 1 digit, 6 has 1 digit, and 7896 has 4 digits. The function should return 2 because only 12 and 7896 have an even number of digits.

Examples

Example 1

Input: nums = [12,345,2,6,7896]

Output: 2

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

Only 1771 contains an even number of digits.

Constraints

1 <= nums.length <= 500
1 <= nums[i] <= 105

Solution Approach

Iterative Digit Counting

Loop through each number and repeatedly divide by 10 to count its digits. Increment a result counter when the digit count is even. This avoids converting numbers to strings and keeps space O(1).

String Conversion Method

Convert each number to a string and check if its length is even. This is straightforward and readable but slightly increases space usage for temporary strings.

Math Shortcut Using Logarithms

Use logarithms to determine the number of digits by calculating floor(log10(num)) + 1. Then check if the result is even to update the count, combining array iteration with a compact math computation.

Complexity Analysis

Metric	Value
Time	O(N)
Space	O(1)

The algorithm scans each element once, giving O(N) time complexity, and uses only a counter and loop variables, maintaining O(1) space.

What Interviewers Usually Probe

Expect you to discuss both string-based and math-based digit counting.
They may ask how to optimize for space without creating new arrays.
Be ready to justify O(N) time as sufficient for array sizes up to 500.

Common Pitfalls or Variants

Common pitfalls

Forgetting to handle single-digit numbers correctly when counting digits.
Using string conversion unnecessarily when division or log10 is faster and cleaner.
Miscounting numbers like 10 or 100 due to off-by-one errors in digit calculation.

Follow-up variants

Return the actual numbers with even digits instead of just the count.
Count numbers with an odd number of digits instead of even digits.
Apply the same logic to a list of negative numbers and zero.

FAQ

How do I efficiently find numbers with an even number of digits?

Iterate through each number, count digits using division or log10, and increment a counter whenever the digit count is even.

Can I use strings to solve Find Numbers with Even Number of Digits?

Yes, converting numbers to strings and checking the length is valid, though math-based counting is more space-efficient.

What is the time complexity of this solution?

Time complexity is O(N) because each number is visited once to compute its digits.

Are there special cases to watch out for?

Single-digit numbers, powers of ten, and the maximum allowed value 105 should be counted correctly using the chosen digit-counting method.

Is using logarithms better than string conversion for this problem?

Logarithms are more space-efficient and slightly faster, but string conversion is simpler to implement and understand.

terminal

Solution

Solution 1: Simulation

We traverse each element $x$ in the array $\textit{nums}$. 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.

1

2

3

class Solution:
    def findNumbers(self, nums: List[int]) -> int:
        return sum(len(str(x)) % 2 == 0 for x in nums)

Continue Practicing

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