# Leetcode – Find Numbers with Even Number of Digits Solution

The problem titled “Find Numbers with Even Number of Digits” on Leetcode is a beginner-friendly challenge that explores array manipulation and basic number theory. The problem appears simple at first glance but provides a useful opportunity to understand Python idioms, language features, and optimization techniques. This comprehensive guide will explore multiple approaches to solve this problem.

1. Problem Description
2. Preliminary Concepts
3. Naive Approach: Using String Conversion
4. Optimized Approach: Mathematical Calculation
5. Hybrid Approach: Combining String and Math
6. Testing and Validation
7. Time and Space Complexity Analysis
8. Conclusion

## 1. Problem Description

Given an array, nums, consisting of integers, the task is to find and return the number of integers in the array that have an even number of digits.

### Constraints:

• 1 <= nums.length <= 500
• 1 <= nums[i] <= 10^5

### Example:

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

Output: 2

Explanation: 12 and 7896 have an even number of digits.

## 2. Preliminary Concepts

### Counting Digits

Two primary ways can count the digits of a number:

1. Converting it to a string and using the len() function.
2. Using mathematical operations like dividing the number repeatedly by 10 until it becomes zero, while keeping a count of the operations.

## 3. Naive Approach: Using String Conversion

### Algorithm:

1. Initialize a counter variable to zero.
2. Loop through each number in the array.
3. Convert the number to a string.
4. Check the length of the string. If it’s even, increment the counter.
5. Return the counter.

### Python Code:

def findNumbers(nums):
count = 0

for num in nums:
if len(str(num)) % 2 == 0:
count += 1

return count

## 4. Optimized Approach: Mathematical Calculation

### Algorithm:

1. Initialize a counter variable to zero.
2. Loop through each number in the array.
3. Use mathematical operations to count the number of digits.
4. If the count is even, increment the counter.
5. Return the counter.

### Python Code:

def findNumbers(nums):
count = 0

for num in nums:
num_digits = 0
while num > 0:
num //= 10
num_digits += 1

if num_digits % 2 == 0:
count += 1

return count

## 5. Hybrid Approach: Combining String and Math

It’s also possible to combine both the string conversion and mathematical calculation methods to solve this problem, although this doesn’t bring any particular advantages in this case.

## 6. Testing and Validation

1. Test with an array that has only one-digit numbers.
2. Test with an array where all numbers have an even number of digits.
3. Test with an array where all numbers have an odd number of digits.
4. Test with an empty array (this is a corner case and doesn’t actually meet the problem’s constraints, but good to check).

## 7. Time and Space Complexity Analysis

1. Naive Approach:
• Time Complexity: O(n)
• Space Complexity: O(1) (excluding the input)
2. Optimized Approach:
• Time Complexity: O(n logā” m) where m is the maximum number in the array
• Space Complexity: O(1) (excluding the input)

## 8. Conclusion

The “Find Numbers with Even Number of Digits” problem provides a nice entry point into array manipulation and digit counting in Python. It offers an opportunity to compare the utility and efficiency of different approaches for what appears to be a simple problem. By exploring both naive and optimized approaches, we can appreciate the trade-offs involved in choosing one method over another, particularly with respect to time and space complexity.