How do you solve Find Numbers with Even Number of Digits on LeetCode?

Find Numbers with Even Number of Digits (LeetCode #1295) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Counting digits can be done using division or string conversion.

What algorithm or data structure is used to solve Find Numbers with Even Number of Digits?

Find Numbers with Even Number of Digits uses the following concepts: Array, Math. The recommended patterns to study are: Array, String Manipulation.

What are the interview tips for Find Numbers with Even Number of Digits?

Always clarify the constraints and edge cases before coding. Think about both time and space complexity when discussing solutions. Practice explaining your thought process clearly while coding.

What are common mistakes when solving Find Numbers with Even Number of Digits?

Not considering single-digit numbers as having odd digits. Overcomplicating the digit counting process.

Find Numbers with Even Number of Digits — LeetCode #1295 (Easy)

Tags: Array, Math

Related patterns: Array, String Manipulation

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute force approach involves checking each number in the array individually to count its digits. For each number, we can repeatedly divide it by 10 until it becomes 0, counting how many times we divide to determine the number of digits.

The time complexity is O(n²) because for each number (n), we may have to count its digits, which can take up to O(log m) time, where m is the maximum number in the array. However, since the maximum number of digits is limited, we can consider it constant for practical purposes.

Step 1: Initialize a counter to zero.
Step 2: For each number in the array, initialize a digit count to zero.
Step 3: While the number is greater than 0, divide the number by 10 and increment the digit count.
Step 4: After counting digits, check if the count is even. If it is, increment the counter.
Step 5: Return the counter after processing all numbers.

Input: nums = [12, 345, 2, 6, 7896] 1. Initialize count = 0. 2. Check 12: digit count = 2 (even), count = 1. 3. Check 345: digit count = 3 (odd), count remains 1. 4. Check 2: digit count = 1 (odd), count remains 1. 5. Check 6: digit count = 1 (odd), count remains 1. 6. Check 7896: digit count = 4 (even), count = 2. Final count = 2.

Optimal Solution approach

Time complexity: O(n). Space complexity: O(1).

The optimal approach leverages the fact that we can convert each number to a string and directly check the length of the string to determine the number of digits. This is efficient and straightforward, allowing us to avoid the repeated division.

The time complexity is O(n) because we only iterate through the array once, and checking the string length is O(1) since the maximum length is constant (at most 6 for the given constraints).

Step 1: Initialize a counter to zero.
Step 2: For each number in the array, convert the number to a string.
Step 3: Check the length of the string. If the length is even, increment the counter.
Step 4: Return the counter after processing all numbers.

Input: nums = [12, 345, 2, 6, 7896] 1. Initialize count = 0. 2. Check 12: length = 2 (even), count = 1. 3. Check 345: length = 3 (odd), count remains 1. 4. Check 2: length = 1 (odd), count remains 1. 5. Check 6: length = 1 (odd), count remains 1. 6. Check 7896: length = 4 (even), count = 2. Final count = 2.

Key Insights

Counting digits can be done using division or string conversion.
Evenness can be checked using the modulus operator.

Common Mistakes

Not considering single-digit numbers as having odd digits.
Overcomplicating the digit counting process.

Interview Tips

Always clarify the constraints and edge cases before coding.
Think about both time and space complexity when discussing solutions.
Practice explaining your thought process clearly while coding.

#1295

Find Numbers with Even Number of Digits

Easy

Array↗Math↗Array↗String Manipulation↗

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

⚙️

Algorithm

4 steps

1Step 1: Initialize a counter to zero.
2Step 2: For each number in the array, convert the number to a string.
3Step 3: Check the length of the string. If the length is even, increment the counter.
4Step 4: Return the counter after processing all numbers.

solution.py4 lines

1# Full working Python code
2
3def count_even_digit_numbers(nums):
4    return sum(1 for num in nums if len(str(num)) % 2 == 0)

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the array once, and checking the string length is O(1) since the maximum length is constant (at most 6 for the given constraints).

1Counting digits can be done using division or string conversion.
2Evenness can be checked using the modulus operator.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.