For each number n, repeatedly divide n by 10 until it becomes 0, while incrementing the counter in each iteration.

After counting the digits, check if the count is even.

Increment the total count whenever a number has an even number of digits.

Advantages of Using Division

No dependency on additional libraries: Only basic arithmetic operations are involved, like division and modulo.
Space efficiency: No conversion to strings or use of logarithms.

Code

[{"language":"Java","code":"class Solution {\n    public int findNumbers(int[] nums) {\n        int ans = 0;\n        for (int n : nums) {\n            int digitCount = 0;\n            while (n > 0) {\n                n /= 10;  // Reduce the number by dividing it by 10\n                digitCount++;  // Increase digit count for every division\n            }\n            if (digitCount % 2 == 0) {  // Check if the digit count is even\n                ans++;\n            }\n        }\n        return ans;\n    }\n}","lang":"java"},{"language":"Python","code":"class Solution:\n    def findNumbers(self, nums: List[int]) -> int:\n        ans: int = 0\n        for n in nums:\n            digit_count = 0\n            while n > 0: \n                n //= 10  # Reduce the number by dividing it by 10\n                digit_count += 1  # Increase digit count for every division\n            if digit_count % 2 == 0:  # Check if the digit count is even\n                ans += 1\n        return ans","lang":"python"}]

Complexity

⏰ Time complexity: O(n*d). Each division reduces the number by a factor of 10, so the number of divisions is proportional to the number of digits d. For an array of size n, it’s O(n * d) where d is the average number of digits.
🧺 Space complexity: O(1)

Method 2 - Convert numbers to string

Here is what we can do:

To determine the number of digits in a number, convert it to a string and check its length, as this is straightforward and avoids mathematical complexity.
Check if the length of the string representation of each number is even.
Increment the count whenever a number satisfies the condition.

Code

[{"language":"Java","code":"class Solution {\n    public int findNumbers(int[] nums) {\n        int ans = 0;\n        for (int n : nums) {\n            if (String.valueOf(n).length() % 2 == 0) {\n                ans++;\n            }\n        }\n        return ans;\n    }\n}","lang":"java"},{"language":"Python","code":"class Solution:\n    def findNumbers(self, nums: List[int]) -> int:\n        ans: int = 0\n        for n in nums:\n            if len(str(n)) % 2 == 0:\n                ans += 1\n        return ans","lang":"python"}]

Complexity

⏰ Time complexity: O(n*d), where n is the length of the array and d is the number of digits in the number. For each number, it takes O(d) time to count, hence O(n*d) in total.
🧺 Space complexity: O(1), as we are not using additional space proportional to the input size, aside from the space required for the count and temporary variables.

Method 3 - Using the logarithm

When working with numbers, finding their digit count can be efficiently determined using logarithm properties. For a positive integer n, the number of digits is calculated as: $$ \text{digits} = \lfloor \log_{10}(n) \rfloor + 1 ) $$ This formula works because logarithm base 10 gives the exponent required to express n as 10^k.

Using this, we can:

For each number in the array, calculate the digit count using logarithms.
Check if the digi

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