Leetcode: Find Numbers with Even Number of Digits in PHP

Given an array nums of integers, return how many of them contain an even number of digits.

Example 1:

Input: nums = [12,345,2,6,7896]
Output: 2
Explanation: 
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 
Explanation: 
Only 1771 contains an even number of digits.

Constraints:

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

Solution:

function findNumbers($nums) {
    $count = 0;
    
    foreach ($nums as $num) {
        // Count the digits in the number
        $digits = strlen((string)$num);
        
        // Check if the number of digits is even
        if ($digits % 2 === 0) {
            $count++;
        }
    }
    
    return $count;
}


$nums = [12, 345, 2, 6, 7896];
echo findNumbers($nums); // Output: 2

Time and Space Complexity

Time Complexity: Loop through the array once: O(N), where N is the number of elements in nums. Converting each number to a string and calculating its length takes O(log⁡M), where M is the number’s value.

Total Time Complexity: O(N⋅log⁡M).

Space Complexity:

The space used is O(1) since we only use variables like $count and $digits.