1295.find-numbers-with-even-number-of-digits.c

/*
 * @lc app=leetcode id=1295 lang=c
 *
 * [1295] Find Numbers with Even Number of Digits
 *
 * https://leetcode.com/problems/find-numbers-with-even-number-of-digits/description/
 *
 * algorithms
 * Easy (82.83%)
 * Likes:    321
 * Dislikes: 51
 * Total Accepted:    113.5K
 * Total Submissions: 137.1K
 * Testcase Example:  '[12,345,2,6,7896]'
 *
 * 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
 *
 *
 */

// @lc code=start

int findNumbers(int *nums, int numsSize)
{
	int count = 0;
	for (int i = 0; i < numsSize; i++)
	{
		int digits = 0;
		int value = nums[i];
		while (value >= 1)
		{
			value = value / 10;
			digits++;
		}

		if (digits % 2 == 0)
		{
			count++;
		}
	}
	return count;
}

// @lc code=end