FastPrep
Problem · Array

Count Even-Digit Numbers

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Given an array of positive integers numbers, calculate how many of its elements have an even number of digits.

Note: You are not expected to provide the most optimal solution, but a solution with time complexity not worse than O(numbers.length2) will fit within the execution time limit.

Examples
01 · Example 1
numbers = [12, 134, 111, 1111, 10]
return = 3
  • numbers[0] = 12 has 2 digits, which is an even number.
  • numbers[1] = 134 has 3 digits, which is not an even number.
  • numbers[2] = 111 has 3 digits, which is not an even number.
  • numbers[3] = 1111 has 4 digits, which is an even number.
  • numbers[4] = 10 has 2 digits, which is an even number.

There are 3 elements, numbers[0], numbers[3], and numbers[4], with an even number of digits.

Constraints
  • 1 <= numbers.length <= 1000
  • 1 <= numbers[i] <= 104
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public int solution(int[] numbers) {
  // write your code here
}
numbers[12, 134, 111, 1111, 10]
expected3
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