Leetcode Array101 Find Numbers with Even Number of Digits by Go

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

Constraints:

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

My code

func findNumbers(nums []int) int {
	count := 0
	for _, val := range nums {
		for i := 10; 10^499/i > 1; {
			if val/i < 1 {
				break
			} else if val/i >= 10 {
				i = i * 100
			} else {
				count++
				break
			}
		}
	}
	return count
}

• go.dev/play/p/2KlCdhBdEB4
• Due to nums length, I put power 499.
  - length 10^1 = 2
  - length 10^2 = 3
  .
  .
  .
  - length 10^499 = 500
• lenght even have patten about 10 * 100^n
• when starts, val / i(10 * 100^n) less than 1 will not be even
  - example
  • 1 / 10 = 0.1 (odd)
    • this num will be passed because it is odd.
  • 100 / 10 = 10 (odd?)
  • 101 / 10 = 10.1 (odd?)
    • to be sure, i = i * 100
  • 100 / 1000 = 0.1(odd)

sample 0 ms submission & Accepted Solutions Memory Distribution

Code

func getNumberOfDigits(n int) int {
    var numDigit int
    for n >= 1 {
        numDigit += 1
        n /= 10
    }
    return numDigit
}

func findNumbers(nums []int) int {
    var numEven int
    
    for _, n := range nums {
        if getNumberOfDigits(n) % 2 == 0 {
            numEven += 1
        }
    }
    
    return numEven
}

• This code can check how many digits does num have.
• By checking number is more than 1, this code counts +1 digit and divide by 10 in getNumberofDigits.
• finally returned num digits were divided by 2 and remainder is 0 then it will be even.
• this code is perfect both sides(speed and size).