To find out whether a number is even or not, we use the modulus operator. Y % X implies that Y is divided by X, and whatever the remainder is will be the result.
If we divide a number by 2, and it's odd we will have a remainder.
6 % 2 == 0 # False
The number being divided is called the dividend, and the number that divides the dividend is called the divisor.
We can also use modulus for other things, such as checking if a number is odd. Or turning a list into a loop. If we have a list that is size 6, and we want to iterate 9 numbers but have it loop back around, we can use modulus for this.
6 % 9 = 3, which means our index will be at 3.
Given an array nums of integers, return how many of them contain an even number of digits.
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.
Solution
For every number in nums, we check to see if its digits are even. If it is, we generate a tuple of these numbers (this is similar to list generation. We could have used [function for i in list]. See my for more info.
We do this by calculating the length of the string of the number. This turns the number into an array (string), which as a length. The number 6 as a string will be length 1. The number 12 as a string will be length 2.
For every number,
Turn it into a string
Calculate its length
Calculate whether it is even
If it is, store True in the tuple.
Sum all the Trues in the tuple to work out how many numbers match this property.
def findNumbers(self, nums: List[int]) -> int:
return sum(len(str(n)) % 2 == 0 for n in nums)
This code is O(n) time complexity, as it touches every list item exactly once.
The space complexity is O(n), but note that because we are creating a tuple / generator of True values our space complexity will be slightly higher than n, but under Big O notational rules it is still O(n).