[LeetCode]#1295. Find Numbers with Even Number of Digits

Environment: Python 3.7

Key technique: for loop

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

Analysis:

1. Convert int to string.
2. Use len() function to find is even number or not.

Solution:

class Solution:
    def findNumbers(self, nums):
        count=0
        for i in nums:
            if len(str(i))%2==0:
                count+=1
        return count

Better Solution:

class Solution:
    def findNumbers(self, nums):
         return sum([len(str(x))%2==0 for x in nums])

Submitted result:

Reference:

https://leetcode.com/problems/find-numbers-with-even-number-of-digits/discuss/576998/Python-One-Liner-91