LeetCode 1295. Find Numbers with Even Number of Digits Solution in Java, C++, Python & More | Explanation + Code

Description

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] <= 105

Solutions

Solution 1: Simulation

We traverse each element x in the array nums. For the current element x, we directly convert it to a string and then check if its length is even. If it is, we increment the answer by one.

After the traversal is complete, we return the answer.

The time complexity is O(n × log M), and the space complexity is O(log M). Here, n is the length of the array nums, and M is the maximum value of the elements in the array nums.

PythonJavaC++GoTypeScriptRustJavaScriptC#

class Solution: def findNumbers(self, nums: List[int]) -> int: return sum(len(str(x)) % 2 == 0 for x in nums)(code-box)

class Solution { public int findNumbers(int[] nums) { int ans = 0; for (int x : nums) { if (String.valueOf(x).length() % 2 == 0) { ++ans; } } return ans; } }(code-box)

class Solution { public: int findNumbers(vector<int>& nums) { int ans = 0; for (int& x : nums) { ans += to_string(x).size() % 2 == 0; } return ans; } };(code-box)

func findNumbers(nums []int) (ans int) { for _, x := range nums { if len(strconv.Itoa(x))%2 == 0 { ans++ } } return }(code-box)

function findNumbers(nums: number[]): number { return nums.filter(x => x.toString().length % 2 === 0).length; }(code-box)

impl Solution { pub fn find_numbers(nums: Vec<i32>) -> i32 { nums.iter().filter(|&x| x.to_string().len() % 2 == 0).count() as i32 } }(code-box)

/** * @param {number[]} nums * @return {number} */ var findNumbers = function (nums) { return nums.filter(x => x.toString().length % 2 === 0).length; };(code-box)

public class Solution { public int FindNumbers(int[] nums) { return nums.Count(x => x.ToString().Length % 2 == 0); } }(code-box)