Question

1295. Find Numbers with Even Number of Digits

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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

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

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

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

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

/**
 * @param {number[]} nums
 * @return {number}
 */
var findNumbers = function (nums) {
  let ans = 0;
  for (const v of nums) {
    ans += String(v).length % 2 == 0;
  }
  return ans;
};