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1295. Find Numbers with Even Number of Digits

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

  • 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;
    }
};

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


  • 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;
};


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


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

All Problems

All Solutions