Find Numbers with Even Number of Digits LeetCode Solution

Problem – Find Numbers with Even Number of Digits LeetCode Solution

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

Find Numbers with Even Number of Digits LeetCode Solution in Java

class Solution {
    public int findNumbers(int[] nums) {
        
        int count=0;
        
        for(int i =0 ; i< nums.length; i++){
            
            if((nums[i]>9 && nums[i]<100) || (nums[i]>999 && nums[i]<10000) || nums[i]==100000){
                count++;
            }
        }
        
        return count;
        
    }
}

Find Numbers with Even Number of Digits LeetCode Solution in C++

class Solution {
public:
    int findNumbers(vector<int>& nums) {
        int n,count=0;
        for(int i=0;i<nums.size();i++)
        {
            n=nums[i];
            if(( 10<=n && n<=99) || (1000<=n && n<=9999 ) || ( n==100000 ))
            {
               count++;
            }
        }
        return count;
    }
};

Find Numbers with Even Number of Digits LeetCode Solution in Python

class Solution:
    def findNumbers(self, nums: List[int]) -> int:
        
        counter = 0
        
        for number in nums:
            
            if len( str(number) ) % 2 == 0:
                
                counter += 1
                
        return counter

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