Find Numbers with Even Number of Digits

Problem Description

Given an array nums of integers, return how many of them contain an even number of digits.

Example

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[i] <= 10^5

Solution Approach

Intuition:

To efficiently determine Find Numbers with Even Number of Digits

Solution Implementation

Code In Different Languages:

JavaScript

Written by @Ishitamukherjee2004

 
class Solution {
findNumbers(nums) {
 let rem = 0, cnt1 = 0, cnt2 = 0;
 const n = nums.length;
 for(let i=0; i<n; i++){
   cnt1 = 0;
   let num = nums[i];
   while(num>0){
     rem = num%10;
     cnt1++;
     num=Math.floor(num/10);
   }
   if(cnt1%2==0) cnt2++;
 }
 return cnt2;
}
}

TypeScript

Written by @Ishitamukherjee2004

 class Solution {
findNumbers(nums: number[]): number {
 let rem = 0, cnt1 = 0, cnt2 = 0;
 const n = nums.length;
 for(let i=0; i<n; i++){
   cnt1 = 0;
   let num = nums[i];
   while(num>0){
     rem = num%10;
     cnt1++;
     num=Math.floor(num/10);
   }
   if(cnt1%2==0) cnt2++;
 }
 return cnt2;
}
}

Python

Written by @Ishitamukherjee2004

 class Solution:
def findNumbers(self, nums: List[int]) -> int:
 rem = 0
 cnt1 = 0
 cnt2 = 0
 n = len(nums)
 for i in range(n):
   cnt1 = 0
   num = nums[i]
   while num>0:
     rem = num%10
     cnt1+=1
     num=num//10
   if cnt1%2==0:
     cnt2+=1
 return cnt2

Java

Written by @Ishitamukherjee2004

 public class Solution {
public int findNumbers(int[] nums) {
 int rem = 0, cnt1 = 0, cnt2 = 0;
 int n = nums.length;
 for(int i=0; i<n; i++){
   cnt1 = 0;
   int num = nums[i];
   while(num>0){
     rem = num%10;
     cnt1++;
     num=num/10;
   }
   if(cnt1%2==0) cnt2++;
 }
 return cnt2;
}
}

C++

Written by @Ishitamukherjee2004

class Solution {
public:
 int findNumbers(vector<int>& nums) {
     int rem = 0, cnt1 = 0, cnt2 = 0;
     int n = nums.size();
     for(int i=0; i<n; i++){
         cnt1 = 0;
         while(nums[i]>0){
         rem = nums[i]%10;
         cnt1++;
         nums[i]/=10;

         }
     if(cnt1%2==0) cnt2++;
     }
     
     return cnt2;
 }
};

Complexity Analysis

• Time Complexity: $O(n*m)$
• Space Complexity: $O(1)$
• The time complexity is $O(n*m)$ where n is the length of the input array nums and m is the maximum number of digits in an element of the array. This is because the function iterates over each element in the array (O(n*m)) and for each element, it performs a while loop that iterates up to the number of digits in the element (O(m)).
• The space complexity is $O(1)$ because we are not using any extra space.