Data Structures & Algorithms using Rust: Find Numbers with Even Number of Digits

As part of Data Structures and Algorithms implementation using Rust, today we will explore Find Numbers with Even Number of Digits problem.

Leetcode problem link: https://leetcode.com/problems/find-numbers-with-even-number-of-digits/

Problem Statement

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

Instructions

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.

Solution

Here is my solution to the above mentioned problem using Rust.

impl Solution {
    pub fn find_numbers(nums: Vec<i32>) -> i32 {
        nums.iter().filter(|&x| {
            let mut i_cnt: i32 = 1;
            let mut i: i32 = *x;
            while i / 10 > 0 {
                i = i / 10;
                i_cnt += 1;
            }
            i_cnt % 2 == 0
        }).count() as i32
    }
}

Please feel free to let me know your feedback.

Happy reading!!!

Comments

[Carter Snook]

You can just use the bitwise AND (&) to check if a number is odd or not. Here is an example:

console.log(1 & 1); // logs `1`
console.log(2 & 1); // logs `0`

The reason this works is because the AND operator checks if the bits (in 2 bit) on the two things are matching. In our case, these are 10-bit numbers. Our 10-bit numbers will be converted into 2 bits. If all present bits are equal, then it returns 1, else 0. Here is a figurative expression of the code above:

// These numbers are in 2-bits
1 AND 1 // returns `1`
10 AND 1 // returns  `0`

In the first case, both numbers have a 1 in the 1s place, so it returns 1 (true). In the second case, the first number has a 0 in the 1s place, but the second number has a 1, therefore it returns 0 (false).

[Sivakumar]

Thanks Coding Carter for your feedback.

As per my understanding, the problem statement is to find an element in Array which is "Even" interms of number of digits. For ex: 342 is an Even number but in terms of number of digits it is not "Even".

Hope this explains.

Please feel free to share your feedback.

[Carter Snook]

Yes, but couldn't you do i_cnt & 1 instead of i_cnt % 2 == 0?

[Sivakumar]

Yes, that can be refactored as you suggested.