Rust for Python Developers - Operators

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Problem 2: Number of Steps to Reduce a Number to Zero

In this problem, you input a non-negative integer num and return the number of steps to reduce it to zero. If the current number is even, you have to divide it by 2, otherwise, you have to subtract 1 from it.

For example:

Input: num = 14
Output: 6
Explanation: 
Step 1) 14 is even; divide by 2 and obtain 7.
Step 2) 7 is odd; subtract 1 and obtain 6.
Step 3) 6 is even; divide by 2 and obtain 3.
Step 4) 3 is odd; subtract 1 and obtain 2.
Step 5) 2 is even; divide by 2 and obtain 1.
Step 6) 1 is odd; subtract 1 and obtain 0.

Input: num = 8
Output: 4
Explanation: 
Step 1) 8 is even; divide by 2 and obtain 4.
Step 2) 4 is even; divide by 2 and obtain 2.
Step 3) 2 is even; divide by 2 and obtain 1.
Step 4) 1 is odd; subtract 1 and obtain 0.

This is a good example that we can use the Modulus/Remainder operator and the compound assignment operators.

Python Solution

Line: 3–10: We use a while loop for num > 0. If the modulus is 0, then it must be an even number so we divide the num by 2 using a compound assignment /=2, otherwise, we subtract 1 using a compound assignment -=1. We increase the steps by 1. Finally, we return the steps.

We adjust the above code to the LeetCode environment.

class Solution:
    def numberOfSteps (self, num: int) -> int:
        steps = 0
        while num > 0:
            if num % 2 == 0:
                num //= 2
            else:
                num -=1
            steps += 1
        return steps

Rust Solution

In Rust, we take the same steps as we did in Python.

Line 7–16: We assign 0 to a mutable variable steps. While self.num is greater than 0, we use the compound assignment /=2 if self.num 's remainder is 0, otherwise, we subtract 1, and increase the number of step by 1.

We adjust the above code to the LeetCode environment.

impl Solution {
    pub fn number_of_steps (mut num: i32) -> i32 {
        let mut steps = 0;
        while num > 0 {
            if num % 2 == 0 {
                num /= 2;
            } else {
                num -=1;
            }
            steps += 1;
        }
        steps
    }
}