Leetcode 1342. Number of Steps to Reduce a Number to Zero

Given an integer `num`, return the number of steps to reduce it to zero. In one step, if the current number is even, you have to divide it by `2`, otherwise, you have to subtract `1` from it.

Easy

Description

Given an integer num, return the number of steps to reduce it to zero.

In one step, if the current number is even, you have to divide it by 2, otherwise, you have to subtract 1 from it.

Example 1:

Input: num = 14
Output: 6

• 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.

Example 2:

Input: num = 8
Output: 4

• 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.

Example 3:

Input: num = 123
Output: 12

Constraints:

• 0 <= num <= 10^6

Solution

class Solution:
    # Iteration
    # Time Complexity: BigO(N)
    # Space Complexity: BigO(1)
    def numberOfSteps(self, num: int) -> int:
        count = 0
        while num > 0:
            if num % 2 == 0:
                num /= 2
            else:
                num -= 1
            count += 1
        return count

/**
 * Iteration
 * Time Complexity: BigO(N)
 * Space Complexity: BigO(1)
 */
function numberOfSteps(num: number): number {
  let count = 0;
  while (num > 0) {
    num = num % 2 == 0 ? num >> 1 : num ^ 1;
    count += 1;
  }
  return count;
}