Number of Steps to Reduce a Number to Zero

Similar Problems:

- CheatSheet: LeetCode For Code Interview
- CheatSheet: Common Code Problems & Follow-ups
- Tag: #simulation, #recursive, #bitmanipulation

Given a non-negative integer num, 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.

Example 1:

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.

Example 2:

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.

Example 3:

Input: num = 123 Output: 12

Constraints:

- 0 <= num <= 10^6

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

- Solution: bit manipulation

## https://code.dennyzhang.com/number-of-steps-to-reduce-a-number-to-zero ## Basic Ideas: simulation ## ## Check the string of binary representation ## 14 -> 1110 ## From right to left ## For any 0, it takes one change (divide by 2) ## for any 1, it takes two changes (substract then divide) ## Complexity: Time ?, Space O(1) class Solution: def numberOfSteps (self, num: int) -> int: s = f'{num:b}' return s.count('1')+len(s)-1

- Solution: loop

## https://code.dennyzhang.com/number-of-steps-to-reduce-a-number-to-zero ## Basic Ideas: simulation ## ## Complexity: Time O(log(n)), Space O(1) class Solution: def numberOfSteps (self, num: int) -> int: res = 0 while num != 0: if num%2 == 0: num = int(num/2) else: num -= 1 res += 1 return res

- Solution: recursive

## https://code.dennyzhang.com/number-of-steps-to-reduce-a-number-to-zero ## Basic Ideas: simulation ## ## Complexity: Time O(log(n)), Space O(1) class Solution: def numberOfSteps (self, num: int) -> int: if num == 0: return 0 if num%2 == 0: return 1+self.numberOfSteps(int(num/2)) else: return 1+self.numberOfSteps(num-1)