EasyNo. 1342

Number of Steps to Reduce a Number to Zero — LeetCode 1342 Python Solution

Topics
Bit Manipulation, Math
Reading time
2 min

The problem

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

Input
num = 14
Output
6
Explanation
Step 1) 14 is even; divide by 2 and obtain 7.

Python solution

PythonSolution
class Solution:
    def numberOfSteps(self, num: int) -> int:
        ans = 0
        while num:
            if num & 1:
                num -= 1
            else:
                num >>= 1
            ans += 1
        return ans

Complexity

Time
O(n)
Space · auxiliary
O(1)

Pattern: Bit Manipulation

Use XOR, masks and the low-bit trick to replace whole data structures with an integer. LeetCode 1342. Number of Steps to Reduce a Number to Zero is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Bit Manipulation.

The bit manipulation guide has the Python template for the pattern and the 194 LeetCode problems that use it.

Related problems

Related problems
No.ProblemDifficulty
0029Divide Two IntegersMedium
0067Add BinaryEasy
0231Power of TwoEasy
0342Power of FourEasy
0371Sum of Two IntegersMedium
0405Convert a Number to HexadecimalEasy

Frequently asked questions

How hard is LeetCode 1342. Number of Steps to Reduce a Number to Zero?

LeetCode 1342. Number of Steps to Reduce a Number to Zero is rated Easy on LeetCode.

What is the time complexity of LeetCode 1342. Number of Steps to Reduce a Number to Zero?

The Python solution on this page runs in O(n).

What is the space complexity of LeetCode 1342. Number of Steps to Reduce a Number to Zero?

The Python solution on this page uses O(1) auxiliary space.

What topics does LeetCode 1342. Number of Steps to Reduce a Number to Zero cover?

LeetCode 1342. Number of Steps to Reduce a Number to Zero is tagged Bit Manipulation and Math on LeetCode.

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