Approach: Simulate the process

def numberOfSteps(n):
    steps = 0
    while n > 0:
        n = n // 2 if n % 2 == 0 else n - 1
        steps += 1
    return steps

Complexity: O(log n) time, O(1) space.

Full Solution — all languages

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

import java.util.*;
class Solution {
    public int numberOfSteps(int n) {
        int steps = 0;
        while (n > 0) {
            if (n % 2 == 0) n /= 2;
            else n -= 1;
            steps++;
        }
        return steps;
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: int numberOfSteps(int n) {
        int steps = 0;
        while (n > 0) {
            if (n % 2 == 0) n /= 2;
            else n -= 1;
            steps++;
        }
        return steps;
    }
};

/**
 * @param {number} n
 * @return {number}
 */ var numberOfSteps = function(n) {
    let steps = 0;
    while (n > 0) {
        if (n % 2 === 0) n /= 2;
        else n -= 1;
        steps++;
    }
    return steps;
};

Number of Steps to Reduce a Number to Zero

Example 1

Example 2

Example 3

Constraints