## Problem – Number of Steps to Reduce a Number to Zero LeetCode Solution

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
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:**

### Number of Steps to Reduce a Number to Zero LeetCode Solution in C++

```
int numberOfSteps (int num) {
return num == 0 ? 0 : log2(num) + bitset<32>(num).count();
}
```

### Number of Steps to Reduce a Number to Zero LeetCode Solution in Python

```
class Solution:
def numberOfSteps (self, num: int) -> int:
bitstring = bin(num)[2:] # [2:] will remove the '0b' that is prepended to each bitstring by bin()
return len(bitstring) - 1 + bitstring.count('1')
```

### Number of Steps to Reduce a Number to Zero LeetCode Solution in Java

```
public int numberOfSteps (int num) {
return Integer.toBinaryString(num).length() - 1 + Integer.bitCount(num);
}
```

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