*This is part of a series of Leetcode solution explanations (index). If you liked this solution or found it useful,* *please like**this post and/or* *upvote**my solution post on Leetcode's forums.*

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Leetcode Problem #1342 (*Easy*): Number of Steps to Reduce a Number to Zero

*Description:*

*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.*

*Examples:*

Example 1: | |
---|---|

Input: |
nums = [1,2,3] |

Output: |
[1,3,2] |

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`

*Idea:*

For this problem, we just have to follow the directions: we can tell if **num** is odd by using **modulo 2**; **num % 2** is **1** if **num** is odd, otherwise it's **0**. Anytime **num** is odd, subtract **1** from **num**. Anytime **num** is even, divide **num** by **2**.

While we do this, we can just increment our counter (**ans**) and then **return** it once **num** reaches **0**.

*Javascript Code:*

The best result for the code below is **68ms / 37.1MB** (beats 99% / 100%).

```
var numberOfSteps = function(num) {
let ans = 0
for (; num; ans++)
if (num % 2) num--
else num /= 2
return ans
};
```

*Python Code:*

The best result for the code below is **24ms / 14.1MB** (beats 96% / 89%).

```
class Solution:
def numberOfSteps (self, num: int) -> int:
ans = 0
while num > 0:
if num % 2: num -= 1
else: num /= 2
ans += 1
return ans
```

*Java Code:*

The best result for the code below is **0ms / 35.5MB** (beats 100% / 95%).

```
class Solution {
public int numberOfSteps (int num) {
int ans = 0;
for (; num > 0; ans++)
if (num % 2 == 1) num--;
else num /= 2;
return ans;
}
}
```

*C++ Code:*

The best result for the code below is **0ms / 5.8MB** (beats 100% / 85%).

```
class Solution {
public:
int numberOfSteps (int num) {
int ans = 0;
for (; num > 0; ans++)
if (num % 2 == 1) num--;
else num /= 2;
return ans;
}
};
```

## Discussion (3)

Hey hi, I cant understand what is happening on this line.

// for (; num; ans++)

Could u explain?

Sure!

The for loop declaration is broken up into three parts. The first part is an expression that's executed once at the very beginning of the loop. Usually it's used to initialize loop variables. Here, I'm using

ansas the iterable, but I also want to be able to refer to it once the loop is over, so I have to initialize it outside the loop. Since I don't have any other loop variables to initialize, the first part is left blank.The middle portion is the "while" conditional. Here, I want to continue looping while

num > 0. Since0evaluates as falsy and any other number evaluates as truthy, I can simplify that conditional to justnum, so that whennumdecrements to0, it triggers an end to the loop.The third part is more standard: it's an expression that will run at the end of each loop, before restarting the next iteration. In this case, it's a fairly standard variable increment.

Thanks a ton ðŸ˜„