Question: I am converting a list to set and back to list. I know that `set(list)` takes O(n) time, but I am converting it back to list in same line `list(set(list))`. Since both these operations take O(n) time, would the time complexity be O(n^2) now?

Logic 1:

```
final = list(set(list1)-set(list2))

```

Logic 2:

```
s = set(list1)-set(list2)
final = list(s)

```

Do both these implementations have different time complexities, and if they do which of them is more efficient?

Answer: 
There are normally a few ways to do this, here are some suggestions:

1. Set menu to null when creating the window:

`const mainWin = new BrowserWindow({menu: null});`
2. Remove menu after window object has been created (Set a blank menu with MacOS):

`const { Menu } = require('electron');`

`process.platform === "win32" && mainWin.removeMenu();`

`process.platform === "darwin" && Menu.setApplicationMenu(Menu.buildFromTemplate([]));`
3. Using the Menu module from electron:

`const { Menu } = require('electron');`

`Menu.setApplicationMenu(null);`
You could try this

```js
const { Menu } = require('electron');
Menu.setApplicationMenu(null);

```