The Top
The 1 at the top is already fulfilled, and we can make progress with the two 2’s, also at the top.
![step1](https://cdn.statically.io/img/i.sstatic.net/3UdBg.png)
If the lower end of the top right 2 goes down, it would make the 2 below it have three borders instead of just two. Therefore, the top right 2 must form a square instead.
![step2](https://cdn.statically.io/img/i.sstatic.net/eHSQA.png)
Same logic can be applied to the downward extension of the 2 at Row 1 Column 5 and the 2 at Row 3 Column 8.
![step3](https://cdn.statically.io/img/i.sstatic.net/99PuO.png)
The Middle Left and the Two Down Under
Let us also turn our attention to the 1 in the middle left. If its border were anywhere other than to its right, the 2 would always be forced to have three borders. Therefore, the border of the 1 must be to its right.
![step4](https://cdn.statically.io/img/i.sstatic.net/3FIw5.png)
The square to the top left of the same 1 must be obtained by the 2 in the top left; otherwise, it will belong to a region that does not have exactly four squares. And there will only be one way that this is fulfilled:
![enter image description here](https://cdn.statically.io/img/i.sstatic.net/yPlnP.png)
The square at Row 3 Column 3 cannot extend rightward, as it would force the 2 to its right to have three borders. Therefore it must extend downward.
![step6](https://cdn.statically.io/img/i.sstatic.net/9Jfpd.png)
Note than the 2 at the bottom must extend fully horizontally. Otherwise, either a group of less than four cells would be trapped, or the 2 would obtain three borders.
![step7](https://cdn.statically.io/img/i.sstatic.net/VbHn3.png)
The Bottom Right, and Finishing Off the Twos
Now the unsolved 1 is limited as to how it can extend: two configurations of the area where the 1 will be yield a square trapped to the bottom left of the 1. Therefore, we are sure that the 1 will extend upward and rightward. It also cannot include the 2 to its right, because of the same problem that arose in previous steps: the 2 will have three borders.
![step8](https://cdn.statically.io/img/i.sstatic.net/5vkch.png)
And quickly we see that the 1 must extend downward, or else it would trap a few squares below it.
![step9](https://cdn.statically.io/img/i.sstatic.net/qXp7U.png)
Here it took me a bit of time to figure out that the 2 at the top cannot form a square, or the 2 at Row 3 Column 4 would be forced into an area of six squares.
![step10](https://cdn.statically.io/img/i.sstatic.net/vZkWR.png)
Therefore it must form an S-shape instead.
![step11](https://cdn.statically.io/img/i.sstatic.net/aTmBy.png)
Finally, the 2 at Row 3 Column 6 cannot extend downward, or else it would trap seven squares to the right. Therefore, it must extend rightward. The puzzle is completed after this.
![step12](https://cdn.statically.io/img/i.sstatic.net/YnvIC.png)