**PROBLEM LINK**

ALUN2021-LAMA

**DIFFICULTY**

Easy-medium

**PREREQUISITES**

Maths

**EXPLANATION**

Lets assume *A1, A2 and A3* be located at points (0,A1), (1,A2) and (2,A3) respectively on a cartesian coordinate. If *A1, A2 and A3* have to form an AP (arithmetic-progression) they must lie on a single straight line because if X-coordinates are in AP then Y -coordinates are also in AP on a straight line(property).

It is also given that we can only increase(not decrease) the Y-coordinates. Let’s fix the point *A1* then two cases arise as follows:

- Case 1: A2>A3

This straight line has optimal solution because if we rotate it about

A1anticlockwise it will cost us more to bring all the points on the same line.

Then the question is reduced to - how much units is needed to bring A3 on the line , and solution is given by-

```
(2*A2 - A1) - A3
```

- Case 2: A3>=A2

This line can bring optimal solution if (A1+A3)%2=0 because all three points must have integral coordinates and A2 = (A1+A3)/2.

If (A1+A3)%2 != 0 then increase A3 by one and redraw the graph. Then the problem is reduced to - how many units is needed to bringA2on the same line and its solution is given by -

```
((A1 + A3) / 2) - A2
```

*Note: - If (A1+A3)%2=1 final answer is one more than the above expression because we used one unit to increase A3*

**TIME COMPLEXITY**

Simple if else conditions, therefore:

*O(1)*

**SOLUTION**

Here is my whole solution code.