# PROBLEM LINK:

Div1, Div2Practice

**Author:**Misha Chorniy

**Tester:**Lewin Gan

**Editorialist:**Adarsh Kumar

# DIFFICULTY:

Easy# PREREQUISITES:

None# PROBLEM:

You are given three integers $A$,$B$ and $C$. You need to find the minimum no. of operations required to make the sequence $A$,$B$,$C$ an arithmetic progression. In one operation you can choose one of the numbers $A$,$B$,$C$ and either add $1$ to it or subtract $-1$ from it.# EXPLANATION:

For $A$,$B$,$C$ to be in arithmetic progression, they must satisfy $2B = A+C$. We need to make both side of the expression equal. Observe that in right hand side, you need to do the same type of operation on $A$,$C$. You can't add $1$ on $A$ and subtract $-1$ from $C$ because it will only increase the number of operations. Observe that left hand side of the expression can be increased by $2$ or reduced by $2$ in one operation. Hence, in order to minimize the number of operation we must first operate on left hand side and try to make the equation as balance as possible.The only case in which we can not make the equation balance by operating on left hand side is when right hand side is odd. In this case we can either add 1 or subtract 1 from right hand side and try to find the minimum operation which results from both the cases. A pseudo-code to illustrate this:

```
def g(B,sum):
return abs(B-sum/2)
def f(A,B,C):
if (A+C)%2 != 0:
return min(g(B,A+C-1),g(B,A+C+1))
else:
return g(B,A+C)
```

Function g(B,sum) here tries to find minimum number of operation required to balance the expression 2.B = sum, where changes can only be done in left hand side. Say, you operate on B by x here, then 2.(B+x) = sum. Now, x can be computed as sum/2-B and the number of operations will be abs(B-sum/2).