Since it was asked to reduce the number of comparisons, Solution with O(3n/2) comparisons and runtime complexity O(n)

```
def solve(self, A):
if len(A) == 1:
return 2 * A[0]
elif len(A) == 2:
return A[0] + A[1]
if A[0] > A[1]:
ma = A[0]
mi = A[1]
else:
mi = A[0]
ma = A[1]
for i in range(2, len(A) - 1):
if A[i] < A[i + 1]:
x = A[i]
y = A[i + 1]
else:
y = A[i]
x = A[i + 1]
if x < mi:
mi = x
if y > ma:
ma = y
return mi + ma
```