Find Min Element in A Sequence w/ $O(\log n)$ Complexity

[evinda]

Hello! (Nerd)

A sequence of numbers $A=(a_1, a_2, \dots, a_n)$ is called convex if there is a $k$ with $1 \leq k \leq n$ such that $\forall 1 \leq i \leq k-1$, we have that $a_i>a_{i+1}$ and $\forall k \leq i \leq n-1$ we have that $a_i<a_{i+1}$.
Write an algorithm with time complexity $O(\log n)$,that finds the minimum element of the sequence $A$.

That's what I have tried:

Minimum(A[1...n], low, high){
   int mid=low+floor((high-low)/2);
   if (high<low) return -1;
   if (A[mid]>A[mid-1] and mid-1>=low) return Minimum(A, low, mid-1);
   else if (A[mid]<A[mid+1] and mid+1<=high) return Minimum(A,mid+1,high);
   return A[mid];

Could you tell me if it's right or if I have done something wrong? (Thinking)

 

[Future Bruno]

Yes, your algorithm looks correct! It has a time complexity of $O(\log n)$ because the recursive calls are halving the size of the input array each time.