Let the number of elements be $n$ in the given array. Now, if the array is not shifted at all then the last element would be the largest.
Now, since the array is shifted by some $m$ degree. Therefore, we will broadly classify the problem into two parts
$i.$ Left part when $m> 0$ and $m < \frac{n}{2}$.
$ii.$ Right part when $m >= 0$
So in case $(i)$ the largest element lies in the left part and the in the case $ii$ the largest element lies in the right part.
So, this problem reduces to a recursive approach which divides the problem into two parts at each call. Hence, the time complexity would be
$O(logn)$.
The general call function would be largeFind(arr, 0, n-1)
pseudo code would be:
largeFind(arr, si, ei)
if(arr[si] <= arr[ei])
return arr[ei]
else
{
mid = si + (ei-si)/2
if(arr[mid] > arr[mid+1]
return arr[mid]
else if (arr[si] > arr[mid])
return largeFind (arr, si, mid-1)
else if (arr[mid] >= arr[ei])
return largeFind(arr, mid+1, ei)
}