The problem falls into identifying the section with monotonically ascending array. If it is found, then the left element is naturally the minimum (Assuming sorted array is ascending array, not descending one).

So, what is a monotonically ascending array? Simply num[left] <= num[mid] <= num[right]. You can actually ignore num[mid] here.

There are 4 possibilities for the combinations with a mid-point selection:

- num[left] <= num[mid] && num[mid] <= num[right] ==> you found it!
- num[left] <= num[mid] && num[mid] > num[right] ==> pivot on the right, left=mid+1
- num[left] > num[mid] && num[mid] <= num[right] ==> pivot on the left, right=mid (NOT right=mid+1, since mid-point could be the pivot
- num[left] > num[mid] && num[mid] > num[right] ==> IMPOSSIBLE! (assume ascending sorted array)

Use the binary search framework, the code can be organized as below:

```
int findMin(vector<int> &num) {
int left = 0;
int right = num.size() - 1;
while (left <= right)
{
int mid = left + (right - left) / 2;
if (num[left] <= num[mid] && num[mid] <= num[right])
{
// is a sorted array
break;
}
// judge where pivot is
if (num[mid] < num[left])
{
// pivot is on the left
right = mid;
}
else if (num[mid] > num[right])
{
// pivot is on the right
left = mid + 1;
}
}
return num[left];
}
```