binSearchMin is a recursive helper function that takes two arguments, l and r, representing the left and right indices of the current search range.

Inside binSearchMin, the code first checks if l is greater than r, which means the search range is invalid or empty, and it returns immediately.

It uses a non-local variable min to keep track of the minimum element found during the search.

The code calculates the middle index mid of the current search range using binary search.

If the element at the middle index nums[mid] is less than the current minimum (min), it updates min with the new minimum value.

The code then checks if nums[mid] is greater than nums[r]. If it is, it means that the minimum element is in the right half of the current search range, so it recursively searches the right half (binSearchMin(mid + 1, r)). Otherwise, it searches the left half (binSearchMin(l, mid - 1)).

The main findMin function initializes the min variable to positive infinity to ensure that it's updated during the search.

It sets the left and right pointers to the beginning and end of the array, respectively.

It calls the binSearchMin function to perform the binary search for the minimum element.

Finally, it returns the min variable, which contains the minimum element found during the search.

The problem statement describes finding the minimum element in a rotated sorted array efficiently in O(log n) time. The code achieves this by adapting the binary search algorithm. It compares elements at the middle of the search range with the rightmost element to determine whether the minimum is in the left or right half. This approach narrows down the search range quickly, ensuring logarithmic time complexity.

The time complexity of this code is O(log n) because it effectively divides the search range in half with each recursive call. The space complexity is O(1) because it uses only a constant amount of additional space regardless of the input size.