No, In this algorithm with a time complexity of O(log n) is not possible for getting the last index for an actually filled element in an array, given the condition that the array may not be filled in a sorted order because a binary search algorithm, which has a time complexity of O(log n), relies on the input data being sorted or structured in a certain way. Since the array may not be filled in a sorted order, such an algorithm would not be applicable. In worst case time complexity for an efficient algorithm to get the last index for an actually filled element in an array, given the condition that the array may not be filled in a sorted order, would be O(n). So algorithm would need to scan through the entire array, one element at a time, until it finds an element that is not filled.