Answer:

Best: O(n) where n is the number of elements in the array occurs if the array is already sorted. The algorithm will store temp, then compare with the element to the left of it in the sorted partition. Since temp is greater than the element in the sorted partition, it will essentially do nothing because it’s already in correct order. It will repeat this for n elements in the array.

Worst: O(n2) where n is the number of elements in the array. Worst case occurs if the array is sorted in reverse order. For every unsorted element, the algorithm will then have to traverse the entire sorted partition from right to left until it finds the correct position for temp (which is at the beginning). If the program has to traverse the entire sorted partition (O(n)) for n elements, we get O(n2)