Show that the average-case time complexity of Interpolation Search is in \(\Theta\) (Ig (Ig \(n\) )), assuming the keys are uniformly distributed and that search key \(x\) is equally probable to be in each of the array slots.

Interpolation Search finds a particular item by computing the probe position. Initially, it starts searching at the beginning and the end of the array and these values are used to find a new mid. If the value of the key being sought, \(x\), is greater than the middle value, \(x\) must lie in the right half of the list, else it lies in the left half. It continues the process until the element is found or the end of the searchable part is reached.

The average-case time complexity of Interpolation search is derived based on the idea that the search keys are uniformly distributed and that search key \(x\) is equally probable to be in each of the array slots. For this scenario, the expected amount of time needed to reach an element is proportional to the logarithm of the logarithm of the number of elements, stated as \(\Theta(\log(\log(n)))\). This is because in each step, the algorithm reduces the size of the problem by a constant factor, allowing it to skip over many elements. Because the number of elements being skipped is logarithmic, and the number of steps required to reduce the number of elements to a constant size is also logarithmic, the total number of steps can be expressed as a logarithm of a logarithm, hence \( \Theta(\log(\log(n))) \).

In an average-case scenario, where the keys are evenly distributed and each key has an equal probability to be in any of the array slots, the complexity of Interpolation Search becomes \(\Theta(\log(\log(n)))\), making it more efficient than a binary search in certain situations. This is largely due to the algorithm's ability to probe the list depending on the actual value of \(x\), rather than just splitting it in half.