How is it possible to modify any sorting algorithm based on comparisons so that it has a “good” best-case running time (ie, Θ(n))? Justify!

I asked a question about the Algorithm of Order Statistics with the recursive select function or the Median of Medians algorithm and if n/6, n/7, n/10 etc would work instead of dividing n into sections of 5 elements and this was the answer. Can you give me examples showing the reasons for this answer?   The algorithm you're studying is known as the Median of Medians algorithm for finding the k-th smallest element in linear time, or O(n). The algorithm relies on dividing the sequence S into groups of 5 because using groups of this size allows for an efficient selection of approximate medians, which are critical to the algorithm's efficiency.   If you try dividing into groups of 6, 7, or 10 elements instead of 5, it would affect the efficiency and accuracy of the median-finding step. Here's why the group size matters: Group of 5 Advantage: By choosing groups of 5, the algorithm ensures that the median of medians (the pivot) is close to the true median of the entire set, which enables…

If there are any, what are they specifically for the binary search algorithm?

The median m of a sequence of n elements is the element that would fall in the middle if the sequence was sorted. That is, e ≤ m for half the elements, and m ≤ e for the others. Clearly, one can obtain the median by sorting the sequence, but one can do quite a bit better with the following algorithm that finds the kth element of a sequence between a (inclusive) and b (exclusive). (For the median, use k = n/2, a = 0, and b = n.) select(k, a, b)Pick a pivot p in the subsequence between a and b.Partition the subsequence elements into three subsequences: the elements <p, =p, >p Let n1, n2, n3 be the sizes of each of these subsequences.if k < n1    return select(k, 0, n1).else if (k > n1 + n2)    return select(k, n1 + n2, n).else    return p. c++