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Which sorting algorithm has minimum number of comparisons?
Merge-insertion sort
Merge-insertion sort is the sorting algorithm with the minimum possible comparisons for n items whenever n ≤ 15 or 20 ≤ n ≤ 22, and it has the fewest comparisons known for n ≤ 46.
What is the minimum number of comparisons required?
The minimum number of comparisons required to find the minimum and the maximum of 100 numbers is ______________. Therefore, we need 3 comparisons for each 2 elements, so total number of required comparisons will be (3n)/2 – 2, because we do not need to update min or max in the very first step.
How many comparisons are needed to sort 5 elements using comparison based?
Figure 5.15 illustrates this fact on a five item list. Part (a) of Figure 5.15 shows that 10 comparisons are required to sort the five items when they are originally arranged in reverse sorted order.
What is the minimum number of comparisons required for a linear search?
A linear search runs in at worst linear time and makes at most n comparisons, where n is the length of the list. If each element is equally likely to be searched, then linear search has an average case of n+12 comparisons, but the average case can be affected if the search probabilities for each element vary.
What are the number of comparisons needed for MaxMin algorithm?
⌉−2 comparisons are necessary for any A to find both max and min from a given list of n elements. Hence our efficient algorithm is optimal.
How many comparisons are needed to sort a list?
The last two steps take two comparisons each; we insert into a three-element list in both cases (for step 4., note that we know from c < d that c is smaller than the last element of the list at hand) and compare with the middle element first. That makes a total of seven comparisons.
What’s the minimum number of comparisons for 4 elements?
for 4 elements the min. number of comparisons is 3. In general, to find largest of N elements you need N-1 comparisons. This gives you 4 for 5 numbers, not 6. there is always a solution with N-1 comparisons: just compare first two and then select the larger and compare with next one, select the larger and compare with next one etc….
How to calculate the minimum number of comparisons?
In this method, the total number of comparisons is 1 + 2 (n-2) in the worst case and 1 + n – 2 in the best case. In the above implementation, the worst case occurs when elements are sorted in descending order and the best case occurs when elements are sorted in ascending order.
What is the number of comparisons for sorting algorithms?
Number of comparisons for sorting algorithms Insertion Sort: Θ(n2) worst case O(kn) if ≤k items out of order Mergesort: Θ(nlgn) worst case Heapsort: Θ(nlgn) worst case Quicksort: Θ(n2) worst case Θ(nlgn) average case Lower Bound: Ω(nlgn) worst case and average case Four ways to apply recursion to sorting
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