¿Cómo se cuenta el número de swaps en un tipo de selección?

How do you count the number of swaps in a selection sort?

So we know that selection sort gives the minimum number of swaps to sort an array. The formula I came up with is given an unsorted array and it’s descending or ascending order. We find the number of elements dissimilar to the sorted array. When we subtract 1 from this number we can get the number of swaps.

Selection sort

In ascending order: In Bubble sort, the largest element moves to the right. So swapping is done, when a smaller element is found on the right side. So to count the number of swaps for an element, just count the number of elements on the right side that are smaller than it.

1. Count swaps while performing Insertion sort. void insertionSort(int *array, int size) { int key, j,swap=0;; for(int i = 1; i 0 && array[j-1]>key) { array[j] = array[j-1]; swap++; j–; } array[j] = key; //insert in right place } }
2. Finding bigger elements.

In general, the average number of comparisons per pass in selection sort will always be one half of the number of items to be sorted. For eight items, we have 1/2(82 + 8) = 1/2(64 + 8) = 1/2(72) = 36 comparisons.

How to implement Selection sort in Java

1. Initially, the whole array is unsorted.
2. Find the minimum of the unsorted part and swap it with the first element.
3. Now, consider the minimum number chosen in the previous step as part of the sorted array.

In a selection sort, the first process in the algorithm is to find the data element that has the largest value.

• Since 5 is not larger than 55, max and maxpos are left alone.
• At this point the array looks like this:
• The process is repeated by changing the lead position to be the second element.

So 14 swaps will be done.

The Bubble Sort algorithm utilizes two loops: an outer loop to iterate over each element in the input list, and an inner loop to iterate, compare and exchange a pair of values in the list. The inner loop takes (N-1) iterations while the outer loop takes N iterations.

In the expected case, insertion sort requires 1/4(N2 – N) comparisons, and thus should require about 1/2 the comparisons needed by selection sort. This result is verified by the eight item list of Figure 5.11 and Figure 5.13.

To determine the average efficiency of insertion sort consider the number of times that the inner loop iterates. As with other loops featuring nested loops, the number of iterations follows a familiar pattern: 1 + 2 + + (n – 2) + (n – 1) = n(n – 1) = O(n2).

So we know that selection sort gives the minimum number of swaps to sort an array. The formula I came up with is given an unsorted array and it’s descending or ascending order. We find the number of elements dissimilar to the sorted array. When we subtract 1 from this number we can get the number of swaps. Using selection sort for descending order:

Given an array of values of length n, is there a way to count the number of swaps that would be performed by insertion sort to sort that array in time better than O (n 2 )? arr []= {2 ,1, 3, 1, 2}; // Answer is 4.

The formula I came up with is given an unsorted array and it’s descending or ascending order. We find the number of elements dissimilar to the sorted array. When we subtract 1 from this number we can get the number of swaps. Using selection sort for descending order: Descending array is [9,4,3,2,1].

It is the worst case as the elements are sorted in descending order. Selection sort finds the minimum element and swap it with the first element and so on.Here 5 and 4 will swap with 1 and 2 resp. It will become 12345. We needed 2 swaps in that.