![]() While this version of the algorithm may appear somewhat more complex than the English version presented above, it is really just a more detailed description of the same process. This is approximately one half of the 36 comparisons needed by selection sort.Ī more formal version of the insertion sort algorithm is presented in. A careful examination of will reveal that 19 comparison operations are required to sort these values using the insertion method. Notice that this is exactly the same list that was used to illustrate the behavior of selection sort in. The insertion sort procedure is illustrated on an eight item list in. Eventually, the unsorted list will be empty since all of the items will have been inserted into the sorted list. The entire process of selecting the first item from the unsorted list and scanning backwards through the sorted list for the insertion point is then repeated. When that happens the algorithm inserts the item at the current insertion point. Eventually it will either reach the beginning of the list or encounter an item that is less than or equal to the item to be inserted. As long as the current item is larger than the item to be inserted, the algorithm continues moving backward through the sorted list. It then works its way from the back to the front of the sorted list, at each step comparing the item to be inserted with the current item. Insertion sort removes the first item from the unsorted list and marks it as the item to be inserted. While this way of thinking about the input list may seem a bit odd at first, it really makes quite a bit of sense, since a list that is one item long certainly cannot be out of order and is thus sorted. ![]() The first item of the input list is considered to be a sorted list one item long, with the rest of the input items (2 through N) forming an unsorted list. To begin to understand the algorithm, think of an unordered input list as two separate lists. Insertion sort is the procedure that most people use to arrange a hand of cards. This characteristic makes the Insertion Sort a stable sorting algorithm.Another approach to sorting is typified by the insertion sort. If there are duplicate elements in the sorting array, the Insertion Sort won’t move one element in front of the other, instead, the key value will be maintained in order. However, there will be no shifting or insertion of elements performed. ![]() In case of a completely sorted list, the Insertion Sort requires the same number of comparisons as that of an unsorted list. You might recognize the similar pattern from Selection Sort. The total number of comparisons required to sort: n = 5 elements is (n-1) + (n-2) + (n-3) + (n-4). ![]() The array size in our example is n = 5 and the total iterations performed will be (n-1). If the current unsorted array element is smaller than the sorted array element, the current element is shifted to one place higher than its current position. Although the first element is unsorted, it becomes the sorted element in the first iteration. Iteration is performed on every element and it is from left to right, growing the sorted array. Well, the obvious observation here is, in each iteration, we compared the first unsorted item against the sorted item to its left to figure out if it’s in the right sorted position. We observe that the unsorted subset shrinks with every iteration. In the next iteration the sorted subset is, the process is repeated for the next number in an unsorted subset. The above steps will be repeated until a valid sorted position is obtained for the same number. The sorting will iterate to the next number (i.e., 8) to compare against a new sorted subset. Meaning, the smaller number is inserted into the sorted subset. If a number in a sorted list is found to be greater than the unsorted list number (i.e., 4), the number is shifted one position right. The first element in the unsorted subset (i.e., 4 in our example) is compared against each element from the sorted subset (i.e., 10 in this case) for its rightful sorted position. Insertion Sort will iterate through each element of the unsorted list and compare it with the sorted subset, to shift the largest number to the right and insert the smallest element in its current rightful sorted position. Now we have two sections, one sorted section, and other unsorted sections. The first element in the array will remain in place and is assumed to be sorted, as there are no further elements on the left side to compare. Two subsets or sub-lists are maintained, the left side is a sorted subset and the right side is an unsorted subset. For a given unsorted array, the iteration will be performed for every array element, from left to right.
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