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In the array implementation of lists, elements are stored into continuous locations. In order to add an element into the list at the end, we can insert it without any problem. But, assume if we desire to add the element at the starting or middle of the list, then we ought to rewrite all the elements after the position where the element ought to be added. We ought to shift (n)th element to (n+1)th position, where 'n' refer to number of elements in the list. The (n-1)thelement to (n)th position and it will continue till the ( r ) thelement to ( r + 1 )th position, where 'r' refer to the position of insertion. For doing this, thecount will be incremented.
From the above instance, if we desire to add element '35' after element '33'. We ought to shift 77 to the 8th position, 66 to the 7th position, so on, 44 to the 5th position.
Before Insertion
Count 1 2 3 4 5 6 7
11
22
33
44
55
66
77
Step 1
Count 1 2 3 4 5 6 7 8
Step 2
Step 3
Step 4
Step 5
35
Open addressing The easiest way to resolve a collision is to start with the hash address and do a sequential search by the table for an empty location.
Best Case: If the list is sorted already then A[i] T (n) = c1n + c2 (n -1) + c3(n -1) + c4 (n -1) = O (n), which indicates that the time complexity is linear. Worst Case:
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