Merge sort is O(n log n) for both best case and average case scenarios.
binary search
Average case complexity for Binary search O(log N). (Big O log n)Habibur Rahman (https://www.facebook.com/mmhabib89)BUBT University Bangladeshhttp://www.bubt.edu.bd/
If it is an unbalanced binary tree, O( ln( n ) / ln( 2 ) ) is best-case. Worst case is O( n ). If it is balanced, worst case is O( ln( n ) / ln( 2 ) ).
Linear search takes linear time with a worst case of O(n) for n items, and an average of O(n/2). Binary search takes logarithmic time, with a worst and average case of O(n log n). Binary search is therefore faster on average.
For a list with n elements, the expected cost is the same as the worst-case cost, which is O(n). The average cost will be O(n/2). However, if the list is ordered by probability and geometrically distributed, the complexity becomes constant, O(1). Compare with a binary search which has a cost of O(log n).
binary search
The best case for a binary search is finding the target item on the first look into the data structure, so O(1). The worst case for a binary search is searching for an item which is not in the data. In this case, each time the algorithm did not find the target, it would eliminate half the list to search through, so O(log n).
Average case complexity for Binary search O(log N). (Big O log n)Habibur Rahman (https://www.facebook.com/mmhabib89)BUBT University Bangladeshhttp://www.bubt.edu.bd/
In the worst case a binary search tree is linear and has a height equal to the number of nodes. so h=O(h).
If it is an unbalanced binary tree, O( ln( n ) / ln( 2 ) ) is best-case. Worst case is O( n ). If it is balanced, worst case is O( ln( n ) / ln( 2 ) ).
Linear search takes linear time with a worst case of O(n) for n items, and an average of O(n/2). Binary search takes logarithmic time, with a worst and average case of O(n log n). Binary search is therefore faster on average.
For a list with n elements, the expected cost is the same as the worst-case cost, which is O(n). The average cost will be O(n/2). However, if the list is ordered by probability and geometrically distributed, the complexity becomes constant, O(1). Compare with a binary search which has a cost of O(log n).
When sequentially searching n items, the best-case is O(1) and the worst-case is O(n). But when the items are sorted, binary search will improve efficiency. The best case is still O(1), but worst case drops to O(log n) where log n is the binary logarithm of n. Binary search starts with the middle element of the set. If the set is empty, the item we're looking for does not exist but if the middle element is the item we are looking for then we are done. If not, a simple comparison will tell us in which half of the set to discard (including the middle element). We repeat the process with the remaining half. If there are no elements remaining, the item does not exist.
When sequentially searching n items, the best-case is O(1) and the worst-case is O(n). But when the items are sorted, binary search will improve efficiency. The best case is still O(1), but worst case drops to O(log n) where log n is the binary logarithm of n. Binary search starts with the middle element of the set. If the set is empty, the item we're looking for does not exist but if the middle element is the item we are looking for then we are done. If not, a simple comparison will tell us in which half of the set to discard (including the middle element). We repeat the process with the remaining half. If there are no elements remaining, the item does not exist.
When sequentially searching n items, the best-case is O(1) and the worst-case is O(n). But when the items are sorted, binary search will improve efficiency. The best case is still O(1), but worst case drops to O(log n) where log n is the binary logarithm of n. Binary search starts with the middle element of the set. If the set is empty, the item we're looking for does not exist but if the middle element is the item we are looking for then we are done. If not, a simple comparison will tell us in which half of the set to discard (including the middle element). We repeat the process with the remaining half. If there are no elements remaining, the item does not exist.
In linear search, the searched key will be compared with each element of the array from the beginning and terminate comparing when the searched key is found or the array is reached. Here time complexity in worst case and average case is O (n). To find an element quickly we use divide and conquer method by using binary search algorithm. Here probed region is reduced from n to n/2. Time complexity is O (log2 n), but here the array should be sorted. But in interpolation search the probed region is reduced from n to n1/2. If the array elements are uniformly distributed the average case complexity is O (log2 (log2n)). Am also searching for hashing to compare & contrast with above.
These are terms given to the various scenarios which can be encountered by an algorithm. The best case scenario for an algorithm is the arrangement of data for which this algorithm performs best. Take a binary search for example. The best case scenario for this search is that the target value is at the very center of the data you're searching. So the best case time complexity for this would be O(1). The worst case scenario, on the other hand, describes the absolute worst set of input for a given algorithm. Let's look at a quicksort, which can perform terribly if you always choose the smallest or largest element of a sublist for the pivot value. This will cause quicksort to degenerate to O(n2). Discounting the best and worst cases, we usually want to look at the average performance of an algorithm. These are the cases for which the algorithm performs "normally."