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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).

Q: What is the worst case and best case for binary search?

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Lower case 'x' is 120 (decimal) or 1111000 (binary) in the ASCII character table.

You can easily convert decimal to binary in the scientific calculator - for example, the scientific calculator found in Windows. In this case, type the number in decimal, then click on "binary" to convert to binary.

It is the binary number 1010100. A binary number represents exponential values of 2, in this case 7 digits for 64, 32, 16, 8, 4, 2, and 1 84 = 64 + (0x32) + 16 + (0x8) +4 + (0x2) + (0x1) = 1010100

The remainder of the division, by 4, is a number between 0 and 3. In the case of binary, this would maintain the last two bits of the original number.

Each octal digit is equivalent to three binary digits; each hexadecimal digit is equal to four binary digits. I think the best way to do this conversion is to convert each octal digit into the binary equivalent (3 digits in each case - don't omit the zeros on the left), then convert the binary to hexadecimal by grouping four binary digits at a time (starting from the right). Note that nowadays, most scientific calculators - including the calculator that comes included in Windows - have the ability to do this sort of conversion. If you want to practice doing it yourself, you can still use the Windows calculator to check your calculations.

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binary search

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 ) ).

Merge sort is O(n log n) for both best case and average case scenarios.

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/

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.

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."

Binary search is a log n type of search, because the number of operations required to find an element is proportional to the log base 2 of the number of elements. This is because binary search is a successive halving operation, where each step cuts the number of choices in half. This is a log base 2 sequence.

If the array is unsorted, the complexity is O(n) for the worst case. Otherwise O(log n) using binary search.

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.