Binary search - ANSWER-A faster algorithm for searching a list if the list's
elements are sorted and directly accessible (such as an array). Binary search
first checks the middle element of the list. If the search key is found, the
algorithm returns the matching location. If the search key is not found, the
algorithm repeats the search on the remaining left sublist (if the search key
was less than the middle element) or the remaining right sublist (if the
search key was greater than the middle element).
Binary search efficiency - ANSWER-For an N element list, the maximum
number of steps required to reduce the search space to an empty sublist is
[ log2 N ] + 1
Selection sort - ANSWER-Sorting algorithm that treats the input as two parts,
a sorted part and an unsorted part, and repeatedly selects the proper next
value to move from the unsorted part to the end of the sorted part.
,Selection sort efficiency - ANSWER-If a list has N elements, the outer loop
executes N times. For each of those N outer loop executions, the inner loop
executes an average of N/2 times. So the total number of comparisons is
proportional to N * (N/2), or O(N^2)
Selection sort (python) - ANSWER-# replace "^\.+" with space
def SelectionSort(numbers):
....for idx in range(len(numbers)):
........min_idx = idx
........for comp in range(idx+1, len(numbers)):
............if numbers[min_idx] > numbers[comp]:
................min_idx = comp
........temp = numbers[idx]
........numbers[idx] = numbers[min_idx]
........numbers[min_idx] = temp
if __name__ == "__main__":
....numlist = [ 99, 77, 33, 55, 11 ]
....print("Before: " + str(numlist))
....SelectionSort(numlist)
, ....print("After: " + str(numlist))
Binary search (python) - ANSWER-# replace "^\.+" with space
def BinarySearch(num, numbers):
....low = 0
....high = len(numbers) - 1
....mid = 0
....while low <= high:
........mid = (high + low) // 2
........if numbers[mid] < num:
............low = mid + 1
........elif numbers[mid] > num:
............high = mid - 1
........else:
............return mid
....return -1
if __name__ == "__main__":
....numlist = [ 11, 33, 55, 77, 99 ]
....for val in (11, 22, 55, 88, 99):
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