Algorithms answers
Binary search✔✔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✔✔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✔✔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✔✔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)✔✔# 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)✔✔# 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):
........print(f"Index of {val}: {BinarySearch(val, numlist)}")
Insertion sort✔✔The list is split into a sorted half and unsorted half. Values from the
unsorted part are picked and placed at the correct position in the sorted part (left) by
shifting all the elements towards the right (unsorted portion).
Insertion sort efficiency✔✔Time complexity of O(n*2)
Insertion sort (python)✔✔# replace "^\.+" with space
def InsertionSort(numbers):
....for idx in range(1, len(numbers)):
........val = numbers[idx]
........comp = idx-1
........while comp >=0 and val < numbers[comp]:
............numbers[comp+1] = numbers[comp]
............comp -= 1
........numbers[comp+1] = val
if __name__ == "__main__":
....numlist = [ 99, 77, 33, 55, 11 ]
....print("Before: " + str(numlist))
....InsertionSort(numlist)
....print("After: " + str(numlist))
Shell sort✔✔Sorting algorithm that splits the input list into a number of lists based on
the gap value. The number of interleaved lists is equal to the gap value and the
value of interleave_a[0] = input_list[0], interleave_b[0] = input_list[gap value],
interleave_c[0] = input_list[gap value * 2] and so on. The interleaved lists are
individually insertion sorted, merged and insertion sorted to get the final sorted list.
Shell sort efficiency✔✔Worst case efficiency is O(N^2), best case is O(N*log N)
Shell sort (python)✔✔# replace "^\.+" with space
def ShellSort(numbers):
....gap = len(numbers) // 2
....while gap > 0:
........for idx in range(gap, len(numbers)):
............temp = numbers[idx]
............comp = idx
............while comp >= gap and numbers[comp - gap] > temp:
................numbers[comp] = numbers[comp - gap]
................comp = comp - gap
............numbers[comp] = temp
........gap = gap // 2