1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
| # Sample implementation of quicksort, mergesort and binary_search in ruby
# Both algorithm sort in O(n * lg(n)) time
# Quicksort works inplace, where mergesort works in a new array
def quicksort(array, from=0, to=nil)
if to == nil
# Sort the whole array, by default
to = array.count - 1
end
if from >= to
# Done sorting
return
end
# Take a pivot value, at the far left
pivot = array[from]
# Min and Max pointers
min = from
max = to
# Current free slot
free = min
while min < max
if free == min # Evaluate array[max]
if array[max] <= pivot # Smaller than pivot, must move
array[free] = array[max]
min += 1
free = max
else
max -= 1
end
elsif free == max # Evaluate array[min]
if array[min] >= pivot # Bigger than pivot, must move
array[free] = array[min]
max -= 1
free = min
else
min += 1
end
else
raise "Inconsistent state"
end
end
array[free] = pivot
quicksort array, from, free - 1
quicksort array, free + 1, to
end
def mergesort(array)
if array.count <= 1
# Array of length 1 or less is always sorted
return array
end
# Apply "Divide & Conquer" strategy
# 1. Divide
mid = array.count / 2
part_a = mergesort array.slice(0, mid)
part_b = mergesort array.slice(mid, array.count - mid)
# 2. Conquer
array = []
offset_a = 0
offset_b = 0
while offset_a < part_a.count && offset_b < part_b.count
a = part_a[offset_a]
b = part_b[offset_b]
# Take the smallest of the two, and push it on our array
if a <= b
array << a
offset_a += 1
else
array << b
offset_b += 1
end
end
# There is at least one element left in either part_a or part_b (not both)
while offset_a < part_a.count
array << part_a[offset_a]
offset_a += 1
end
while offset_b < part_b.count
array << part_b[offset_b]
offset_b += 1
end
return array
end
# Search a sorted array in O(lg(n)) time
def binary_search(array, value, from=0, to=nil)
if to == nil
to = array.count - 1
end
mid = (from + to) / 2
if value < array[mid]
return binary_search array, value, from, mid - 1
elsif value > array[mid]
return binary_search array, value, mid + 1, to
else
return mid
end
end
a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15].shuffle
# Quicksort operates inplace (i.e. in "a" itself)
# There's no need to reassign
quicksort a
puts "quicksort"
puts a
b = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16].shuffle
# Mergesort operates in new array
# So we need to reassign
b = mergesort b
puts "mergesort"
puts b
offset_3 = binary_search a, 3
puts "3 is at offset #{offset_3} in a"
offset_15 = binary_search b, 15
puts "15 is at offset #{offset_15} in b"
|