forked from keon/algorithms
-
Notifications
You must be signed in to change notification settings - Fork 0
/
next_greatest_letter.py
60 lines (51 loc) · 1.36 KB
/
next_greatest_letter.py
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
'''
Given a list of sorted characters letters containing only lowercase letters,
and given a target letter target, find the smallest element in the list that
is larger than the given target.
Letters also wrap around. For example, if the target is target = 'z' and
letters = ['a', 'b'], the answer is 'a'.
Input:
letters = ["c", "f", "j"]
target = "a"
Output: "c"
Input:
letters = ["c", "f", "j"]
target = "c"
Output: "f"
Input:
letters = ["c", "f", "j"]
target = "d"
Output: "f"
Reference: https://leetcode.com/problems/find-smallest-letter-greater-than-target/description/
'''
import bisect
"""
Using bisect libarary
"""
def next_greatest_letter(letters, target):
index = bisect.bisect(letters, target)
return letters[index % len(letters)]
"""
Using binary search: complexity O(logN)
"""
def next_greatest_letter_v1(letters, target):
if letters[0] > target:
return letters[0]
if letters[len(letters) - 1] <= target:
return letters[0]
left, right = 0, len(letters) - 1
while left <= right:
mid = left + (right - left) // 2
if letters[mid] > target:
right = mid - 1
else:
left = mid + 1
return letters[left]
"""
Brute force: complexity O(N)
"""
def next_greatest_letter_v2(letters, target):
for index in letters:
if index > target:
return index
return letters[0]