-
Notifications
You must be signed in to change notification settings - Fork 22
Expand file tree
/
Copy path169.majority-element.py
More file actions
executable file
·76 lines (64 loc) · 1.98 KB
/
169.majority-element.py
File metadata and controls
executable file
·76 lines (64 loc) · 1.98 KB
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
# Tag: Array, Hash Table, Divide and Conquer, Sorting, Counting
# Time: O(N)
# Space: O(1)
# Ref: -
# Note: -
# Given an array nums of size n, return the majority element.
# The majority element is the element that appears more than ⌊n / 2⌋ times. You may assume that the majority element always exists in the array.
#
# Example 1:
# Input: nums = [3,2,3]
# Output: 3
# Example 2:
# Input: nums = [2,2,1,1,1,2,2]
# Output: 2
#
#
# Constraints:
#
# n == nums.length
# 1 <= n <= 5 * 104
# -109 <= nums[i] <= 109
#
#
# Follow-up: Could you solve the problem in linear time and in O(1) space?
class Solution:
def majorityElement(self, nums: List[int]) -> int:
count = 0
res = 0
for x in nums:
if count == 0:
res = x
if x == res:
count += 1
else:
count -= 1
return res
from collections import Counter
class Solution:
def majorityElement(self, nums: List[int]) -> int:
counter = Counter(nums)
for key in counter:
if counter[key] > len(nums)//2:
return key
class Solution:
def majorityElement(self, nums: List[int]) -> int:
nums.sort()
mid = len(nums)//2
return nums[mid]
class Solution:
def majorityElement(self, nums):
return self.majorityElementRec(nums, 0, len(nums) - 1)
def helper(self, nums, left, right):
if left == right:
return nums[left]
# divide
mid = (left + right) // 2
left_majority = self.helper(nums, left, mid)
right_majority = self.helper(nums, mid + 1, right)
# conquer
if left_majority == right_majority:
return left_majority
left_count = nums[left:right + 1].count(left_majority)
right_count = nums[left:right + 1].count(right_majority)
return left_majority if left_count > right_count else right_majority