|
| 1 | +class Solution: |
| 2 | + def majorityElement(self, nums: List[int]) -> int: |
| 3 | + length = len(nums)//2 |
| 4 | +# Time limit will exceed in case of long arrays [5k elements] |
| 5 | +# majority = 0 |
| 6 | +# element = 0 |
| 7 | + |
| 8 | +# for i in nums: |
| 9 | +# numCount = nums.count(i) |
| 10 | +# if numCount > length and numCount > majority: |
| 11 | +# majority = numCount |
| 12 | +# element = i |
| 13 | +# return element |
| 14 | + |
| 15 | +# Best solution |
| 16 | + # nums.sort() |
| 17 | + # element = nums[length] |
| 18 | + # return element |
| 19 | + |
| 20 | +# if not nums: |
| 21 | +# return None |
| 22 | + |
| 23 | +# Boyce-Moore algorithm |
| 24 | + count = 0 |
| 25 | + candidate = None |
| 26 | + |
| 27 | + for num in nums: |
| 28 | + if count == 0: |
| 29 | + candidate = num |
| 30 | + count += (1 if num == candidate else -1) |
| 31 | + |
| 32 | + return candidate |
0 commit comments