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"""
Question
Duplicate Integer
Easy
Given an integer array nums, return true if any value appears more than once in the array, otherwise return false.
Example 1:
Input: nums = [1, 2, 3, 3]
Output: true
Example 2:
Input: nums = [1, 2, 3, 4]
Output: false
"""
class Solution:
# 1. brute force
def brute_force(self, nums: list[int]) -> bool:
return False
# 2. sorting
def sorting(self, nums: list[int]) -> bool:
return False
# 3. hashset
def hashset(self, nums: list[int]) -> bool:
hs = set()
for i in range(len(nums)):
if nums[i] in hs:
return True
hs.add(nums[i])
return False
case1 = [1, 2, 3, 4, 5]
case2 = [2, 2, 3, 4, 5]
case3 = [3, 4, 1, 6, 3]
solution = Solution()
print(f"hashset case1: {solution.hashset(case1)}")
print(f"hashset case2: {solution.hashset(case2)}")
print(f"hashset case3: {solution.hashset(case3)}")
"""
Solution
url: https://neetcode.io/problems/duplicate-integer
video: https://youtu.be/3OamzN90kPg
1. brute force
time: O(n^2)
space: O(1)
2. sort
time: O(nlogn)
space: O(1)
3. hashset
time: O(n)
space: O(n)
code:
class Solution:
def hasDuplicate(self, nums: List[int]) -> bool:
hashset = set()
for n in nums:
if n in hashset:
return True
hashset.add(n)
return False
"""
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