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[EGON] Week 01 Solutions #322
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from unittest import TestCase, main | ||
from typing import List | ||
from collections import Counter | ||
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class Solution: | ||
def containsDuplicate(self, nums: List[int]) -> bool: | ||
return self.solve_3(nums=nums) | ||
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""" | ||
Runtime: 412 ms (Beats 75.17%) | ||
Analyze Complexity: O(n) | ||
Memory: 31.92 MB (Beats 45.93%) | ||
""" | ||
def solve_1(self, nums: List[int]) -> bool: | ||
return len(nums) != len(set(nums)) | ||
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""" | ||
Runtime: 423 ms (Beats 39.66%) | ||
Analyze Complexity: O(n) | ||
Memory: 34.54 MB (Beats 14.97%) | ||
""" | ||
def solve_2(self, nums: List[int]) -> bool: | ||
counter = {} | ||
for num in nums: | ||
if num in counter: | ||
return True | ||
else: | ||
counter[num] = True | ||
else: | ||
return False | ||
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""" | ||
Runtime: 441 ms (Beats 16.59%) | ||
Analyze Complexity: O(n) | ||
Memory: 34.57 MB (Beats 14.97%) | ||
""" | ||
def solve_3(self, nums: List[int]) -> bool: | ||
return Counter(nums).most_common(1)[0][1] > 1 | ||
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class _LeetCodeTCs(TestCase): | ||
def test_1(self): | ||
nums = [1, 2, 3, 1] | ||
output = True | ||
self.assertEqual(Solution.containsDuplicate(Solution(), nums=nums), output) | ||
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def test_2(self): | ||
nums = [1, 2, 3, 4] | ||
output = False | ||
self.assertEqual(Solution.containsDuplicate(Solution(), nums=nums), output) | ||
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def test_3(self): | ||
nums = [1, 1, 1, 3, 3, 4, 3, 2, 4, 2] | ||
output = True | ||
self.assertEqual(Solution.containsDuplicate(Solution(), nums=nums), output) | ||
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if __name__ == '__main__': | ||
main() |
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import Foundation | ||
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class Solution { | ||
func containsDuplicate(_ nums: [Int]) -> Bool { | ||
return solve_2(nums) | ||
} | ||
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/* | ||
Runtime: 246 ms (Beats 68.44%) | ||
Analyze Complexity: O(n) | ||
Memory: 19.94 MB (Beats 76.01%) | ||
*/ | ||
func solve_1(_ nums: [Int]) -> Bool { | ||
return nums.count != Set(nums).count | ||
} | ||
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/* | ||
Runtime: 240 ms (Beats 90.56%) | ||
Analyze Complexity: O(n) | ||
Memory: 22.56 MB (Beats 33.43%) | ||
*/ | ||
func solve_2(_ nums: [Int]) -> Bool { | ||
var counter: [Int: Bool] = [:] | ||
for num in nums { | ||
if counter[num] != nil { | ||
return true | ||
} else { | ||
counter[num] = true | ||
} | ||
} | ||
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return false | ||
} | ||
} |
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@@ -0,0 +1,111 @@ | ||
from typing import Optional, List | ||
from unittest import TestCase, main | ||
from heapq import heappush, heappop | ||
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class TreeNode: | ||
def __init__(self, val=0, left=None, right=None): | ||
self.val = val | ||
self.left = left | ||
self.right = right | ||
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class Solution: | ||
def kthSmallest(self, root: Optional[TreeNode], k: int) -> int: | ||
return self.solve_2(root, k) | ||
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""" | ||
Runtime: 50 ms (Beats 25.03%) | ||
Analyze Complexity: O(n * log n), 순회에 n, heap push에 log n | ||
Memory: 19.55 MB (Beats 15.91%) | ||
""" | ||
def solve_1(self, root: Optional[TreeNode], k: int) -> int: | ||
visited = [] | ||
stack = [root] | ||
while stack: | ||
curr_node = stack.pop() | ||
heappush(visited, curr_node.val) | ||
if curr_node.left is not None: | ||
stack.append(curr_node.left) | ||
if curr_node.right is not None: | ||
stack.append(curr_node.right) | ||
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result = visited[0] | ||
for _ in range(k): | ||
result = heappop(visited) | ||
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return result | ||
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""" | ||
Runtime: 43 ms (Beats 69.91%) | ||
Analyze Complexity: O(n) | ||
Memory: 19.46 MB (Beats 60.94%) | ||
""" | ||
def solve_2(self, root: Optional[TreeNode], k: int) -> int: | ||
vals = [] | ||
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def inorder_traverse(root: Optional[TreeNode]): | ||
if root is None: | ||
return | ||
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if len(vals) >= k: | ||
return | ||
Comment on lines
+51
to
+52
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 이 조건으로 빠르게 탐색을 마칠 수 있어 정말 좋네요! 👍 |
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inorder_traverse(root.left) | ||
vals.append(root.val) | ||
inorder_traverse(root.right) | ||
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inorder_traverse(root) | ||
return vals[k - 1] | ||
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class _LeetCodeTCs(TestCase): | ||
def test_1(self): | ||
root = TreeNode( | ||
val=3, | ||
left=TreeNode( | ||
val=1, | ||
left=None, | ||
right=TreeNode( | ||
val=2, | ||
left=None, | ||
right=None, | ||
) | ||
), | ||
right=TreeNode( | ||
val=4, | ||
left=None, | ||
right=None | ||
) | ||
) | ||
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k = 1 | ||
output = 1 | ||
self.assertEqual(Solution.kthSmallest(Solution(), root, k), output) | ||
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def test_2(self): | ||
root = TreeNode( | ||
val=5, | ||
left=TreeNode( | ||
val=3, | ||
left=TreeNode( | ||
val=2, | ||
left=TreeNode( | ||
val=1 | ||
) | ||
), | ||
right=TreeNode( | ||
val=4 | ||
) | ||
), | ||
right=TreeNode( | ||
val=6 | ||
) | ||
) | ||
k = 3 | ||
output = [3] | ||
self.assertEqual(Solution.kthSmallest(Solution(), root, k), output) | ||
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if __name__ == '__main__': | ||
main() |
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Original file line number | Diff line number | Diff line change |
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@@ -0,0 +1,41 @@ | ||
import Foundation | ||
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public class TreeNode { | ||
public var val: Int | ||
public var left: TreeNode? | ||
public var right: TreeNode? | ||
public init() { self.val = 0; self.left = nil; self.right = nil; } | ||
public init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; } | ||
public init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) { | ||
self.val = val | ||
self.left = left | ||
self.right = right | ||
} | ||
} | ||
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class Solution { | ||
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/* | ||
Runtime: 28 ms (Beats 46.95%) | ||
Analyze Complexity: O(n) | ||
Memory: 16.52 MB (Beats 53.05%) | ||
*/ | ||
func kthSmallest(_ root: TreeNode?, _ k: Int) -> Int { | ||
func inorderTraverse(_ root: TreeNode?, _ k: Int) { | ||
guard let root = root, visited.count < k else { | ||
return | ||
} | ||
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inorderTraverse(root.left, k) | ||
if visited.count < k { | ||
visited.append(root.val) | ||
} | ||
inorderTraverse(root.right, k) | ||
} | ||
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var visited: [Int] = [] | ||
inorderTraverse(root, k) | ||
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return visited[k - 1] | ||
} | ||
} |
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@@ -0,0 +1,72 @@ | ||
from unittest import TestCase, main | ||
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class Solution: | ||
def hammingWeight(self, n: int) -> int: | ||
return self.solve_4(n=n) | ||
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""" | ||
Runtime: 26 ms (Beats 97.13%) | ||
Analyze Complexity: O(log n), bin(int)가 O(log(n)) | ||
Memory: 16.56 MB (Beats 22.67%) | ||
""" | ||
def solve_1(self, n: int) -> int: | ||
return bin(n).count('1') | ||
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""" | ||
Runtime: 31 ms (Beats 85.00%) | ||
Analyze Complexity: O(log n) | ||
Memory: 16.60 MB (Beats 22.67%) | ||
""" | ||
def solve_2(self, n: int) -> int: | ||
hamming_weight = 0 | ||
while n: | ||
hamming_weight += n % 2 | ||
n = n >> 1 | ||
return hamming_weight | ||
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""" | ||
Runtime: 30 ms (Beats 88.73%) | ||
Analyze Complexity: O(k), k는 2진수의 1의 갯수 (== O(log(n))) | ||
Memory: 16.56 MB (Beats 22.67%) | ||
""" | ||
# Brian Kernighan's Algorithm | ||
def solve_3(self, n: int) -> int: | ||
hamming_weight = 0 | ||
while n: | ||
n &= (n - 1) | ||
hamming_weight += 1 | ||
return hamming_weight | ||
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""" | ||
Runtime: 36 ms (Beats 53.56%) | ||
Analyze Complexity: O(k), k는 2진수의 1의 갯수 (== O(log(n))) | ||
Memory: 16.55 MB (Beats 22.67%) | ||
""" | ||
def solve_4(self, n: int) -> int: | ||
hamming_weight = 0 | ||
while n: | ||
hamming_weight += n & 1 | ||
n >>= 1 | ||
return hamming_weight | ||
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class _LeetCodeTCs(TestCase): | ||
def test_1(self): | ||
n = 11 | ||
output = 3 | ||
self.assertEqual(Solution.hammingWeight(Solution(), n=n), output) | ||
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def test_2(self): | ||
n = 128 | ||
output = 1 | ||
self.assertEqual(Solution.hammingWeight(Solution(), n=n), output) | ||
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def test_3(self): | ||
n = 2147483645 | ||
output = 30 | ||
self.assertEqual(Solution.hammingWeight(Solution(), n=n), output) | ||
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if __name__ == '__main__': | ||
main() |
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Original file line number | Diff line number | Diff line change |
---|---|---|
@@ -0,0 +1,32 @@ | ||
import Foundation | ||
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class Solution { | ||
func hammingWeight(_ n: Int) -> Int { | ||
return solve_2(n) | ||
} | ||
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/* | ||
Runtime: 5 ms (Beats 38.73%) | ||
Analyze Complexity: O(k), k는 2진수의 1의 갯수 | ||
Memory: 14.88 MB (Beats 98.27%) | ||
*/ | ||
func solve_1(_ n: Int) -> Int { | ||
return n.nonzeroBitCount | ||
} | ||
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/* | ||
Runtime: 3 ms (Beats 65.90%) | ||
Analyze Complexity: O(k), k는 2진수의 1의 갯수 | ||
Memory: 15.08 MB (Beats 89.02%) | ||
*/ | ||
func solve_2(_ n: Int) -> Int { | ||
var num = n | ||
var hammingWeight = 0 | ||
while 0 < num { | ||
num &= num - 1 | ||
hammingWeight += 1 | ||
} | ||
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return hammingWeight | ||
} | ||
} |
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다양한 풀이로 하셔서 좋네요 :) 요부분 조금 궁금한게 있는데, bst의 heappush의 Time complexity가 제 기억엔 logn 이었던거로 기억하는데, 혹시 어떻게 n으로 도출하셨는지 알려주실수 있을까요?
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n개의 노드를 순회하며 각각 heap push를 하니 말씀하신 것 처럼 n * log n 이 맞습니다. 주석 복붙하다 빼먹은 것 같네요. 리뷰 감사합니다.