[wozlsla] WEEK 02 solutions

3sum/wozlsla.py

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+"""
+# Intuition
+-
+
+# Approach
+index가 모두 다를것,
+합 = 0 -> -a = b+c
+중복 X -> set : dic와 다르게 가변객체 삽입 X
+Two Pointer -> 정렬된 배열을 활용
+
+# Complexity
+- Time complexity
+  - Brute-force : O(N^3)
+  - Set : O(N^2) - Time Limit Exceeded (memory??)
+  - Two Pointer : O(N^2)
+
+- Space complexity
+  - Brute-force : O(N)
+  - Set : O(N)
+  - Two Pointer : O(1)
+"""
+
+from typing import List
+
+
+class Solution:
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+
+        triplets = set()
+        nums.sort()  # O(NlogN), return None
+
+        for i in range(len(nums) - 2):  # O(N)
+            low, high = i + 1, len(nums) - 1
+
+            while low < high:  # O(N)
+                three_sum = nums[i] + nums[low] + nums[high]
+
+                if three_sum < 0:
+                    low += 1
+                elif three_sum > 0:
+                    high -= 1
+                else:
+                    triplets.add((nums[i], nums[low], nums[high]))
+                    low, high = low + 1, high - 1
+
+        return list(triplets)
+
+
+""" Set
+class Solution:
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+
+        triplets = set()
+
+        for i in range(len(nums) - 2):
+            seen = set()
+
+            for j in range(i + 1, len(nums)):
+                complement = -(nums[i] + nums[j])
+
+                if complement in seen:
+                    triplets.add(tuple(sorted([nums[i], nums[j], complement])))
+
+                seen.add(nums[j])
+
+        return list(triplets)
+"""
+
+""" Brute-force
+class Solution:
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+
+        n = len(nums)
+        result = []
+
+        for i in range(n-2):
+            for j in range(i+1, n-1):
+                for k in range(j+1, n):
+                    if i != j and i != k and j != k:
+
+                        if nums[i] + nums[j] + nums[k] == 0:
+                            li = sorted([nums[i], nums[j], nums[k]])
+
+                            if li not in result: # O(L)
+                                result.append(li)
+
+        return result
+"""
+
+nums = [-1, 0, 1, 2, -1, -4]
+sol = Solution()
+
+print(sol.threeSum(nums))
+# [[-1,-1,2],[-1,0,1]]
+
+# print(len({1, 1, 2})) # 2

climbing-stairs/wozlsla.py

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+"""
+# Intuition
+계단 정상에 오를 수 있는 방법의 수 - 1,2 - 순열? -> X. 자리수가 일정하지 않음
+
+# Approach
+접근 1) 1 or 2 로만 이동 가능
+
+풀이 참고
+- 계단을 한 번에 최대 2칸 밖에 올라갈 수 없으므로, 3번째 칸에 발을 딛기 위해서는 바로 아래 칸인 2번째 칸이나 적어도 1번째 칸에 반드시 먼저 올라와있어야 함.
+- 즉, n 칸에 발을 딛기위해서는 그 전에 n - 1 칸이나 n - 2 칸까지 올라와왔어야 한다.
+
+접근 2)
+4 -> 5
+1 1 1 1
+2 1 1
+1 2 1
+1 1 2
+2 2
+
+5 -> 8
+1 1 1 1 1
+2 2 1
+1 2 2
+2 1 2
+1 1 1 2
+1 1 2 1
+1 2 1 1
+2 1 1 1
+
+n=5까지만 봤을 때 늘어나는 규칙이 피보나치수열과 같음. (hint)
+
+# Complexity
+- Time complexity
+  - DP 1 : O(N)
+  - DP 2 : O(N)
+
+  - Recursive 1 : O(2^N)
+  - Recursive 2 (caching): O(N)
+
+- Space complexity
+  - DP 1 : O(N)
+  - DP 2 : O(1)
+
+  - Recursive 1 : O(N)
+  - Recursive 2 (caching): O(N)
+
+"""
+
+
+# DP 2 (공간 최적화)
+class Solution:
+    def climbStairs(self, n: int) -> int:
+
+        if n < 3:
+            return n
+
+        pre, cur = 1, 2
+
+        for _ in range(n - 2):
+            pre, cur = cur, pre + cur  # 순서 !
+
+        return cur
+
+
+""" DP 1
+class Solution:
+    def climbStairs(self, n: int) -> int:
+
+        dp = {1:1, 2:2}
+
+        for i in range(3, n+1):
+            dp[i] = dp[i-1] + dp[i-2]
+
+        return dp[n]
+"""
+
+""" Memoization (재귀 + 캐싱)
+class Solution:
+    def climbStairs(self, n: int) -> int:
+        memo = {}   
+
+        def _climb(n):
+
+            if n not in memo:
+                if n < 3:
+                    memo[n] = n
+                else:
+                    memo[n] = _climb(n - 1) + _climb(n - 2)
+
+            return memo[n]
+
+        return _climb(n)
+"""
+
+
+""" Recursive
+class Solution:
+    def climbStairs(self, n: int) -> int:
+
+        if n < 3:
+            return n
+
+        return self.climbStairs(n-1) + self.climbStairs(n-2)
+"""

house-robber/wozlsla.py

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+"""
+# Intuition
+<!-- Describe your first thoughts on how to solve this problem. -->
+
+# Approach
+<!-- Describe your approach to solving the problem. -->
+
+# Complexity
+- Time complexity:
+
+- Space complexity:
+
+"""
+
+
+# DP
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+
+        # initialize dp array
+        dp = [0] * (len(nums) + 1)
+
+        if nums:
+            dp[1] = nums[0]
+
+        for n in range(2, len(dp)):
+            # current house : nums[n-1]
+            rob_current = nums[n - 1] + dp[n - 2]
+            skip_current = dp[n - 1]
+
+            dp[n] = max(rob_current, skip_current)
+
+        return dp[-1]
+
+
+""" Recursive + Memoization
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+
+        memo = {}
+
+        def dfs(start):
+            if start in memo:
+                return memo[start]
+            
+            if start >= len(nums):
+                memo[start] = 0
+            else:
+                memo[start] = max(nums[start] + dfs(start + 2), dfs(start + 1))
+
+            return memo[start]
+
+        return dfs(0)
+"""
+
+""" Recursive
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+
+        def dfs(start):
+            if start >= nums(len):
+                return 0
+
+            return max(nums[start] + dfs[start+2], dfs[start+1])
+
+        return dfs(0)
+"""

product-of-array-except-self/wozlsla.py

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+"""
+# Intuition
+하나씩 순회하면서 나머지 값들을 곱함
+
+# Approach
+접근 1) 원형 큐를 사용하면?
+현재 제외하려는 요소를 popleft()로 제거하고, 나머지 요소들의 곱을 계산한 뒤, 다시 그 요소를 append()하여 다음 반복을 위해 덱의 순서를 회전
+
+접근 2)
+1 b c d --> 1 x bcd
+a 1 c d --> a x  cd
+a b 1 d --> ab x  d
+a b c 1 --> abc x 1
+
+
+# Complexity
+- Time complexity : O(N)
+- Space complexity : O(N) / O(1)
+"""
+
+from typing import List
+
+
+class Solution:
+    def productExceptSelf(self, nums: List[int]) -> List[int]:
+
+        n = len(nums)
+        res = [1] * n
+
+        prefix_product = 1
+        for i in range(n):
+            res[i] = prefix_product
+            prefix_product *= nums[i]
+
+        suffix_product = 1
+        for i in range(n - 1, -1, -1):
+            res[i] *= suffix_product
+            suffix_product *= nums[i]
+
+
+"""
+class Solution:
+    def productExceptSelf(self, nums: List[int]) -> List[int]:
+
+        # 중복 계산을 피하는 방법?
+        res_1 = [1] * len(nums) 
+        res_2 = [1] * len(nums)
+
+        for n in range(1, len(nums)):  # [1, 1(a), 2(ab), 6(abc)]
+            res_1[n] = res_1[n - 1] * nums[n - 1]
+
+        # reverse X -> range를 반대로
+        for n in range(len(nums) - 1, -1, -1):  # [1, 4(d), 12(cd), 24(bcd)]
+            res_2[n] = res_2[n + 1] * nums[n + 1]
+
+        res_2.reverse()
+        return [res_1[i] * res_2[i] for i in range(len(nums))]
+
+
+sol = Solution()
+print(sol.productExceptSelf([1, 2, 3, 4]))
+"""
+
+
+"""
+# deque : O(N^2)
+from collections import deque
+import math
+
+class Solution:
+    def productExceptSelf(self, nums: List[int]) -> List[int]:
+
+        nums = deque(nums)
+        result = []
+
+        for _ in range(len(nums)): # O(N)
+            tmp = nums.popleft()
+            result.append(math.prod(nums)) # O(N)
+            nums.append(tmp)
+        
+        return result
+"""