Merge pull request #378 from lymchgmk/feat/week3

[EGON] Week 03 Solutions

Author: EgonD3V

coin-change/EGON.py

@@ -1,38 +1,48 @@
 from typing import List
 from unittest import TestCase, main
-from collections import defaultdict
 
 
 class Solution:
     def coinChange(self, coins: List[int], amount: int) -> int:
-        return self.solve_with_dp(coins, amount)
+        return self.solveWithDP(coins, amount)
 
     # Unbounded Knapsack Problem
-    def solve_with_dp(self, coins: List[int], amount: int) -> int:
+    """
+    Runtime: 801 ms (Beats 48.54%)
+    Time Complexity: O(n)
+        - coins 길이를 c, amount의 크기를 a라고 하면
+        - coins를 정렬하는데 O(log c)
+        - dp 배열 조회에 O((n + 1) * c)
+        > c의 최대 크기는 12라서 무시가능하므로 O((n + 1) * c) ~= O(n * c) ~= O(n) 
+
+    Memory: 16.94 MB (Beats 50.74%)
+    Space Complexity: O(n)
+        > 크기가 n + 1인 dp를 선언하여 사용했으므로 O(n + 1) ~= O(n)
+    """
+    def solveWithDP(self, coins: List[int], amount: int) -> int:
         if amount == 0:
             return 0
 
         coins.sort()
 
         if amount < coins[0]:
             return -1
+          
+        dp = [float('inf')] * (amount + 1)
 
-        dp = [[0] * (amount + 1) for _ in range(len(coins) + 1)]
-        for curr_r in range(1, len(coins) + 1):
-            coin_index = curr_r - 1
-            curr_coin = coins[coin_index]
-            if amount < curr_coin:
-                continue
+        for coin in coins:
+            if coin <= amount:
+                dp[coin] = 1
 
-            dp[curr_r][curr_coin] += 1
-            for curr_amount in range(curr_coin + 1, amount + 1):
-                for coin in coins:
-                    if 0 < dp[curr_r][curr_amount - coin]:
-                        dp[curr_r][curr_amount] = max(dp[curr_r - 1][curr_amount], dp[curr_r][curr_amount - coin] + 1)
-                    else:
-                        dp[curr_r][curr_amount] = dp[curr_r - 1][curr_amount]
+        for curr_amount in range(amount + 1):
+            for coin in coins:
+                if 0 <= curr_amount - coin:
+                    dp[curr_amount] = min(
+                        dp[curr_amount],
+                        dp[curr_amount - coin] + 1
+                    )
 
-        return dp[-1][-1] if 0 < dp[-1][-1] else -1
+        return dp[-1] if dp[-1] != float('inf') else -1
 
 
 class _LeetCodeTestCases(TestCase):
@@ -57,7 +67,7 @@ def test_3(self):
     def test_4(self):
         coins = [1, 2147483647]
         amount = 2
-        output = -1
+        output = 2
         self.assertEqual(Solution.coinChange(Solution(), coins, amount), output)
 
 

combination-sum/EGON.py

@@ -5,16 +5,23 @@
 
 class Solution:
     def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
-        return self.solve_with_dfs(candidates, target)
+        return self.solveWithBackTracking(candidates, target)
 
     """
     Runtime: 2039 ms (Beats 5.01%)
-    Time Complexity: ?
-
+    Time Complexity: O(c * c * log c)
+        - 처음 stack의 크기는 c에 비례 O(c)
+        - 중복 제거에 사용하는 변수인 curr_visited_checker 생성에 O(c' log c')
+        - stack의 내부 로직에서 c에 비례한 for문을 순회하는데 O(c)
+        > O(c) * O(c' log c') + O(c) * O(c) ~= O(c * c * log c)
+    
     Memory: 16.81 MB (Beats 11.09%)
-    Space Complexity: ?
+    Space Complexity: O(c * c)
+        - curr_combination의 크기가 c에 비례  
+        - stack의 크기는 curr_combination의 크기와 c에 비례
+        > O(c * c)
     """
-    def solve_with_dfs(self, candidates: List[int], target: int) -> List[List[int]]:
+    def solveWithDFS(self, candidates: List[int], target: int) -> List[List[int]]:
         result = []
         stack = []
         visited = defaultdict(bool)
@@ -38,6 +45,38 @@ def solve_with_dfs(self, candidates: List[int], target: int) -> List[List[int]]:
 
         return result
 
+    """
+    Runtime: 58 ms (Beats 32.30%)
+    Time Complexity: O(c * c)
+        - candidates 정렬에 O(log c)
+        - 첫 depte에서 dfs 함수 호출에 O(c)
+        - 그 후 candidates의 길이에 비례해서 재귀적으로 dfs를 호출하는데 O(c)
+            - lower_bound_idx에 따라 range가 감소하기는 하나 일단은 비례 O(c')
+        > O(log c) + O(c * c') ~= O(c * c), 단 c' <= c 이므로 이 복잡도는 upper bound
+    Memory: 16.59 MB (Beats 75.00%)
+    Space Complexity: O(c)
+        - result를 제외하고 모두 nonlocal 변수를 call by reference로 참조
+        - dfs 함수 호출마다 메모리가 증가하는데, 호출횟수는 candidates의 길이에 비례 O(c)
+            - lower_bound_idx에 따라 range가 감소하기는 하나 일단은 비례
+        > O(c), 단 이 복잡도는 upper bound
+    """
+    def solveWithBackTracking(self, candidates: List[int], target: int) -> List[List[int]]:
+        def dfs(stack: List[int], sum: int, lower_bound_idx: int):
+            nonlocal result, candidates, target
+
+            if target < sum:
+                return
+            elif sum < target:
+                for idx in range(lower_bound_idx, len(candidates)):
+                    dfs(stack + [candidates[idx]], sum + candidates[idx], idx)
+            else:  # target == sum
+                result.append(stack)
+                return
+
+        result = []
+        candidates.sort()
+        dfs([], 0, 0)
+        return result
 
 class _LeetCodeTestCases(TestCase):
     def test_1(self):

two-sum/EGON.py

@@ -10,10 +10,11 @@ def twoSum(self, nums: List[int], target: int) -> List[int]:
     """
     Runtime: 3762 ms (Beats 5.00%)
     Time Complexity: O(n ** 2)
-        - 크기가 n인 nums 배열을 2중으로 조회하므로 O(n ** 2)
+        > 크기가 n인 nums 배열을 2중으로 조회하므로 O(n ** 2)
 
     Memory: 17.42 MB (Beats 61.58%)
-    Space Complexity: ?
+    Space Complexity: O(1)
+        > 딱히 저장하는 변수 없음 (반환하는 list 제외)
     """
     def solveWithBruteForce(self, nums: List[int], target: int) -> List[int]:
         for i in range(len(nums)):