[BEMELON] WEEK 3 Solution

climbing-stairs/bemelon.py

@@ -0,0 +1,17 @@
+class Solution:
+    # Space complexity: O(1)
+    # Tiem complexity: O(n)
+    def climbStairs(self, n: int) -> int:
+        # dp[0] is n - 2
+        # dp[1] is n - 1
+        dp = [1, 2]
+
+        if n <= 2: 
+            return dp[n - 1]
+
+        for i in range(3, n + 1): 
+            # dp[n] = dp[n - 1] + dp[n - 2]
+            #       = dp[1] + dp[0]
+            dp[(i - 1) % 2] = sum(dp)
+
+        return dp[(n - 1) % 2]

coin-change/bemelon.py

@@ -0,0 +1,16 @@
+class Solution:
+    # Space complexity: O(n)
+    # Time complexity: O(n * m)
+    #  - n: amount
+    #  - m: len(coins)
+    def coinChange(self, coins: list[int], amount: int) -> int:
+        INIT_VALUE = 999999999
+        dp = [INIT_VALUE] * (amount + 1) 
+        dp[0] = 0
+
+        for x in range(1, amount + 1):
+            for coin in coins: 
+                if x - coin >= 0:
+                    dp[x] = min(dp[x], dp[x - coin] + 1)
+
+        return dp[amount] if dp[amount] != INIT_VALUE else -1

combination-sum/bemelon.py

@@ -0,0 +1,25 @@
+class Solution:
+    # Space complexity: O(n)
+    #  - n: len(candidates)
+    #  - Stack Frame -> O(n) 
+    #  - list_of_combination -> O(n) ? 
+    # Time complexity: O(n!)
+    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
+        list_of_combination = [] 
+        n = len(candidates)
+
+        def backtracking(curr: int, curr_combination: List[int], curr_sum: int): 
+            if curr_sum == target:  # 목표값에 도달했을 경우
+                list_of_combination.append(list(curr_combination))
+                return
+
+            if curr_sum > target:  # 목표값을 초과한 경우
+                return
+
+            for i in range(curr, n):
+                curr_combination.append(candidates[i])
+                backtracking(i, curr_combination, curr_sum + candidates[i])
+                curr_combination.pop()  # 백트래킹 과정에서 마지막 요소 제거
+
+        backtracking(0, [], 0)
+        return list_of_combination

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

@@ -0,0 +1,46 @@
+class Solution:
+    # Space complexity: O(n)
+    # Time complexity: O(n)
+    def naive(self, nums: list[int]) -> list[int]:
+        prefix = [1]
+        for num in nums[:-1]:
+            prefix.append(prefix[-1] * num)
+
+        reverse_nums = nums[::-1]        
+        postfix = [1]
+        for num in reverse_nums[:-1]:
+            postfix.append(postfix[-1] * num)
+        postfix = postfix[::-1]
+
+        return [prefix[i] * postfix[i] for i in range(len(nums))]
+
+    # Space complexity: O(1)
+    # Time complexity: O(n)
+    def with_constant_space(self, nums: list[int]) -> list[int]:
+        n = len(nums)
+        answer = [1] * n
+
+        # 1. save prefix product to temp 
+        temp = 1
+        for i in range(1, n):
+            temp *= nums[i - 1]
+            answer[i] *= temp 
+
+        # 2. save postfix product to temp 
+        temp = 1
+        for i in range(n - 2, -1, -1):
+            temp *= nums[i + 1]
+            answer[i] *= temp
+
+        return answer
+
+
+    def productExceptSelf(self, nums: List[int]) -> List[int]:
+        # index -> product
+        # 0 ->  - [1, 2, 3] 
+        # 1 -> [0] - [2, 3]
+        # 2 -> [0, 1] - [3] 
+        # 3 -> [0, 1, 2] -
+        return self.with_constant_space(nums)
+
+

two-sum/bemelon.py

@@ -0,0 +1,13 @@
+class Solution:
+    # Space complexity: O(n)
+    # Time complexity: O(n)
+    def twoSum(self, nums: list[int], target: int) -> list[int]:
+        num_index = {} 
+        for curr, num in enumerate(nums): 
+            rest = target - num 
+            if rest in num_index: 
+                return [num_index[rest], curr]
+            else: 
+                num_index[num] = curr 
+        return [0, 0]
+