chore: clean up content phase organization

work/1/Medium/Array/198.house-robber.py

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+"""
+198. House Robber
+Difficulty: Medium
+https://leetcode.com/problems/house-robber/
+
+──────────────────────────────────────────────────
+
+You are a professional robber planning to rob houses along a street.
+Each house has a certain amount of money stashed, the only constraint
+stopping you from robbing each of them is that adjacent houses have
+security systems connected and it will automatically contact the
+police if two adjacent houses were broken into on the same night.
+
+Given an integer array nums representing the amount of money of each
+house, return the maximum amount of money you can rob tonight without
+alerting the police.
+
+
+
+Example 1:
+
+Input: nums = [1,2,3,1]
+Output: 4
+Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
+Total amount you can rob = 1 + 3 = 4.
+
+Example 2:
+
+Input: nums = [2,7,9,3,1]
+Output: 12
+Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob
+house 5 (money = 1).
+Total amount you can rob = 2 + 9 + 1 = 12.
+
+
+
+Constraints:
+
+        • 1 <= nums.length <= 100
+
+        • 0 <= nums[i] <= 400
+"""
+
+
+class Solution:
+    def rob(self, nums: list[int]) -> int:
+        n = len(nums)
+        if n == 0:
+            return 0
+        if n <= 2:
+            return max(nums)
+
+        dp = [0] * n
+        dp[0] = nums[0]
+        dp[1] = max(nums[0], nums[1])
+        for i in range(2, n):
+            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
+        return dp[-1]

work/1/Medium/Array/322.coin-change.py

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+"""
+322. Coin Change
+Difficulty: Medium
+https://leetcode.com/problems/coin-change/
+
+──────────────────────────────────────────────────
+
+You are given an integer array coins representing coins of different
+denominations and an integer amount representing a total amount of
+money.
+
+Return the fewest number of coins that you need to make up that
+amount. If that amount of money cannot be made up by any combination
+of the coins, return -1.
+
+You may assume that you have an infinite number of each kind of coin.
+
+
+
+Example 1:
+
+Input: coins = [1,2,5], amount = 11
+Output: 3
+Explanation: 11 = 5 + 5 + 1
+
+Example 2:
+
+Input: coins = [2], amount = 3
+Output: -1
+
+Example 3:
+
+Input: coins = [1], amount = 0
+Output: 0
+
+
+
+Constraints:
+
+        • 1 <= coins.length <= 12
+
+        • 1 <= coins[i] <= 2^31 - 1
+
+        • 0 <= amount <= 10^4
+"""
+
+from math import inf
+
+
+class Solution:
+    def coinChange(self, coins: list[int], amount: int) -> int:
+        n = len(coins)
+        memo = [[-1] * (amount + 1) for _ in range(n + 1)]
+
+        def min_coins(coins, i, amount, memo):
+            if amount == 0:
+                return 0
+            if i == 0 or amount < 0:
+                return inf
+            if memo[i][amount] != -1:
+                return memo[i][amount]
+
+            skip = min_coins(coins, i - 1, amount, memo)
+            use = min_coins(coins, i, amount - coins[i - 1], memo)
+            if use != inf:
+                use += 1
+
+            memo[i][amount] = min(skip, use)
+            return memo[i][amount]
+
+        result = min_coins(coins, n, amount, memo)
+        return result if result != inf else -1  # pyright: ignore[reportReturnType]