chore: clean up content phase organization

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]