Solve coin-change with python

Author: BEMELON

coin-change/bemelon.py

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+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