DP-2

coinChange2.py

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+# Time Complexity : O(n * amount) 
+# Space Complexity : O(n * amount)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach:
+# Use 2D DP Count ways to make each amount using available coins.
+# For every coin, either don’t use it or use it (can reuse many times).
+
+class Solution:
+    def change(self, amount: int, coins: List[int]) -> int:
+        dp = [[0 for i in range(amount + 1)] for j in range(len(coins) + 1)]
+
+        for i in range(len(coins) + 1):
+            dp[i][0] = 1
+
+        for i in range(1, len(coins) + 1):
+            for j in range(amount + 1):
+                if j < coins[i - 1]:
+                    dp[i][j] = dp[i - 1][j]
+                else:
+                    dp[i][j] = dp[i - 1][j] + dp[i][j - coins[i - 1]]
+
+        return dp[len(coins)][amount]

paintHouse.py

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+# Time Complexity : O(n)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach: Start from the last house. For each house, add its cost to the minimum of the other two colors.
+# Return the minimum final cost.
+
+class Solution:
+    def minCost(self, costs):
+        if not costs:
+            return 0
+        r, g, b = costs[-1]   
+        for i in range(len(costs) - 2, -1, -1):
+            nr = costs[i][0] + min(g, b)
+            ng = costs[i][1] + min(r, b)
+            nb = costs[i][2] + min(r, g)
+            r, g, b = nr, ng, nb
+
+        return min(r, g, b)