DP-2 complete

Problem1.py

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+'''
+The problem is same as house robber. We just have 3 choices to make.
+Time complexity: O(n)
+Space complexity: O(1)
+'''
+
+class Solution:
+    def paint(self, costs: List[int][int]) -> int:
+        dp = [math.inf, math.inf, math.inf]
+
+        for cost in costs:
+            chooseRed = cost[0] + min(dp[2], dp[1])
+            chooseGreen = cost[1] + min(dp[0], dp[2])
+            chooseBlue = cost[2] + min(dp[0], dp[1])
+
+            dp = [chooseRed, chooseGreen, chooseBlue]
+
+        return max(dp)

Problem2.py

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+'''
+The brute force is to make a tree of choosing a coin or not and keep traversing
+until the amount is reached. The branches that reach the amount are the total
+number of ways to reach the amount.
+Space complexity: O(amount/min(coins))
+Time complexity: O(2^(amount/min(coins)))
+
+This produces repeated nodes and hence dp can be used.
+
+The number of ways to reach 3 using coins[1,2] is
+1. 1+1+1
+2. 1+2
+
+Which can be also seen as:
+1. Number of ways to reach 3 using 1s
+2. Number of ways to reach 3 using 2s and 1s.
+    - This can be further broken down into:
+        1) Number of ways to reach 1 using 1s and 2s and then using a coin of 2
+        This keeps the number of ways to reach 3 same as number of ways to reach 1.
+
+We formulate this into a table and iterate over the whole table to find the final number
+of ways to reach the amount using all the given coins.
+
+Time complexity: O(amount * len(coins))
+Space complexity: O(amount * len(coins))
+'''
+
+class Solution:
+    def change(self, amount: int, coins: List[int]) -> int:
+        dp = [ 0 for _ in range(amount+1)]
+        dp[0] = 1
+
+        for coin in coins:
+            for total in range(amount+1):
+                if total >= coin:
+                    dp[total] = dp[total] + dp[total-coin]
+
+        return dp[-1]