DP-2 complete

256. Paint House.py

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+# EXHAUSTIVE SOLUTION
+
+# Time: O(2^n) n=> number of houses
+# Space: O(n) As the recursion depth is only till the length of costs i.e the number of rows which determines the houses
+
+# Loop through the number of colors
+# check in the helper function which color we use and use the remaining colors
+# Check whats the cost of minimum between the two colors and add the current cost in it
+class Solution:
+    def minCost(self, costs: List[List[int]]) -> int:
+        min_val = float('inf')
+        for n in range(len(costs[0])):
+            min_val = min(min_val, self.helper(costs, 0, n))
+        return min_val
+
+    def helper(self, costs, idx, n):
+        # base
+        if idx == len(costs):
+            return 0
+
+        # Logic
+        if n == 0:
+            choose_blue = self.helper(costs, idx + 1, n + 1)
+            choose_green = self.helper(costs, idx + 1, n + 2)
+
+            return costs[idx][n] + min(choose_blue, choose_green)
+        elif n == 1:
+            choose_red = self.helper(costs, idx + 1, n - 1)
+            choose_green = self.helper(costs, idx + 1, n + 1)
+
+            return costs[idx][n] + min(choose_red, choose_green)
+        else:
+            choose_red = self.helper(costs, idx + 1, n - 2)
+            choose_blue = self.helper(costs, idx + 1, n - 1)
+
+            return costs[idx][n] + min(choose_red, choose_blue)
+
+
+
+# MEMOISATION SOLUTION
+# Time: O(3n) => which is O(n) because we have 3 colors and n number of houses
+# Space: O(3n) => which is O(n) we crate a matrix of n * number of colors
+
+# Just put everything in a memo matrix and use it to optimise on time
+
+class Solution:
+    def minCost(self, costs: List[List[int]]) -> int:
+        memo = [[-1 for _ in range(3)] for _ in range(len(costs))]
+        for m in range(len(costs) -1, -1, -1 ):
+                if m == len(costs) - 1:
+                    memo[m][0] = costs[m][0]
+                    memo[m][1] = costs[m][1]
+                    memo[m][2] = costs[m][2]
+                else:
+                    memo[m][0] = costs[m][0] + min(memo[m+1][1], memo[m+1][2])
+                    memo[m][1] = costs[m][1] + min(memo[m+1][0], memo[m+1][2])
+                    memo[m][2] = costs[m][2] + min(memo[m+1][0], memo[m+1][1])
+
+        return min(memo[0][0], memo[0][1],memo[0][2])

518. Coin Change II.py

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+# Time: O(2^m+n) => m is the amount and n is the no. of elements in coins array
+# The time complexity is when we choose until amount is 1 and then we don't choose until the idx is not out of range
+# Space: O(m+n) => That's the recursive space that will be used at the end
+
+# Everything is similar to coin change except returning choose + not_choose because we have to count all the possible paths
+# Hence we have also changes the base cases accordingly so if amount is 0 it's valid and returns 1 and rest is invalid path which returns 0
+class Solution:
+    def change(self, amount: int, coins: List[int]) -> int:
+        return self.helper(coins, amount, 0)
+    def helper(self,coins,amount,idx):
+        # Base case
+        if amount == 0:
+            return 1
+        if idx > (len(coins) - 1) or amount < 0:
+            return 0
+
+        # choose
+        choose = self.helper(coins,amount-coins[idx],idx)
+        # Not choose
+        not_choose = self.helper(coins,amount,idx+1)
+
+        return choose + not_choose
+
+# MEMOIZATION SOLUTION
+# Time and Space: O(m * n) where m is the len(coins) and n is the amount
+
+# Just add the values calculate (choose + not_choose) in a memo matrix and use the same amount for repeated sub problems
+# Add the result in the memo matrix i.e the additioin of choose and not_choose
+# return self.memo[idx][amount] from the function 
+class Solution:
+    def change(self, amount: int, coins: List[int]) -> int:
+        self.memo = [[-1 for _ in range(amount + 1)] for _ in range(len(coins) +1)] 
+
+        return self.helper(coins, amount, 0)
+    def helper(self,coins,amount,idx):
+        # Base case
+        if amount == 0:
+            return 1
+        if idx > (len(coins) - 1) or amount < 0:
+            return 0
+        if self.memo[idx][amount] != -1:
+            return self.memo[idx][amount]
+
+        # choose
+        choose = self.helper(coins,amount-coins[idx],idx)
+        # Not choose
+        not_choose = self.helper(coins,amount,idx+1)
+
+        self.memo[idx][amount] = choose + not_choose
+        return self.memo[idx][amount]
+
+
+# TABULARISATION SOLUTION
+# Time and Space: O(m * n) where m is the len(coins) and n is the amount
+class Solution:
+    def change(self, amount: int, coins: List[int]) -> int:
+        rows = len(coins)
+        cols = amount
+        dp_matrix = [[0 for _ in range(cols + 1)] for _ in range(rows + 1)]
+
+        for i in range(1,rows + 1):
+            for j in range(0,cols + 1):
+                if j == 0:
+                    dp_matrix[i][j] = 1
+                elif j < coins[i-1]:
+                    dp_matrix[i][j] = dp_matrix[i - 1][j]
+                else:
+                    dp_matrix[i][j] = dp_matrix[i - 1][j] + dp_matrix[i][j-coins[i - 1]]
+
+        return dp_matrix[rows][cols]