From 3d9d6e07feadac47aaa8b0c81f810c5f0ffad2eb Mon Sep 17 00:00:00 2001
From: Mayank Padshala <mayankpadshala@gmail.com>
Date: Sat, 29 Nov 2025 22:40:39 +0000
Subject: [PATCH] Complete DP-2

---
 coin_change_two.py | 48 ++++++++++++++++++++++++++++++++++++++++++++++
 paint_house.py     | 45 +++++++++++++++++++++++++++++++++++++++++++
 2 files changed, 93 insertions(+)
 create mode 100644 coin_change_two.py
 create mode 100644 paint_house.py

diff --git a/coin_change_two.py b/coin_change_two.py
new file mode 100644
index 00000000..2174b603
--- /dev/null
+++ b/coin_change_two.py
@@ -0,0 +1,48 @@
+class Solution:
+    """
+    Time Complexity: O(m * n) 
+        m = number of coins, n = amount
+        We fill an m x n DP table.
+
+    Space Complexity: O(m * n)
+        Using a 2D DP array dp[m+1][n+1].
+
+    Approaches:
+    1. Recursion + Memoization:
+       Try using or skipping each coin and recursively compute ways 
+       to reach the remaining amount. Cache results to avoid recomputation.
+
+    2. Bottom-up DP (2D Tabulation):
+       Create a DP table where dp[i][j] represents the number of ways 
+       to make amount j using the first i coins.
+       For each coin, we decide:
+         - Not using the coin (dp[i-1][j])
+         - Using the coin (dp[i][j - coin value])
+
+    3. Bottom-up DP (Optimized to 1D array):
+       Iterate coins first, then amount—which ensures combinations 
+       (no permutation counting) and reduces space to O(n).
+    """
+    def change(self, amount: int, coins: List[int]) -> int:
+        m = len(coins)
+        n = amount
+
+        # dp[i][j] = number of ways to form amount j using first i coins
+        dp = [[0] * (n + 1) for _ in range(m + 1)]
+
+        # Base case: One way to form amount 0 (choose nothing)
+        dp[0][0] = 1
+
+        for i in range(1, m + 1):
+            for j in range(n + 1):
+                # If current amount is less than the coin value,
+                # we cannot use this coin
+                if j < coins[i - 1]:
+                    dp[i][j] = dp[i - 1][j]
+                else:
+                    # Two choices:
+                    # 1. Don't use the coin -> dp[i - 1][j]
+                    # 2. Use the coin -> dp[i][j - coin_value]
+                    dp[i][j] = dp[i - 1][j] + dp[i][j - coins[i - 1]]
+
+        return dp[m][n]
diff --git a/paint_house.py b/paint_house.py
new file mode 100644
index 00000000..43f410c6
--- /dev/null
+++ b/paint_house.py
@@ -0,0 +1,45 @@
+class Solution:
+    """
+    Time Complexity: O(n) -> We process each house once.
+    Space Complexity: O(1) -> We only store costs for 3 colors per step.
+
+    Approaches:
+    1. Top-down DP (Recursion + Memoization):
+       Recursively explore all ways to paint each house while ensuring no two adjacent
+       houses share the same color. Cache results of subproblems to avoid recomputation.
+
+    2. Bottom-up DP (Tabulation with DP Table):
+       Build a dp[n][3] table where each row stores the minimum cost to paint up to
+       that house with each color. Each entry depends on the minimum of the other two
+       colors from the previous house.
+
+    3. Bottom-up DP with Space Optimization (Tabulation -> O(1) Space):
+       Instead of a full dp table, keep only the previous house's three minimum costs.
+       Update these values iteratively to compute the final minimum cost.
+    """
+    def minCost(self, costs: List[List[int]]) -> int:
+        n = len(costs)
+        
+        # If there are no houses, total cost is 0
+        if n == 0:
+            return 0
+        
+        # Initialize minimum costs for painting the first house
+        chooseRedPrev = costs[0][0]
+        chooseBluePrev = costs[0][1]
+        chooseGreenPrev = costs[0][2]
+
+        # Iterate through each subsequent house
+        for i in range(1, n):
+            # Compute the new costs if we paint this house each color
+            chooseRed = costs[i][0] + min(chooseBluePrev, chooseGreenPrev)
+            chooseBlue = costs[i][1] + min(chooseRedPrev, chooseGreenPrev)
+            chooseGreen = costs[i][2] + min(chooseRedPrev, chooseBluePrev)
+            
+            # Update previous values for the next iteration
+            chooseRedPrev, chooseBluePrev, chooseGreenPrev = (
+                chooseRed, chooseBlue, chooseGreen
+            )
+
+        # The answer is the minimum of the last computed color choices
+        return min(chooseRedPrev, chooseBluePrev, chooseGreenPrev)