Add implementation for leetcode problems 256, 518

leetcode_256.py

@@ -0,0 +1,37 @@
+#!/usr/bin/env python3
+# -*- coding: utf-8 -*-
+"""
+Created on Sun Feb 22 13:19:10 2026
+
+@author: rishigoswamy
+
+    Problem:
+    Each house can be painted Red, Green, or Blue.
+    No two adjacent houses can have the same color.
+    Return the minimum total cost.
+
+    Approach:
+    Dynamic Programming.
+    For each house:
+        cost[color] = current cost +
+                      minimum of previous house's other two colors
+
+    Time Complexity: O(n)
+    Space Complexity: O(1)
+
+"""
+
+from typing import List
+
+#https://leetcode.com/problems/paint-house/
+class Solution:
+    def minCost(self, costs: List[List[int]]) -> int:
+        currCost = costs[0]
+        val = min(costs[0])
+        for i in range(1, len(costs)):
+            costR = costs[i][0] + min(currCost[1], currCost[2])
+            costG = costs[i][1] + min(currCost[0], currCost[2])
+            costB = costs[i][2] + min(currCost[1], currCost[0])
+            val = min(costR, costG, costB)
+            currCost = [costR, costG, costB]
+        return val

leetcode_518.py

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+#!/usr/bin/env python3
+# -*- coding: utf-8 -*-
+"""
+Created on Sun Feb 22 13:21:36 2026
+
+@author: rishigoswamy
+
+
+    Problem:
+    #https://leetcode.com/problems/coin-change-ii/description/
+
+    Count number of combinations to make 'amount'
+    using unlimited supply of given coins.
+
+    Approach:
+    2D Dynamic Programming
+
+    dp[i][j] =
+        number of ways to make amount j
+        using first i coins
+
+    Time Complexity: O(n * amount)
+    Space Complexity: O(n * amount)
+
+"""
+
+
+from typing import List
+class Solution:
+    def change(self, amount: int, coins: List[int]) -> int:
+        dp = [[0] * (amount + 1) for _ in range(len(coins)+1)]
+
+        for i in range(0, len(dp)):
+            dp[i][0] = 1
+
+        for i in range(1,len(dp)):
+            for j in range(len(dp[0])):
+                currCoin = coins[i-1]
+                dp[i][j] = dp[i-1][j]
+                if j-currCoin >= 0:
+                    dp[i][j] += dp[i][j-currCoin]
+        return dp[-1][-1]