Implement Paint House and Coin Change

CoinChange2.java

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+// Time Complexity : O(n * amount)
+// Space Complexity : O(amount)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : Needed to avoid counting duplicate permutations, so coins are processed one by one.
+
+class CoinChange2 {
+    public int change(int amount, int[] coins) {
+        int[] dp = new int[amount + 1];
+
+        // Base case: There is 1 way to make amount 0
+        dp[0] = 1;
+
+        // Process each coin
+        for (int i = 1; i <= coins.length; i++) {
+            int coin = coins[i - 1];
+
+            // Update ways for all amounts from coin to target amount
+            for (int j = coin; j <= amount; j++) {
+                dp[j] += dp[j - coin];
+            }
+        }
+
+        return dp[amount];
+    }
+}

PaintHouse.java

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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 : Needed to store old dp values before updating them.
+
+
+class PaintHouse{
+    public int minCost(int[][] costs) {
+
+        int n = costs.length;
+
+        // dp[0], dp[1], dp[2] store min cost if current house is painted
+        // red, blue, or green respectively
+        int dp [] = new int[3];
+
+        // Start from the last house
+        dp[0] = costs[n-1][0];
+        dp[1] = costs[n-1][1];
+        dp[2] = costs[n-1][2];
+
+         // Move from second last house to first house
+        for(int i = n-2; i >= 0; i++){
+            int tempR = dp[0];
+            dp[0] = costs[i][0] + Math.min(dp[1], dp[2]);
+            int tempB = dp[1];
+            dp[1] = costs[i][1] + Math.min(tempR, dp[2]);
+            dp[2] = costs[i][2] + Math.min(tempR, tempB);
+        }
+
+
+        return Math.min(dp[0], Math.min(dp[2], dp[1]));
+
+
+    }
+}