completed DP-2

CoinChangeII.java

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+// Time Complexity : O(N)
+// Space Complexity : O(N)
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this :
+
+
+// Your code here along with comments explaining your approach
+
+class Solution {
+    public int change(int amount, int[] coins) {
+        // // 2D DP
+        // int[][] dp = new int[coins.length+1][amount+1];
+        // dp[0][0] = 1;
+        // for(int i = 1; i <= coins.length; i++)
+        // {
+        //     for(int j = 0; j <= amount; j++)
+        //     {
+        //         if(coins[i-1] > j)
+        //         {
+        //             dp[i][j] = dp[i-1][j];
+        //         }
+        //         else
+        //         {
+        //             dp[i][j] = dp[i-1][j] + dp[i][j - coins[i-1]];
+        //         }
+        //     }
+        // }
+
+        // return dp[coins.length][amount];
+
+        // 1D DP
+        int dp[] = new int[amount+1];
+        dp[0] = 1;
+        for(int i = 1; i <= coins.length; i++)
+        {
+            for(int j = 0; j <= amount; j++)
+            {
+                if(coins[i-1] > j)
+                {
+                    dp[j] = dp[j];
+                }
+                else
+                {
+                    dp[j] = dp[j] + dp[j - coins[i - 1]];
+                }
+            }
+        }
+        return dp[amount];
+    }
+}

PaintHouse.java

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+// Time Complexity : O(1)
+// Space Complexity : O(M*N)
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this :
+
+
+// Your code here along with comments explaining your approach
+
+class Solution {
+    public int minCost(int[][] costs) {
+        // 2D DP
+        int dp[][] = new int[costs.length][3];
+        dp[0][0] = costs[0][0];
+        dp[0][1] = costs[0][1];
+        dp[0][2] = costs[0][2];
+
+        for(int i = 1; i < costs.length; i ++)
+        {
+            dp[i][0] = costs[i][0] + Math.min(dp[i-1][1], dp[i-1][2]);
+            dp[i][1] = costs[i][1] + Math.min(dp[i-1][0], dp[i-1][2]);
+            dp[i][2] = costs[i][2] + Math.min(dp[i-1][0], dp[i-1][1]);
+        }
+
+        return Math.min(dp[costs.length- 1][0] , Math.min(dp[costs.length- 1][1], dp[costs.length- 1][2]));
+
+        // 1D DP
+    }
+}