DP-2 Homework

CoinChange2.java

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+// Time Complexity : O(m * n)
+// Space Complexity : O(m * n)
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : no
+
+
+// Your code here along with comments explaining your approach
+class Solution {
+    public int change(int amount, int[] coins) {
+
+        //coin change 1 we had to find min no of ways to make amt
+        //Here we have to find total num of ways to make amt
+        int m = coins.length;
+        int n = amount;
+
+        int[][] dp = new int[m + 1][n + 1];
+
+        //amount = 0 will always be reached in 1 way
+        dp[0][0] = 1;
+
+        //case 1:- dp[idx][amount] = dp[idx][amount - coins[idx]] (Choose coin)
+        //case 2:- dp[idx][amount] = dp[idx - 1][amount] (Do not choose coin)
+
+        //Traversing through coins
+        for(int i = 1; i <= m; i++) {
+
+            //Traversing through each amount
+            for(int j = 0; j <= n; j++) {
+
+                //if the coin is greater than amount then we avoid that coin
+                if(coins[i - 1] > j)
+                    dp[i][j] = dp[i - 1][j];
+
+                else {
+                    dp[i][j] = dp[i][j - coins[i - 1]] + dp[i - 1][j];
+                }
+            }
+        }
+
+        //last index
+        return dp[m][n];
+    }
+}

PaintHouse.java

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+// Time Complexity :   O(2 * N) where N = number of houses
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No
+
+
+// Your code here along with comments explaining your approach
+class Solution {
+    public int minCost(int[][] costs) {
+
+        //This is the Paint house pattern in DP
+        //As the input is 1 we will use 1D array
+        int dp[] = new int[3];
+        int n = costs.length;
+
+        //instead of dp[] we can also use varR, varB, varG
+        dp[0] = costs[n - 1][0];
+        dp[1] = costs[n - 1][1];
+        dp[2] = costs[n - 1][2];
+
+        //we will start from bottom
+        for(int i = n - 2; i >= 0; i--) {
+
+            int tempRed = dp[0];
+            dp[0] = costs[i][0] + Math.min(dp[1], dp[2]);
+
+            int tempBlue = dp[1];
+            dp[1] = costs[i][1] + Math.min(tempRed, dp[2]);
+
+            dp[2] = costs[i][2] + Math.min(tempRed, tempBlue);
+        }
+
+        //the min cost of painting house with each color
+        return Math.min(dp[0], Math.min(dp[1], dp[2]));
+    }
+}