Day8 - Coin-change, paint-house

Coin-change-ii.java

@@ -0,0 +1,45 @@
+class Solution {
+    //recusion approach- exhaustive
+    public int change(int amount, int[] coins) {
+        if(coins.length==0 || coins==null) return -1;
+        return recurse(coins, amount,0);
+    }
+    private int recurse(int[] coins, int amount, int index){
+        //base
+        if(index == coins.length || amount < 0 ) return 0;
+        if(amount==0) return 1;
+        //case0
+        int case0 = recurse(coins, amount,index+1);
+        //case1
+        int case1 = recurse(coins, amount-coins[index], index);
+
+        return case0 + case1;
+
+
+    }
+
+
+
+    //dp approach
+    public int change(int amount, int[] coins) {
+        //if(coins.length==0 || coins==null) return -1;
+        int n = coins.length;
+        int m = amount;
+        int[][] dp=new int[n+1][m+1];
+        dp[0][0] = 1;
+        for(int i=1;i<=n;i++){
+            for(int j=0;j<=m;j++){
+                //denomination > amount
+                if(coins[i-1]>j){
+                    dp[i][j] = dp[i-1][j];
+                }else{
+                    //case0 + case1
+                    dp[i][j] = dp[i-1][j] + dp[i][j-coins[i-1]];
+                }
+            }
+        }
+        return dp[n][m];
+
+    }
+
+}

Paint-house.java

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+class Solution {
+    public int minCost(int[][] costs) {
+        //final top row elements
+        int colorR = helper(costs,0, 0);
+        int colorB = helper(costs,0, 1);
+        int colorG = helper(costs,0, 2);
+        return Math.min(colorR, Math.min(colorB, colorG));
+    }
+    private int helper(int[][] costs, int idx, int col){ //idx is for house, col is for which color
+        //base
+        if(idx == costs.length) return 0; //rows or if there are no houses
+
+        //logic
+        if(col == 0) return costs[idx][0] + Math.min(helper(costs, idx+1, 1), helper(costs, idx+1, 2));
+        if(col == 1) return costs[idx][1] + Math.min(helper(costs, idx+1, 0), helper(costs, idx+1, 2));
+        if(col == 2) return costs[idx][2] + Math.min(helper(costs, idx+1, 0), helper(costs, idx+1, 1));
+
+        return 868789;
+    }
+}

README.md

@@ -1,8 +1,13 @@
 # DP-2
 
 ## Problem1(https://leetcode.com/problems/paint-house/)
-
+* paint houses with colors R, B, G such that no 2 houses have same color.
+* Greedy - choose min from first row, then choose min from other 2. This fails when 1st choice is wrong. Try exhaustive.
+* Keep the last row as it is, the elements will be that sum of that element + min of element of other 2 colors  in the below row.
 
 
 ## Problem2 (https://leetcode.com/problems/coin-change-2/)
+* indefinite - Return the number of combinations that make up that amount
+* if denomination > amount, return from top row
+* else that element + smae row element & denomination steps back