DP-2 completed

Coin change 2

@@ -0,0 +1,30 @@
+Time complexity - O(nm) where n is number of coins and m is the amount
+Space complexity - O(nm)
+
+class Solution {
+public:
+    int change(int amount, vector<int>& coins) {
+        int numberOfCoins=coins.size();
+
+        int matrix[numberOfCoins+1][amount+1];
+
+        for (int i=0;i<=amount;i++) {
+            matrix[0][i]=0;
+        }
+        for (int i=0;i<=coins.size();i++) {
+            matrix[i][0]=1;
+        }
+
+        for (int i=1;i<=numberOfCoins;i++) {
+            for (int j=1;j<amount+1;j++) {
+                if (coins[i-1]>j) {
+                    matrix[i][j]=matrix[i-1][j];
+                } else if (coins[i-1]<=j){
+                    matrix[i][j]=matrix[i-1][j] + matrix[i][j-coins[i-1]];
+                }
+            }
+        }
+        return matrix[numberOfCoins][amount];
+
+    }
+};

paint house

@@ -0,0 +1,33 @@
+// Time Complexity - O(3n) where n is the number of houses
+// Space Complexity - O(3) 
+class Solution {
+public:
+    int minCost(vector<vector<int>>& costs) {
+        int rows = costs.size();
+        int cols = costs[0].size();
+
+        vector<vector<int>> dp(rows, vector<int>(cols, 0));
+
+        for (int i=0;i<rows;i++) {
+            for(int j=0;j<cols;j++) {
+                if (i == 0) {
+                    dp[i][j] = costs[i][j];
+                } else if ( j == 0){
+                    dp[i][j] = costs[i][j] + min(dp[i-1][j+1], dp[i-1][j+2]);
+                } else if ( j == 1) {
+                    dp[i][j] = costs[i][j] + min(dp[i-1][j+1], dp[i-1][j-1]);
+                } else if ( j == 2) {
+                    dp[i][j] = costs[i][j] + min(dp[i-1][j-1], dp[i-1][j-2]);
+                }
+            }
+        }
+
+        int mini = 100000;
+        for (int i = 0; i < cols; i++)
+        {
+            mini = min(dp[rows-1][i], mini);
+        }
+
+        return mini;
+    }
+};