Completed DP-1

CoinChange.java

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+// Time Complexity: O(mn)
+// Space Complexity: O(mn)
+
+// 1: We maintain a dp table where dp[i][j] contains the total coins we have used to make amount j using coins i so far available to us
+// 2: We check to make sure that the denomination is greater than the amount, else we will only have case0
+// 3: If not, we add case0 and case1 to have the total running sum through which we can return the last element of the dp table
+class Solution {
+    public int coinChange(int[] coins, int amount) {
+        if(coins.length == 0 || coins == null) return -1;
+        int n = coins.length;
+        int m = amount;
+        int[][]dp = new int[n+1][m+1];
+        for(int i = 1;i<=m; i++){
+            dp[0][i] = amount+1;
+        }
+        for(int i = 1; i<=n;i++){
+            for(int j=1; j<=m; j++){
+                // when denomination > amount
+                if(coins[i-1] > j){
+                    dp[i][j] = dp[i-1][j];
+                }
+                else
+                {
+                    dp[i][j] = Math.min(dp[i-1][j], 1 + dp[i][j - coins[i-1]]);
+                }
+            }
+        }
+        int res = dp[n][m];
+        if(res > amount){
+            return -1;
+        }
+        return res;
+    }
+}

HouseRobber.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 : No
+
+
+// Your code here along with comments explaining your approach
+// 1: We maintain variables to calculate the max amount the robber can make
+// 2: We have prev and curr pointers that hold the max amount while iterating through the array
+// 3: Once we have completed iteration, we return the curr amount which holds the total max value that the robber could make
+class Solution {
+    public int rob(int[] nums) {
+        if(nums.length == 1) return nums[0];
+
+        int n = nums.length;
+        int prev = nums[0];
+        int curr = Math.max(nums[0], nums[1]);
+
+        for(int i = 2; i<n; i++){
+            int temp = curr;
+            curr = Math.max(curr, prev + nums[i]);
+            prev = temp;
+        }
+
+        return curr;
+    }
+}