diff --git a/coin_change.java b/coin_change.java
new file mode 100644
index 00000000..32ce1a2f
--- /dev/null
+++ b/coin_change.java
@@ -0,0 +1,45 @@
+// class Solution {
+//     public int coinChange(int[] coins, int amount) {
+//         //Brute force - Exhaustive
+//         int re = helper(coins,amount,0);
+//         if(re<Integer.MAX_VALUE-10) return re;
+//         return -1;
+//     }
+//     private int helper(int[] coins, int amount, int idx){
+//         //base 
+//         if(idx==coins.length || amount<0) return Integer.MAX_VALUE-10;
+//         if(amount==0) return 0;
+//         //logic
+//         //Not choose
+//         int case1= helper(coins,amount,idx+1);
+//         //choose 
+//         int case2=1+helper(coins,amount-coins[idx],idx);
+//         return Math.min(case1,case2);
+//     }
+// }
+
+
+// Time Complexity : O(m*n) where m is the number of coins and n is the amount
+// Space Complexity : O(n) where n is the amount
+// Did this code successfully run on Leetcode : yes 
+class Solution {
+    public int coinChange(int[] coins, int amount) {
+        int m= coins.length;
+        int n= amount;
+        int[] dp = new int[n+1];
+        dp[0]=0;
+        for(int i=1;i<=n;i++){
+            dp[i]= Integer.MAX_VALUE -10;
+        }
+        for(int i=1; i<=m;i++){
+            for(int j=coins[i-1];j<=n;j++){
+                dp[j] = Math.min(dp[j],1+dp[j-coins[i-1]]); 
+            }
+        }
+        if(dp[n]<Integer.MAX_VALUE -10) return dp[n];
+        return -1;
+    }
+}
+
+// Bottom up apprach - Matrix as the decision making variable are the coins and the amount.
+// then i reduced it to tabluation and then to space optimization. 
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diff --git a/house_robber.java b/house_robber.java
new file mode 100644
index 00000000..68a02283
--- /dev/null
+++ b/house_robber.java
@@ -0,0 +1,74 @@
+// Exhaustive method 
+// class Solution {
+//     public int rob(int[] nums) {
+//         return helper(nums,0);
+//     }
+//     private int helper(int[] nums, int idx){
+//         //base
+//         if(idx>=nums.length) return 0;
+//         //logic
+//         //dont choose 
+//         int case1 = helper(nums,idx+1);
+//         //choose 
+//         int case2 = nums[idx]+helper(nums,idx+2);
+//         return Math.max(case1, case2);
+//     }
+// }
+
+// Top Dowm approach 
+// import java.util.Arrays;
+// class Solution {
+//     int[] memo;
+//     public int rob(int[] nums) {
+//         this.memo= new int[nums.length];
+//         Arrays.fill(memo,-1);
+//         return helper(nums,0);
+//     }
+//     private int helper(int[] nums, int idx){
+//         //base
+//         if(idx>=nums.length) return 0;
+//         if(memo[idx]!=-1) return memo[idx];
+//         //logic
+//         //dont choose 
+//         int case1 = helper(nums,idx+1);
+//         //choose 
+//         int case2 = nums[idx]+helper(nums,idx+2);
+//         memo[idx]= Math.max(case1, case2);
+//         return memo[idx];
+//     }
+// }
+
+
+//Bottom up approach 
+// class Solution {
+//     public int rob(int[] nums) {
+//         int n = nums.length;
+//         if(n == 1) return nums[0];
+//         int[] dp = new int[n];
+//         dp[0] = nums[0];
+//         dp[1] = Math.max(nums[0], nums[1]);
+//         for(int i=2; i<nums.length;i++){
+//             dp[i] = Math.max(dp[i-1],nums[i]+dp[i-2]);
+//         }
+//         return dp[n-1];
+//     }
+// }
+
+//Bottom up approach - memory optimised 
+// Time Complexity : O(n) where n is the length of the input array
+// Space Complexity : O(1) as we are using constant space for the variables
+class Solution {
+    public int rob(int[] nums) {
+        int n = nums.length;
+        if(n == 1) return nums[0];
+        //int[] dp = new int[n];
+        int prev = nums[0];
+        int curr = Math.max(nums[0], nums[1]);
+        for(int i=2; i<nums.length;i++){
+            int temp = curr;
+            curr = Math.max(curr,nums[i]+prev);
+            prev = temp;
+        }
+        return curr;
+    }
+}
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