diff --git a/coin-change.java b/coin-change.java
new file mode 100644
index 00000000..d861f3a7
--- /dev/null
+++ b/coin-change.java
@@ -0,0 +1,60 @@
+// Solution - 1
+
+// TimeComplexity: O(m*n) where m is coins length and n is amount
+// SpaceComplexity: O(n)
+// Explanation: I am using dynamic programming to build the answer from smaller subproblems. I create a 1D array matrix where matrix[k] represents the minimum number of coins needed to make amount k.
+// For each coin, I iterate through all amounts and update the minimum coins required using the relation matrix[k] = min(matrix[k], 1 + matrix[k - coin]). 
+// In the end, if the value at matrix[amount] is still very large, it means the amount cannot be formed, so I return -1.
+class Solution {
+    public int coinChange(int[] coins, int amount) {
+        int m = coins.length;
+        int[] matrix = new int[amount+1];
+        for(int i=1; i<amount+1; i++) {
+            matrix[i] = Integer.MAX_VALUE-5;
+        }
+
+        for(int j=1; j<m+1; j++) {
+            for(int k=1; k<amount+1; k++) {
+                if(k>=coins[j-1]) {
+                    matrix[k] = Math.min(matrix[k], 1+matrix[k-coins[j-1]]);
+                }
+            }
+        }
+
+        if(matrix[amount] > Integer.MAX_VALUE-100) return -1;
+        return matrix[amount];
+
+        
+    }
+}
+
+
+// Solution -2
+
+// TimeComplexity:(2^(m+n)) where m is coins length and n is amount
+// SpaceComplexity: O(m+n)
+// Explnation: Here I use recursion to try all possible combinations. At each step, I have two choices: either take the current coin (reduce amount) or skip it (move to next coin).
+// I recursively compute both choices and return the minimum. If the amount becomes negative or I exhaust all coins, I return a large value to indicate invalid.
+class Solution {
+    public int coinChange(int[] coins, int amount) {
+        int result= helper(coins, amount, 0);
+        if (result>=Integer.MAX_VALUE-10) return -1;
+        return result;
+        
+    }
+
+    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;
+
+        // choose
+        int choice1 = 1+ helper(coins, amount-coins[idx], idx);
+
+
+        // no choose
+        int choice2 = helper(coins, amount, idx+1);
+
+        return Math.min(choice1, choice2);
+    }
+}
\ No newline at end of file
diff --git a/house-robber.java b/house-robber.java
new file mode 100644
index 00000000..e2524b60
--- /dev/null
+++ b/house-robber.java
@@ -0,0 +1,82 @@
+// Solution - 1
+
+// TimeComplexity: O(n)
+// SpaceComplexity: O(n)
+// Explanation: I am using dynamic programming to store the maximum money that can be robbed up to each house. 
+// The array dp[i] represents the maximum money I can rob from the first i houses.
+// For each house, I decide whether to skip it (dp[i-1]) or rob it (nums[i-1] + dp[i-2]) and take the maximum of the two. Finally, dp[nums.length] gives the answer.
+class Solution {
+    public int rob(int[] nums) {
+        int [] dp = new int[nums.length+1];
+        dp[1] = nums[0];
+
+        for(int i=2; i<nums.length+1; i++) {
+            int max= Math.max(dp[i-1], nums[i-1]+dp[i-2]);
+            dp[i] = max;
+        }
+        return dp[nums.length];
+    }
+}
+
+// Solution - 2
+// TimeComplexity: O(n)
+// SpaceComplexity: O(n)
+// Explanation : Here I use recursion with memoization. At each index, I have two choices: skip the house or rob it and move two steps ahead.
+//  I store results in a dp array to avoid recomputation. This ensures each index is computed only once.
+class Solution {
+    public int rob(int[] nums) {
+        Integer[] dp = new Integer[nums.length];
+        return helper(nums, 0, dp);
+        
+    }
+
+    private int helper (int[] nums, int idx, Integer[] dp) {
+        // base 
+        if(idx>=nums.length) return 0;
+        if(dp[idx]!=null) {
+            return dp[idx];
+        }
+
+
+        // no choose
+        int case1 = helper(nums, idx+1, dp);
+        
+
+        // choose
+        int case2= nums[idx] + helper(nums, idx+2, dp);
+
+        int max = Math.max(case1, case2);
+        dp[idx] = max;
+
+        return max;
+
+    }
+}
+
+// Solution - 3
+// TimeComplexity: O(2ⁿ)
+// SpaceComplexity: O(n) (recursion stack)
+// Explanation: In this version, I recursively try both choices (rob or skip) at every house without storing results. 
+// This causes repeated calculations of the same subproblems, leading to exponential growth in calls.
+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;
+
+
+        // no choose
+        int case1 = helper(nums, idx+1);
+        
+
+        // choose
+        int case2= nums[idx] + helper(nums, idx+2);
+
+        return Math.max(case1, case2);
+
+    }
+}