diff --git a/Problem1.java b/Problem1.java
index e69de29b..1bd70288 100644
--- a/Problem1.java
+++ b/Problem1.java
@@ -0,0 +1,34 @@
+import java.util.HashMap;
+import java.util.Map;
+// Time Complexity : O(n) where n is the number of elements in the array
+// Space Complexity : O(n) where n is the number of elements in the array
+// Did this code successfully run on Leetcode :yes
+// Any problem you faced while coding this : none
+
+
+// Your code here along with comments explaining your approach
+// If we go by brute force approach, we can check for all the combinations of numbers to get the target,
+//  but that will be O(n^2) time complexity.
+// So we can use hashmap or hashset 
+// Here we used hashmap to store the numbers and their indices,
+//  and check for the complement of the current number in the hashmap,
+//  if it exists, we return the indices of the current number and the complement.
+
+class Solution {
+    public int[] twoSum(int[] nums, int target) {
+        Map<Integer, Integer> hashnum = new HashMap<>();
+
+        for (int i = 0; i < nums.length; i++) {
+            int num = target - nums[i];
+
+            if (hashnum.containsKey(num)) {
+                return new int[] { hashnum.get(num), i };
+            } else {
+                if (!hashnum.containsKey(nums[i])) {
+                    hashnum.put(nums[i], i);
+                }
+            }
+        }
+        return null;
+    }
+}
\ No newline at end of file
diff --git a/Problem2.java b/Problem2.java
index e69de29b..44bb6741 100644
--- a/Problem2.java
+++ b/Problem2.java
@@ -0,0 +1,43 @@
+// Time Complexity : O(n*W) where n is the number of items and W is the capacity of the knapsack
+// Space Complexity : O(n*W) where n is the number of items and W is the capacity of the knapsack
+// Did this code successfully run on Leetcode :yes
+// Any problem you faced while coding this : A bit of confusion in buildin the dp table, but got it after some time.
+
+
+// Your code here along with comments explaining your approach
+//  Greedy or selective approach will not work here, 
+// We need to go exhaustive and check for all the combinations of items to get the maximum value.
+// We can use a 2D dp table to store the items and the capacity of the knapsack, and fill the table in a bottom up manner.
+// We fill with the value of the item if we pick it, and the value of the previous item if we don't pick it, 
+// Add the value of the item to the value of the remaining capacity if we pick it, and take the maximum of both.
+class Knapsack {
+
+    static int knapsack(int W, int[] val, int[] wt) {
+        int n = wt.length;
+        int[][] dp = new int[n + 1][W + 1];
+
+        // Build table dp[][] in bottom-up manner
+        for (int i = 0; i <= n; i++) {
+            for (int j = 0; j <= W; j++) {
+
+                // If there is no item or the knapsack's capacity is 0
+                if (i == 0 || j == 0)
+                    dp[i][j] = 0;
+                else {
+                    int pick = 0;
+
+                    // Pick ith item if it does not exceed the capacity of knapsack
+                    if (wt[i - 1] <= j)
+                        pick = val[i - 1] + dp[i - 1][j - wt[i - 1]];
+
+                    // Don't pick the ith item
+                    int notPick = dp[i - 1][j];
+
+                    dp[i][j] = Math.max(pick, notPick);
+                }
+            }
+        }
+        return dp[n][W];
+    }
+}
+