From 125191703a083fe45d2035ed22b386a7f8a25dce Mon Sep 17 00:00:00 2001
From: Samiksha Manjunath <ssamma@amazon.com>
Date: Sat, 16 May 2026 17:41:05 -0700
Subject: [PATCH] Done Competitive-Coding-2

---
 Problem1.java | 24 ++++++++++++++++++++++++
 Problem2.java | 37 +++++++++++++++++++++++++++++++++++++
 2 files changed, 61 insertions(+)

diff --git a/Problem1.java b/Problem1.java
index e69de29b..cca8deb7 100644
--- a/Problem1.java
+++ b/Problem1.java
@@ -0,0 +1,24 @@
+// Time Complexity : O(n) because we traverse the array once and HashMap operations take O(1) on average
+// Space Complexity : O(n) for storing numbers and their indices in the HashMap
+// 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 in three sentences only
+// We use a HashMap to store each number along with its index as we iterate through the array.
+// For every number, we calculate the required complement needed to reach the target and check whether it already exists in the map.
+// If the complement exists, we return the stored index and current index as the required pair; otherwise, we store the current number and continue.
+
+class Solution {
+    public int[] twoSum(int[] nums, int target) {
+        Map<Integer, Integer> hashMap = new HashMap<>();
+        for (int i=0; i<nums.length; i++) {
+            int diff = target - nums[i];
+            if (hashMap.containsKey(diff)) {
+                return new int[] {hashMap.get(diff), i};
+            }
+            hashMap.put(nums[i], i);
+        }
+        return new int[] {};
+    }
+}
\ No newline at end of file
diff --git a/Problem2.java b/Problem2.java
index e69de29b..b593d05a 100644
--- a/Problem2.java
+++ b/Problem2.java
@@ -0,0 +1,37 @@
+// Time Complexity : O(n * W) where n is number of items and W is the capacity of the knapsack
+// Space Complexity : O(n * W)
+// Did this code successfully run on GeeksForGeeks : Yes
+// Any problem you faced while coding this : No
+
+
+// Your code here along with comments explaining your approach in three sentences only
+// We use dynamic programming where dp[i][w] stores the maximum value we can get using the first i items with capacity w.
+// For every item, we have two choices: skip it, or take it if its weight is less than or equal to the current capacity.
+// The answer is stored in dp[n][W], which represents the best value possible using all items and full capacity.
+
+class Solution {
+
+    static int knapSack(int W, int wt[], int val[], int n) {
+
+        int[][] dp = new int[n + 1][W + 1];
+
+        for (int i = 1; i <= n; i++) {
+
+            for (int w = 1; w <= W; w++) {
+
+                // skip current item
+                dp[i][w] = dp[i - 1][w];
+
+                // take current item if it fits
+                if (wt[i - 1] <= w) {
+                    dp[i][w] = Math.max(
+                        dp[i][w],
+                        val[i - 1] + dp[i - 1][w - wt[i - 1]]
+                    );
+                }
+            }
+        }
+
+        return dp[n][W];
+    }
+}
\ No newline at end of file