Done with competitive coding-2

Problem1.java

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+import java.util.HashMap;
+
+class Solution {
+    public int[] twoSum(int[] nums, int target) {
+
+
+        int n=nums.length;
+
+        HashMap<Integer,Integer> map = new HashMap<>();
+
+        for(int i=0;i<n;i++){
+            if(!map.containsKey(target-nums[i])){
+                map.put(nums[i],i);
+            }
+            else{
+                return new int []{i,map.get(target-nums[i])};
+            }
+
+
+        }
+        return new int []{-1,-1};
+
+    }
+}

Problem2.java

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+class Solution {
+    public static int findMax(int[] weights, int[] profit, int totalCapacity) {
+        return helper(weights, profit, 0, totalCapacity);  // Start from index 0 with full capacity
+    }
+
+    private static int helper(int[] weights, int[] profit, int i, int capacity) {
+        // Base case: if we've considered all items, return 0 profit
+        if (i >= weights.length) return 0;
+
+        // Case 0: skip the current item
+        int case0 = helper(weights, profit, i + 1, capacity);
+
+        int case1 = 0;
+        // Case 1: include the current item if it fits in the remaining capacity
+        if (capacity - weights[i] >= 0) {
+            case1 = profit[i] + helper(weights, profit, i + 1, capacity - weights[i]);
+        }
+
+        // Return the maximum profit between including or excluding the current item
+        return Math.max(case0, case1);
+    }
+
+    // Main method to test the solution
+    public static void main(String args[]) {
+        int[] items = {10, 20, 30, 40};       // Weights of the items
+        int[] profit = {130, 110, 170, 190};  // Corresponding profits
+        int capacity = 50;                    // Total capacity of the knapsack
+
+        // Print the maximum profit that can be obtained
+        System.out.println(findMax(items, profit, capacity));  // Expected: 300
+    }
+}