Completed Competitive coding 2

Problem1.java

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+//Use a HashMap to store each number’s value -> index as you scan.
+// For each nums[i], compute complement = target - nums[i]; if it’s already in the map, return the stored index and i.
+// O(n) time with O(n) extra space.
+
+class Solution {
+    public int[] twoSum(int[] nums, int target) {
+        HashMap<Integer, Integer> map = new HashMap<>();
+
+        for (int i = 0; i < nums.length; i++) {
+            int cmp = target - nums[i];
+            if (map.containsKey(cmp)) {
+                return new int[]{map.get(cmp),i};
+            }
+            map.put(nums[i], i);
+        }
+
+        return new int[]{};
+    }
+}

Problem2.java

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+// 0-1 Knapsack via bottom-up DP: dp[i][j] = max value using first i items within capacity j.
+// For each item i and capacity j: if wt[i-1] > j -> can't take -> carry dp[i-1][j];
+// else choose max of skipping (dp[i-1][j]) vs taking once (profit[i-1] + dp[i-1][j - wt[i-1]]).
+// Time O(m*n)
+// Space O(m*n)
+
+class Solution {
+
+	public static int findMax(int[] weights, int[] profit, int totalCapacity)
+	{
+		int n = totalCapacity;
+		int m = weights.length;
+		int[][]  dp = new int[m+1][n+1];
+
+		for(int i = 1; i <= m; i++ )
+		{
+			for(int j = 1; j <= n ; j++)
+			{
+				 if(weights[i-1] > j)
+				{
+					dp[i][j] = dp[i-1][j];
+				}else {							
+					dp[i][j] = Math.max(dp[i-1][j] , profit[i-1]+dp[i-1][j-weights[i-1]]);
+				}
+			}
+		}
+		return dp[m][n];
+
+	}
+}