adding interview-2 solutions

W2_01knapsack.py

@@ -0,0 +1,30 @@
+def knapsack(W, val, wt):
+    n = len(wt)
+    dp = [[0 for _ in range(W + 1)] for _ in range(n + 1)]
+
+    for i in range(n + 1):
+        for j in range(W + 1):
+
+            # If there are no items or capacity is 0
+            if i == 0 or j == 0:
+                dp[i][j] = 0
+            else:
+                choose = 0
+
+                # Choice 1: choose the current item
+                if wt[i - 1] <= j:
+                    choose = val[i - 1] + dp[i - 1][j - wt[i - 1]]
+
+                # Choice 2: don't choose the current item
+                dont_choose = dp[i - 1][j]
+
+                dp[i][j] = max(choose, dont_choose)
+
+    return dp[n][W]
+
+
+# Time complexity: O(n * W), where n is the number of items and W is the knapsack capacity.
+# We fill a DP table of size (n + 1) * (W + 1), and each cell takes O(1) time.
+
+# Space complexity: O(n * W), because we store a 2D DP table with
+# (n + 1) rows and (W + 1) columns.

W2_two_sum.py

@@ -0,0 +1,19 @@
+class Solution:
+    def twoSum(self, nums: List[int], target: int) -> List[int]:
+        result = []
+        hashmap = {}
+
+        for i in range(len(nums)):
+            complement = target - nums[i]
+
+            if complement not in hashmap:
+                hashmap[nums[i]] = i
+            else:
+                result.append(i)
+                result.append(hashmap[complement])
+
+        return result
+
+# Time complexity: O(N), we visit each element once.
+# Space complexity: O(N), if the pair is found at the last index, then the hashmap will have n-1 keys worst case
+