SDE

You have been given a Binary Tree of distinct integers and two nodes ‘X’ and ‘Y’. You are supposed to return the LCA (Lowest Common Ancestor) of ‘X’ and ‘Y’.

The LCA of ‘X’ and ‘Y’ in the binary tree is the shared ancestor of ‘X’ and ‘Y’ that is located farthest from the root.

Note :

You may assume that given ‘X’ and ‘Y’ definitely exist in the given binary tree.

For example :

For the given binary tree

LCA of ‘X’ and ‘Y’ is highlighted in yellow colour.

You have been given weights and values of ‘N’ items. You are also given a knapsack of size ‘W’.

Your task is to put the items in the knapsack such that the total value of items in the knapsack is maximum.

Note:

You are allowed to break the items.

Example:

If 'N = 4' and 'W = 10'. The weights and values of items are weights = [6, 1, 5, 3] and values = [3, 6, 1, 4]. 
Then the best way to fill the knapsack is to choose items with weight 6, 1 and  3. The total value of knapsack = 3 + 6 + 4 = 13.00   

You are given an array 'nums' of ‘n’ integers.

Return all subset sums of 'nums' in a non-decreasing order.

Note:

Here subset sum means sum of all elements of a subset of 'nums'. A subset of 'nums' is an array formed by removing some (possibly zero or all) elements of 'nums'.

For example

Input: 'nums' = [1,2]

Output: 0 1 2 3

Explanation:
Following are the subset sums:
0 (by considering empty subset)
1 
2
1+2 = 3
So, subset sum are [0,1,2,3].

Given a singly linked list of 'N' nodes. The objective is to determine the middle node of a singly linked list. However, if the list has an even number of nodes, we return the second middle node.

Note:

1. If the list is empty, the function immediately returns None because there is no middle node to find.
2. If the list has only one node, then the only node in the list is trivially the middle node, and the function returns that node.

We have a collection of 'N' stones, each stone has a positive integer weight.

On each turn, we choose the two heaviest stones and smash them together. Suppose the stones have weights 'x' and 'y' with 'x' <= 'y'. The result of this smash will be:

1. If 'x' == 'y', both stones are totally destroyed;

2. If 'x' != 'y', the stone of weight 'x' is totally destroyed, and the stone of weight 'y' has a new weight equal to 'y - x'.

In the end, there is at most 1 stone left. Return the weight of this stone or 0 if there are no stones left.