SDE - 1

You have been given a square chessboard of size ‘N x N’. The position coordinates of the Knight and the position coordinates of the target are also given.

Your task is to find out the minimum steps a Knight will take to reach the target position.

Example:

knightPosition: {3,4}
targetPosition: {2,1}

The knight can move from position (3,4) to positions (1,3), (2,2) and (4,2). Position (4,2) is selected and the ‘stepCount’ becomes 1. From position (4,2), the knight can directly jump to the position (2,1) which is the target point and ‘stepCount’ becomes 2 which is the final answer. 

Note:

1. The coordinates are 1 indexed. So, the bottom left square is (1,1) and the top right square is (N, N).

2. The knight can make 8 possible moves as given in figure 1.

3. A Knight moves 2 squares in one direction and 1 square in the perpendicular direction (or vice-versa).

A thief is robbing a store and can carry a maximum weight of ‘W’ into his knapsack. There are 'N' items available in the store and the weight and value of each item is known to the thief. Considering the constraints of the maximum weight that a knapsack can carry, you have to find the maximum profit that a thief can generate by stealing items.

Note: The thief is not allowed to break the items.

For example, N = 4, W = 10 and 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.

You are given a list of “N” strings A. Your task is to check whether you can form a given target string using a combination of one or more strings of A.

Note :

You can use any string of A multiple times.

Examples :

A =[“coding”, ”ninjas”, “is”, “awesome”]  target = “codingninjas”
Ans = true as we use “coding” and “ninjas” to form “codingninjas”

You are given a linked list of 'n' nodes and an integer 'k', where 'k' is less than or equal to 'n'.

Your task is to reverse the order of each group of 'k' consecutive nodes, if 'n' is not divisible by 'k', then the last group of nodes should remain unchanged.

For example, if the linked list is 1->2->3->4->5, and 'k' is 3, we have to reverse the first three elements, and leave the last two elements unchanged. Thus, the final linked list being 3->2->1->4->5.

Implement a function that performs this reversal, and returns the head of the modified linked list.

Example:

Input: 'list' = [1, 2, 3, 4], 'k' = 2

Output: 2 1 4 3

Explanation:
We have to reverse the given list 'k' at a time, which is 2 in this case. So we reverse the first 2 elements then the next 2 elements, giving us 2->1->4->3.

Note:

All the node values will be distinct.

Two players 'X' and 'Y', are playing a coin game. Initially, there are 'N' coins. Each player can pick exactly 'A' coins or 'B' coins or 1 coin. A player loses the game if he is not able to pick any coins. 'X' always starts the game, and each player plays optimally. You are supposed to find which player wins the coin game.

You are given a set of ‘n’ types of rectangular 3-D boxes. The height, width, and length of each type of box are given by arrays, ‘height’, ‘width’, and ‘length’ respectively, each consisting of ‘n’ positive integers. The height, width, length of the i^th type box is given by ‘height[i]’, ‘width[i]’ and ‘length[i]’ respectively.

You need to create a stack of boxes that is as tall as possible using the given set of boxes.

Below are a few allowances:

You can only stack a box on top of another box if the dimensions of the 2-D base of the lower box ( both length and width ) are strictly larger than those of the 2-D base of the higher box. 

You can rotate a box so that any side functions as its base. It is also allowed to use multiple instances of the same type of box. This means, a single type of box when rotated, will generate multiple boxes with different dimensions, which may also be included in stack building.

Return the height of the highest possible stack so formed.

Note:

The height, Width, Length of the type of box will interchange after rotation.

No two boxes will have all three dimensions the same.

Don’t print anything, just return the height of the highest possible stack that can be formed.