SDE - 1

You are given an 'N * N' matrix of integers where each row and each column is sorted in increasing order. You are given a target integer 'X'.

Find the position of 'X' in the matrix. If it exists then return the pair {i, j} where 'i' represents the row and 'j' represents the column of the array, otherwise return {-1,-1}

For example:

If the given matrix is:
[ [1, 2, 5],
  [3, 4, 9],
  [6, 7, 10]] 
We have to find the position of 4. We will return {1,1} since A[1][1] = 4.

Kevin has two sorted packs of cards. The first pack has N cards and the second one has M cards. Every card has an integer written on it. Now, you have to tell two things. First, how many different types of cards does Kevin have with both packs(union). Second, how many types both the packs have in common(intersection).

Note :

Two cards are said to be of different types if they have different numbers written on them.

You have been given a board where there are '2' rows and 'N' columns. You have an infinite supply of 2x1 tiles, and you can place a tile in the following ways:

1. Horizontally as 1x2 tile
2. Vertically as 2x1 tile

Count the number of ways to tile the given board using the available tiles.

Note :

The number of ways might be large so output your answer modulo 10^9 + 7.

Here an example of tile and board for 'N' = 4 :

Given an array/list of integer numbers 'CHOCOLATES' of size 'N', where each value of the array/list represents the number of chocolates in the packet. There are ‘M’ number of students and the task is to distribute the chocolate to their students. Distribute chocolate in such a way that:

1. Each student gets at least one packet of chocolate.

2. The difference between the maximum number of chocolate in a packet and the minimum number of chocolate in a packet given to the students is minimum.

Example :

Given 'N' : 5 (number of packets) and 'M' : 3 (number of students)

And chocolates in each packet is : {8, 11, 7, 15, 2}

All possible way to distribute 5 packets of chocolates among 3 students are -

( 8,15, 7 ) difference of maximum-minimum is ‘15 - 7’ = ‘8’
( 8, 15, 2 ) difference of maximum-minimum is ‘15 - 2’ = ‘13’ 
( 8, 15, 11 ) difference of maximum-minimum is ‘15 - 8’ = ‘7’
( 8, 7, 2 ) difference of maximum-minimum is ‘8 - 2’ = ‘6’
( 8, 7, 11 ) difference of maximum-minimum is ‘11 - 7’ = ‘4’
( 8, 2, 11 ) difference of maximum-minimum is ‘11 - 2’ = ‘9’
( 15, 7, 2 ) difference of maximum-minimum is ‘15 - 2’ = 13’
( 15, 7, 11 ) difference of maximum-minimum is ‘15 - 7’ = ‘8’
( 15, 2, 11 ) difference of maximum-minimum is ‘15 - 2’ = ‘13’
( 7, 2, 11 ) difference of maximum-minimum is ‘11 - 2’ = ‘9’

Hence there are 10 possible ways to distribute ‘5’ packets of chocolate among the ‘3’ students and difference of combination (8, 7, 11) is ‘maximum - minimum’ = ‘11 - 7’ = ‘4’ is minimum in all of the above.

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