Yasir Javed Kiani's Blog: 2019

There is a valuable treasure that is hidden below a complex maze. The maze is made of rooms and are square in shape, and a maze of size N will have N × N rooms with all of them closed initially. When a room is open, one can enter into it from any of its adjacent open rooms; two rooms are adjacent if they share a common wall.

The maze was built in a way that it opens itself by opening its rooms randomly. A maze is said to be open if there is at least one path from any one of the rooms facing the top of the maze to any room on the bottom side facing the treasure. Anyone, who attempts to enter the maze without being able to reach the treasure and return, will be cruelly killed by the maze.

The local government has spent years researching the maze and figured out a way to determine the sequence of rooms being opened in almost real time. Based on this data, the government has posed the following challenge, with a small percentage of the treasure to whomever solves the problem:

Given the sequence of room openings, determine when the maze becomes open, or if it remains closed throughout.

Input Format

Input begins with a single integer N, which denotes the size of maze.

All of the next lines (except the last one) denotes the sequence of the rooms the maze is opening. Each line contains 2 integers X and Y which denotes the row and column of the room opened by the maze. The last line just includes -1 and marks the end of input.

Constraints

1 <= X, Y <= N <= 1000

Output Format

Output a single integer R based on the final status of the maze. R denotes the number of room openings that occur before the maze first becomes open, or -1 if the maze remains closed.

Sample Input

Sample Output

-1

Explanation

It is easy to understand if you plot the maze. The following is the state of the maze at the end of the inputs. Xindicates that a room is closed and O that a room is open. Note that there is no path from the top of the maze to the bottom of the maze.

O X X O
O X O X
X O X X
O X O X

Consider the second input sample (which is available if you click on the Run Code button):

Below is a figure with the maze after 7 rooms are open. Note that there is no path from the top of the maze to the bottom of the maze, and therefore the maze is closed.

O X X O
O X O X
X O X X
O X O X

However, after 8th room is open, there is a path, as shown below:

O X X O
O X O X
O O X X
O X O X

Thus, the expected output is: 8

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Solution

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Explanation

IEEEXtreme9.0 Problem : Digit Fun

Xtreme9.0 - Digit Fun! (Link )

or try this link

Recurrence relations are an important tool for the computer scientist. Many algorithms, particularly those that use divide and conquer, have time complexities best modeled by recurrence relations. A recurrence relation allows us to recursively define a sequence of values by defining the nth value in terms of certain of its predecessors.

Many natural functions, such as factorials and the Fibonacci sequence, can easily be expressed as recurrences.

The function of interest for this problem is described below.

Let |An| denote the number of digits in the decimal representation of An. Given any number A0, we define a sequence using the following recurrence:

Ai = |Ai-1| for i > 0

The goal of this problem is to determine the smallest positive i such that Ai = Ai-1.

Input Format

Input consists of multiple lines, each terminated by an end-of-line character. Each line (except the last) contains a value for A0, where each value is non-negative and no more than a million digits. The last line of input contains the word END.

Output Format

For each value of A0 given in the input, the program should output one line containing the smallest positive i such that Ai = Ai-1.

Sample Input

Sample Output

Explanation

The first input value is A0 = 9999, resulting in A1 = |9999| = 4. Because 4 does not equal 9999, we find A2 = |A1| = |4| = 1. Since 1 is not equal to 4, we find A3 = |A2| = |1| = 1. A3 is equal to A2, making 3 the smallest positive i such that Ai = Ai-1.

The second input value is A0 = 0, resulting in A1 = |0| = 1. Because 0 does not equal 1, we find A2 = |A1| = |1| = 1. A2 is equal to A1, making 2 the smallest positive i such that Ai = Ai-1.

The third input value is A0 = 1, resulting in A1 = |1| = 1. A1 is equal to A0, making 1 the smallest positive i such that Ai = Ai-1.

The last input value is A0 = 9999999999, resulting in A1 = |9999999999| = 10. Because 10 does not equal 9999999999, we find A2 = |A1| = |10| = 2. Since 2 is not equal to 10, we find A3 = |A2| = |2| = 1. Since 1 is not equal to 2, we find A4 = |A3| = |1| = 1. A4 is equal to A3, making 4 the smallest positive i such that Ai = Ai-1.

----------------------* Solution *------------------------------------------------

Explanation ( Get the Code and Explanation)