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Facebook Hacker Cup

Permutations solution: Facebook hacker cup 2013 Round 2

February 17, 2013 by puzzler Leave a Comment

From 2011 facebook is organizing a Facebook hacker cup every year around feb, visit Facebook hacker cup homepage for more details

This question is the third question(Carried 45 marks out of 100) for the Round 2 of 2013 hacker cup. only top 100 contestants will be moved to Round 3.

Problem Statement

In this problem you need to count number of possible permutations p of the first N integers, given N-1 constraints of the form pi < pj.

Input

The first line contains an integer T, T ≤ 20, followed by T test cases. Each test case begins with an integer N, N ≤ 1000, which is the number of integers in the permutation. The next N – 1 lines each contain a single constraint in the following format: “i sign j“, where 0 ≤ i, j ≤ N – 1 and sign is either “<” or “>“, which denotes whether the i-th element of the permutation should be less than or greater than the j-th element.

It is guaranteed that it is not possible to partition indices into two disjoint sets A and B such that there is no constraint involving elements from both A and B.
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Checkout more Interview Questions Tags: Algorithm, Facebook Hacker Cup, Interview, Java, Solved Puzzles, Very Difficult

Dead Pixels solution: Facebook hacker cup 2013 Round 1

February 8, 2013 by puzzler Leave a Comment

From 2011 facebook is organizing a Facebook hacker cup every year around feb, visit Facebook hacker cup homepage for more details

This question is the third question(Carried 45 marks out of 100) for the Round 1 of 2013 hacker cup. only top 500 contestants will be moved to Round 2.

Problem Statement

John’s friend Peter purchases a new high resolution monitor with dimension W * H where W is the number of pixels in each row (i.e. width) and H is the number of pixels in each column (i.e. height).

However, there are N dead pixels on the monitor. The i-th dead pixel is located at (x[i], y[i]). (0, 0) is the top-left pixel and (W – 1, H – 1) is the bottom-right pixel. The locations of the dead pixels could be generated by 6 given integers X, Y, a, b, c and d by the following rules. If 2 pixels are at the same location, they are considered the same. It is possible that there are less than N distinct dead pixels.

  • x[0] = X
  • y[0] = Y
  • x[i] = (x[i – 1] * a + y[i – 1] * b + 1) % W (for 0 < i < N)
  • y[i] = (x[i – 1] * c + y[i – 1] * d + 1) % H (for 0 < i < N)

Peter connects his monitor to his computer and opens an image with dimension P (width) * Q (height). How many unique positions can the image be placed so that it can be displayed perfectly (i.e. all pixels of the picture are shown on the monitor)? The image cannot be rotated.
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Checkout more Interview Questions Tags: Algorithm, Difficult, Facebook Hacker Cup, Interview, Java, Solved Puzzles

Security problem solution: Facebook hacker cup 2013 Round 1

February 4, 2013 by puzzler Leave a Comment

From 2011 facebook is organizing a Facebook hacker cup every year around feb, visit Facebook hacker cup homepage for more details

This question is the second question(Carried 35 marks out of 100) for the Round 1 of 2013 hacker cup. only top 500 contestants will be moved to Round 2.

Problem Statement

You are designing a new encryption system that works in the following way:

For server-client communication you need a key k, composed of m sections, each of length l, and the key consists only of lowercase characters in the set {a, b, c, d, e, f}. The server has a key k1 and the client has a key k2 where:

  • k1 = f(k). f is a function that receives a key and replace some random letters by ? indicating that those characters can be any lowercase letter of the set described before.
  • k2 = f(g(k)). g is a function that takes a key and produces a random permutation of its m sections. And f is the function defined above.

For example: let m = 3, l = 2

  • f(‘abacbc’) = ‘?ba??c’
  • g(‘abacbc’) = ‘acbcab’ (each section was moved one place to the left).

Your task is given k1 and k2, find key k. If there are several solutions, print the lexicographically smallest key. And if there is no solution at all, print “IMPOSSIBLE” (without the quotes).

Input description:

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Checkout more Interview Questions Tags: Difficult, Facebook Hacker Cup, Interview, Java, Solved Puzzles

Card Game solution: Facebook hacker cup 2013 Round1

February 4, 2013 by puzzler 1 Comment

From 2011 facebook is organizing a Facebook hacker cup every year around feb, visit Facebook hacker cup homepage for more details

This question is the first question(Carried 20 marks out of 100) for the Round 1 of 2013 hacker cup. only top 500 contestants will be moved to Round 2.

Problem Statement

John is playing a game with his friends. The game’s rules are as follows: There is deck of N cards from which each person is dealt a hand of K cards. Each card has an integer value representing its strength. A hand’s strength is determined by the value of the highest card in the hand. The person with the strongest hand wins the round. Bets are placed before each player reveals the strength of their hand.

John needs your help to decide when to bet. He decides he wants to bet when the strength of his hand is higher than the average hand strength. Hence John wants to calculate the average strength of ALL possible sets of hands. John is very good at division, but he needs your help in calculating the sum of the strengths of all possible hands.

Problem

You are given an array a with N ≤ 10 000 different integer numbers and a number, K, where 1 ≤ K ≤ N. For all possible subsets of a of size K find the sum of their maximal elements modulo 1 000 000 007.

Input

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Find the Min solution: Facebook hacker Cup 2013 Qual Round

January 29, 2013 by puzzler 4 Comments

From 2011 facebook is organizing a Facebook hacker cup every year around feb, visit Facebook hacker cup homepage for more details

This question is the third question(Carried 45 marks out of 100) for the qualification round 2013 hacker cup.

Problem Statement

After sending smileys, John decided to play with arrays. Did you know that hackers enjoy playing with arrays? John has a zero-based index array, m, which contains n non-negative integers. However, only the first k values of the array are known to him, and he wants to figure out the rest.

John knows the following: for each index i, where k <= i < n, m[i] is the minimum non-negative integer which is *not* contained in the previous *k* values of m.
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