6 4 4 N
4 1 L 7
5 U 6 10
? 1 14 2
Check your answer:-
See ExplanationMost of us are familiar with google codejam, those who don’t visit https://code.google.com/codejam for more information. This is the first problem from Online Round 1C 2013, top 1000 from this round will be eligible for next online round.
In English, there are 26 letters that are either vowels or consonants. In this problem, we consider a, e, i, o, and u to be vowels, and the other 21 letters to be consonants.
A tribe living in the Greatest Colorful Jungle has a tradition of naming their members using English letters. But it is not easy to come up with a good name for a new member because it reflects the member’s social status within the tribe. It is believed that the less common the name he or she is given, the more socially privileged he or she is.
The leader of the tribe is a professional linguist. He notices that hard-to-pronounce names are uncommon, and the reason is that they have too many consecutive consonants. Therefore, he announces that the social status of a member in the tribe is determined by its n-value, which is the number of substrings with at least n consecutive consonants in the name. For example, when n = 3, the name “quartz” has the n-value of 4 because the substrings quartz, uartz, artz, and rtz have at least 3 consecutive consonants each. A greater n-value means a greater social status in the tribe. Two substrings are considered different if they begin or end at a different point (even if they consist of the same letters), for instance “tsetse” contains 11 substrings with two consecutive consonants, even though some of them (like “tsetse” and “tsetse“) contain the same letters.
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guess the phone brands from images
minimum wins to qualify for playoffs
There are 9 team participating in Indian Premier League, every team will play two matches with every other team. Top 4 teams will be qualified for playoffs.
How many minimum wins required for a team to ensure a place in playoffs?
See SolutionCase 1: Lets say if Team A and B and C wins 14 matches each(they will loose 1 game each from other two teams), then team D can win only 16 – 3*2 = 10 matches.
Case 2: Lets take a case where Top 4 team wins 13 games each
Team A: wins 13 matches, looses 3 games from B,C and D
Team B: wins 13 matches, looses 3 games from A,C and D
Team C: wins 13 matches, looses 3 games from A,B and D
Team D wins 13 matches, looses 3 games from A, B and C
Now Team E, can only win maximum of 8 matches as it A,B,C and D will win all matches played by E.
Case 3: Lets say top 5 teams wins 12 matches each
Team A: wins 12 matches, looses 3 games from B, C, D and E
Team B: wins 12 matches, looses 3 games from A, C, D and E
Team C: wins 12 matches, looses 3 games from A, B, D and E
Team D wins 12 matches, looses 3 games from A, B , C and E
Team E wins 12 matches, looses 4 games from A, B, C and D
In this case, team E though wins 12 matches but still don’t qualify for playoffs(assuming lowest run rate).
Thus from case 2 we can say that a team has to win at least 13 matches to ensure a place in playoffs.
Most of us are familiar with google codejam, those who don’t visit https://code.google.com/codejam for more information. This is the second problem from Online Round 1B 2013, top 1000 from this round will be eligible for next online round.
Diamonds are falling from the sky. People are now buying up locations where the diamonds can land, just to own a diamond if one does land there. You have been offered one such place, and want to know whether it is a good deal.
Diamonds are shaped like, you guessed it, diamonds: they are squares with vertices (X-1, Y), (X, Y+1), (X+1, Y) and (X, Y-1) for some X, Y which we call the center of the diamond. All the diamonds are always in the X-Y plane. X is the horizontal direction, Y is the vertical direction. The ground is at Y=0, and positive Y coordinates are above the ground.
The diamonds fall one at a time along the Y axis. This means that they start at (0, Y) with Y very large, and fall vertically down, until they hit either the ground or another diamond.
When a diamond hits the ground, it falls until it is buried into the ground up to its center, and then stops moving. This effectively means that all diamonds stop falling or sliding if their center reaches Y=0.
When a diamond hits another diamond, vertex to vertex, it can start sliding down, without turning, in one of the two possible directions: down and left, or down and right. If there is no diamond immediately blocking either of the sides, it slides left or right with equal probability. If there is a diamond blocking one of the sides, the falling diamond will slide to the other side until it is blocked by another diamond, or becomes buried in the ground. If there are diamonds blocking the paths to the left and to the right, the diamond just stops.
falling diamonds problem google codejam round 1b 2013
Consider the example in the picture. The first diamond hits the ground and stops when halfway buried, with its center at (0, 0). The second diamond may slide either to the left or to the right with equal probability. Here, it happened to go left. It stops buried in the ground next to the first diamond, at (-2, 0). The third diamond will also hit the first one. Then it will either randomly slide to the right and stop in the ground, or slide to the left, and stop between and above the two already-placed diamonds. It again happened to go left, so it stopped at (-1, 1). The fourth diamond has no choice: it will slide right, and stop in the ground at (2, 0).
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Most of us are familiar with google codejam, those who don’t visit https://code.google.com/codejam for more information. This is the first problem from Online Round 1B 2013, top 1000 from this round will be eligible for next online round.
Armin is playing Osmos, a physics-based puzzle game developed by Hemisphere Games. In this game, he plays a “mote”, moving around and absorbing smaller motes.
A “mote” in English is a small particle. In this game, it’s a thing that absorbs (or is absorbed by) other things! The game in this problem has a similar idea to Osmos, but does not assume you have played the game.
When Armin’s mote absorbs a smaller mote, his mote becomes bigger by the smaller mote’s size. Now that it’s bigger, it might be able to absorb even more motes. For example: suppose Armin’s mote has size 10, and there are other motes of sizes 9, 13 and 19. At the start, Armin’s mote can only absorb the mote of size 9. When it absorbs that, it will have size 19. Then it can only absorb the mote of size 13. When it absorbs that, it’ll have size 32. Now Armin’s mote can absorb the last mote.
Note that Armin’s mote can absorb another mote if and only if the other mote is smaller. If the other mote is the same size as his, his mote can’t absorb it.
You are responsible for the program that creates motes for Armin to absorb. The program has already created some motes, of various sizes, and has created Armin’s mote. Unfortunately, given his mote’s size and the list of other motes, it’s possible that there’s no way for Armin’s mote to absorb them all.
You want to fix that. There are two kinds of operations you can perform, in any order, any number of times: you can add a mote of any positive integer size to the game, or you can remove any one of the existing motes. What is the minimum number of times you can perform those operations in order to make it possible for Armin’s mote to absorb every other mote?
For example, suppose Armin’s mote is of size 10 and the other motes are of sizes [9, 20, 25, 100]. This game isn’t currently solvable, but by adding a mote of size 3 and removing the mote of size 100, you can make it solvable in only 2 operations. The answer here is 2.
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Guess the actor and actresses name from whatsapp emoticons and smileys
Guess the actor and actresses name from whatsapp emoticons and smileys
These English words have indian city or state names in them
Can you guess those places names?
1,16,36,__,81,100,144
See SolutionThere is a peculiar two digit number which is three times the sum of its digits.
Can you find the number?
ab = 3(a+b) We can also write it as 10a + b = 3*(a+b) 7a = 2b a/b = 2/7
As it is given that it is a two digit number, so it can only be 27.
