What is the output expected from the problem? If the goal is to find the earliest time when the frog can jump to the other side of the river How the answer for given sample comes out to be 6???
A small frog wants to get to the other side of a river. The frog is
initially located on one bank of the river (position 0) and wants to
get to the opposite bank (position X+1). Leaves fall from a tree onto
the surface of the river.
You are given a zero-indexed array A consisting of N integers
representing the falling leaves. A[K] represents the position where
one leaf falls at time K, measured in seconds.
The goal is to find the earliest time when the frog can jump to the
other side of the river. The frog can cross only when leaves appear at
every position across the river from 1 to X (that is, we want to find
the earliest moment when all the positions from 1 to X are covered by
leaves). You may assume that the speed of the current in the river is
negligibly small, i.e. the leaves do not change their positions once
they fall in the river.
For example, you are given integer X = 5 and array A such that:
A[0] = 1 A[1] = 3 A[2] = 1 A[3] = 4 A[4] = 2 A[5] = 3
A[6] = 5 A[7] = 4 In second 6, a leaf falls into position 5. This is
the earliest time when leaves appear in every position across the
river.
Time needed to reach the other end of river
or
The index in the array where the X is located
or
The index which contains the highest number?
To cross the river, the frog needs to have leaves in all positions 1 through X (5). The river is initially empty; one leaf per second falls into the river, at the location indicated by A[K], where K is the time-tick at which the leaf falls.
The given sequence for leaf positions, starting at time 0, is [1, 3, 1, 4, 2, 3, 5, 4]. Coverage of the river (given as 5 units wide) as time progresses is like this, where 0 denotes a leaf, - denotes open water:
0 0 - - - -
1 0 - 0 - -
2 0 - 0 - - There are now 2 leaves in position 1
3 0 - 0 0 -
4 0 0 0 0 -
5 0 0 0 0 - ... and a second leaf at 3
6 0 0 0 0 0 ... and now, the frog can cross.
RUBY
I am trying to implement this problem using Ruby with N complexity.. I will run a loop on falling leaves from tree and each time a falling leaf gets new position I will delete that position from the path array which I have created on the basis of steps required.
def frogJump(x,arr)
path_array =(1..x).to_a
count = 0
for position in arr
if path_array.size > 0
count+= 1
path_array.delete(position) if path_array.include?(position)
end
end
puts count
end
x= 5
arr = [1 ,2, 3 , 1 , 4 , 2 , 3 , 5 , 4]
frogJump(x,arr)
You start with a stack of n sticks. Each player removes 2 or 3 sticks each turn. The player who removes the last stick wins. If their is only one stick left, the game is a draw.
I need to determine who will win a game of n stacks in general, expressed as a function of n. However, unless we start the game with 2, 3 or 5 sticks, it is always possible to direct the game so that it ends in a draw. I drew the game tree for 9 sticks and more and it is always possible to prevent a loss by making choices that lead to one stick remaining. How can I write a winning rule for this given problem?
However, unless we start the game with 2, 3 or 5 sticks, it is always possible to direct the game so that it ends in a draw.
I don't think this is true.
For example, suppose we start with 10 sticks. If you remove x sticks, I will always remove 5-x sticks. This will mean that after one turn each there are 5 sticks left, and after two turns each I have won.
The same will apply for any multiple of 5.
Now consider other possible values modulo 5.
Here's the rule:
Losing Position: n = 5k
Draw: n = 5k+1 or n = 5k+4
Winning Position: n = 5k+2 or n = 5k+3
You can observe the pattern by building a table like the one shown below:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 .....
L D W W D L D W W D L D W W D L D
Steps for Building the table:
Observe that you lose if n = 0, mark it as L.
When n = 1, it is draw, mark it as D.
When n = 2, you can only draw 2 sticks. So, your opponent happens to face n=0 which is losing, so you win, mark it as W.
For n = 3, you can take 2 or 3 sticks, so, your opponent can end up at 1 or 0 sticks. 1 is a draw and 0 is loss for him and therefore win for us. So, we will chose win, mark it as W.
For n=4, similarily, opponent can end up at 2 or 1. So, we can draw or lose. We will choose draw, mark it as D.
For n = 5, we can either make our opponent end up at 2 or 3. Both of them are win for him. So, we lose. Mark it as L.
Basically, to determine the state(L, W or D) of a number n, we need to look at states of already computed states n-2 and n-3.
Observe that this pattern (LDWWD) repeats after that.
I was asked this question in a test and I need help with regards to how I should approach the solution, not the actual answer. The question is
You have been given a 7 digit number(with each digit being distinct and 0-9). The number has this property
product of first 3 digits = product of last 3 digits = product of central 3 digits
Identify the middle digit.
Now, I can do this on paper by brute force(trial and error), the product is 72 and digits being
8,1,9,2,4,3,6
Now how do I approach the problem in a no brute force way?
Let the number is: a b c d e f g
So as per the rule(1):
axbxc = cxdxe = exfxg
more over we have(2):
axb = dxe and
cxd = fxg
This question can be solved with factorization and little bit of hit/trial.
Out of the digits from 1 to 9, 5 and 7 can rejected straight-away since these are prime numbers and would not fit in the above two equations.
The digits 1 to 9 can be factored as:
1 = 1, 2 = 2, 3 = 3, 4 = 2X2, 6 = 2X3, 8 = 2X2X2, 9 = 3X3
After factorization we are now left with total 7 - 2's, 4 - 3's and the number 1.
As for rule 2 we are left with only 4 possibilities, these 4 equations can be computed by factorization logic since we know we have overall 7 2's and 4 3's with us.
1: 1X8(2x2x2) = 2X4(2x2)
2: 1X6(3x2) = 3X2
3: 4(2x2)X3 = 6(3x2)X2
4: 9(3x3)X2 = 6(3x2)X3
Skipping 5 and 7 we are left with 7 digits.
With above equations we have 4 digits with us and are left with remaining 3 digits which can be tested through hit and trial. For example, if we consider the first case we have:
1X8 = 2X4 and are left with 3,6,9.
we have axbxc = cxdxe we can opt c with these 3 options in that case the products would be 24, 48 and 72.
24 cant be correct since for last three digits we are left with are 6,9,4(=216)
48 cant be correct since for last three digits we are left with 3,9,4(=108)
72 could be a valid option since the last three digits in that case would be 3,6,4 (=72)
This question is good to solve with Relational Programming. I think it very clearly lets the programmer see what's going on and how the problem is solved. While it may not be the most efficient way to solve problems, it can still bring desired clarity and handle problems up to a certain size. Consider this small example from Oz:
fun {FindDigits}
D1 = {Digit}
D2 = {Digit}
D3 = {Digit}
D4 = {Digit}
D5 = {Digit}
D6 = {Digit}
D7 = {Digit}
L = [D1 D2 D3] M = [D3 D4 D5] E= [D5 D6 D7] TotL in
TotL = [D1 D2 D3 D4 D5 D6 D7]
{Unique TotL} = true
{ProductList L} = {ProductList M} = {ProductList E}
TotL
end
(Now this would be possible to parameterize furthermore, but non-optimized to illustrate the point).
Here you first pick 7 digits with a function Digit/0. Then you create three lists, L, M and E consisting of the segments, as well as a total list to return (you could also return the concatenation, but I found this better for illustration).
Then comes the point, you specify relations that have to be intact. First, that the TotL is unique (distinct in your tasks wording). Then the next one, that the segment products have to be equal.
What now happens is that a search is conducted for your answers. This is a depth-first search strategy, but could also be breadth-first, and a solver is called to bring out all solutions. The search strategy is found inside the SolveAll/1 function.
{Browse {SolveAll FindDigits}}
Which in turns returns this list of answers:
[[1 8 9 2 4 3 6] [1 8 9 2 4 6 3] [3 6 4 2 9 1 8]
[3 6 4 2 9 8 1] [6 3 4 2 9 1 8] [6 3 4 2 9 8 1]
[8 1 9 2 4 3 6] [8 1 9 2 4 6 3]]
At least this way forward is not using brute force. Essentially you are searching for answers here. There might be heuristics that let you find the correct answer sooner (some mathematical magic, perhaps), or you can use genetic algorithms to search the space or other well-known strategies.
Prime factor of distinct digit (if possible)
0 = 0
1 = 1
2 = 2
3 = 3
4 = 2 x 2
5 = 5
6 = 2 x 3
7 = 7
8 = 2 x 2 x 2
9 = 3 x 3
In total:
7 2's + 4 3's + 1 5's + 1 7's
With the fact that When A=B=C, composition of prime factor of A must be same as composition of prime factor of B and that of C, 0 , 5 and 7 are excluded since they have unique prime factor that can never match with the fact.
Hence, 7 2's + 4 3's are left and we have 7 digit (1,2,3,4,6,8,9). As there are 7 digits only, the number is formed by these digits only.
Recall the fact, A, B and C must have same composition of prime factors. This implies that A, B and C have same number of 2's and 3's in their composition. So, we should try to achieve (in total for A and B and C):
9 OR 12 2's AND
6 3's
(Must be product of 3, lower bound is total number of prime factor of all digits, upper bound is lower bound * 2)
Consider point 2 (as it has one possibility), A has 2 3's and same for B and C. To have more number of prime factor in total, we need to put digit in connection digit between two product (third or fifth digit). Extract digits with prime factor 3 into two groups {3,6} and {9} and put digit into connection digit. The only possible way is to put 9 in connection digit and 3,6 on unconnected product. That mean xx9xx36 or 36xx9xx (order of 3,6 is not important)
With this result, we get 9 x middle x connection digit = connection digit x 3 x 6. Thus, middle = (3 x 6) / 9 = 2
My answer actually extends #Ansh's answer.
Let abcdefg be the digits of the number. Then
ab=de
cd=fg
From these relations we can exclude 0, 5 and 7 because there are no other multipliers of these numbers between 0 and 9. So we are left with seven numbers and each number is included once in each answer. We are going to examine how we can pair the numbers (ab, de, cd, fg).
What happens with 9? It can't be combined with 3 or 6 since then their product will have three times the factor 3 and we have at total 4 factors of 3. Similarly, 3 and 6 must be combined at least one time together in response to the two factors of 9. This gives a product of 18 and so 9 must be combined at least once with 2.
Now if 9x2 is in a corner then 3x6 must be in the middle. Meaning in the other corner there must be another multiplier of 3. So 9 and 2 are in the middle.
Let's suppose ab=3x6 (The other case is symmetric). Then d must be 9 or 2. But if d is 9 then f or g must be multiplier of 3. So d is 2 and e is 9. We can stop here and answer the middle digit is
2
Now we have 2c = fg and the remaining choices are 1, 4, 8. We see that the only solutions are c = 4, f = 1, g = 8 and c = 4, f = 8, g = 1.
So if is 3x6 is in the left corner we have the following solutions:
3642918, 3642981, 6342918, 6342981
If 3x6 is in the right corner we have the following solutions which are the reverse of the above:
8192463, 1892463, 8192436, 1892436
Here is how you can consider the problem:
Let's note the final solution N1 N2 N3 N4 N5 N6 N7 for the 3 numbers N1N2N3, N3N4N5 and N5N6N7
0, 5 and 7 are to exclude because they are prime and no other ciphers is a multiple of them. So if they had divided one of the 3 numbers, no other number could have divided the others.
So we get the 7 remaining ciphers : 1234689
where the product of the ciphers is 2^7*3^4
(N1*N2*N3) and (N5*N6*N7) are equals so their product is a square number. We can then remove, one of the number (N4) from the product of the previous point to find a square number (i.e. even exponents on both numbers)
N4 can't be 1, 3, 4, 6, 9.
We conclude N4 is 2 or 8
If N4 is 8 and it divides (N3*N4*N5), we can't use the remaining even numbers (2, 4, 6) to divides
both (N1*N2*N3) and (N6*N7*N8) by 8. So N4 is 2 and 8 does not belong to the second group (let's put it in N1).
Now, we have: 1st grp: 8XX, 2nd group: X2X 3rd group: XXX
Note: at this point we know that the product is 72 because it is 2^3*3^2 (the square root of 2^6*3^4) but the result is not really important. We have made the difficult part knowing the 7 numbers and the middle position.
Then, we know that we have to distribute 2^3 on (N1*N2*N3), (N3*N4*N5), (N5*N6*N7) because 2^3*2*2^3=2^7
We already gave 8 to N1, 2 to N4 and we place 6 to N6, and 4 to N5 position, resulting in each of the 3 numbers being a multiple of 8.
Now, we have: 1st grp: 8XX, 2nd group: X24 3rd group: 46X
We have the same way of thinking considering the odd number, we distribute 3^2, on each part knowing that we already have a 6 in the last group.
Last group will then get the 3. And first and second ones the 9.
Now, we have: 1st grp: 8X9, 2nd group: 924 3rd group: 463
And, then 1 at N2, which is the remaining position.
This problem is pretty easy if you look at the number 72 more carefully.
We have our number with this form abcdefg
and abc = cde = efg, with those digits 8,1,9,2,4,3,6
So, first, we can conclude that 8,1,9 must be one of the triple, because, there is no way 1 can go with other two numbers to form 72.
We can also conclude that 1 must be in the start/end of the whole number or middle of the triple.
So now we have 819defg or 918defg ...
Using some calculations with the rest of those digits, we can see that only 819defg is possible, because, we need 72/9 = 8,so only 2,4 is valid, while we cannot create 72/8 = 9 from those 2,4,3,6 digits, so -> 81924fg or 81942fg and 819 must be the triple that start or end our number.
So the rest of the job is easy, we need either 72/4 = 18 or 72/2 = 36, now, we can have our answers: 8192436 or 8192463.
7 digits: 8,1,9,2,4,3,6
say XxYxZ = 72
1) pick any two from above 7 digits. say X,Y
2) divide 72 by X and then Y.. you will get the 3rd number i.e Z.
we found XYZ set of 3-digits which gives result 72.
now repeat 1) and 2) with remaining 4 digits.
this time we found ABC which multiplies to 72.
lets say, 7th digit left out is I.
3) divide 72 by I. result R
4) divide R by one of XYZ. check if result is in ABC.
if No, repeat the step 3)
if yes, found the third pair.(assume you divided R by Y and the result is B)
YIB is the third pair.
so... solution will be.
XZYIBAC
You have your 7 numbers - instead of looking at it in groups of 3 divide up the number as such:
AB | C | D | E | FG
Get the value of AB and use it to get the value of C like so: C = ABC/AB
Next you want to do the same thing with the trailing 2 digits to find E using FG. E = EFG/FG
Now that you have C & E you can solve for D
Since CDE = ABC then D = ABC/CE
Remember your formulas - instead of looking at numbers create a formula aka an algorithm that you know will work every time.
ABC = CDE = EFG However, you have to remember that your = signs have to balance. You can see that D = ABC/CE = EFG/CE Once you know that, you can figure out what you need in order to solve the problem.
Made a quick example in a fiddle of the code:
http://jsfiddle.net/4ykxx9ve/1/
var findMidNum = function() {
var num = [8, 1, 9, 2, 4, 3, 6];
var ab = num[0] * num[1];
var fg = num[5] * num[6];
var abc = num[0] * num[1] * num[2];
var cde = num[2] * num[3] * num[4];
var efg = num[4] * num[5] * num[6];
var c = abc/ab;
var e = efg/fg;
var ce = c * e
var d = abc/ce;
console.log(d); //2
}();
You have been given a 7 digit number(with each digit being distinct and 0-9). The number has this property
product of first 3 digits = product of last 3 digits = product of central 3 digits
Identify the middle digit.
Now, I can do this on paper by brute force(trial and error), the product is 72 and digits being
8,1,9,2,4,3,6
Now how do I approach the problem in a no brute force way?
use linq and substring functions
example var item = array.Skip(3).Take(3) in such a way that you have a loop
for(f =0;f<charlen.length;f++){
var xItemSum = charlen[f].Skip(f).Take(f).Sum(f => f.Value);
}
// untested code
I want to convert a number in base 10 into a special base form like this:
A*2^2 + B*3^1 + C*2^0
A can take on values of [0,1]
B can take on values of [0,1,2]
C can take on values of [0,1]
For example, the number 8 would be
1*2^2 + 1*3 + 1.
It is guaranteed that the given number can be converted to this specialized base system.
I know how to convert from this base system back to base-10, but I do not know how to convert from base-10 to this specialized base system.
In short words, treat every base number (2^2, 3^1, 2^0 in your example) as weight of an item, and the whole number as the capacity of a bag. This problem wants us to find a combination of these items which they fill the bag exactly.
In the first place this problem is NP-complete. It is identical to the subset sum problem, which can also be seen as a derivative problem of the knapsack problem.
Despite this fact, this problem can however be solved by a pseudo-polynomial time algorithm using dynamic programming in O(nW) time, which n is the number of bases, and W is the number to decompose. The details can be find in this wikipedia page: http://en.wikipedia.org/wiki/Knapsack_problem#Dynamic_programming and this SO page: What's it called when I want to choose items to fill container as full as possible - and what algorithm should I use?.
Simplifying your "special base":
X = A * 4 + B * 3 + C
A E {0,1}
B E {0,1,2}
C E {0,1}
Obviously the largest number that can be represented is 4 + 2 * 3 + 1 = 11
To figure out how to get the values of A, B, C you can do one of two things:
There are only 12 possible inputs: create a lookup table. Ugly, but quick.
Use some algorithm. A bit trickier.
Let's look at (1) first:
A B C X
0 0 0 0
0 0 1 1
0 1 0 3
0 1 1 4
0 2 0 6
0 2 1 7
1 0 0 4
1 0 1 5
1 1 0 7
1 1 1 8
1 2 0 10
1 2 1 11
Notice that 2 and 9 cannot be expressed in this system, while 4 and 7 occur twice. The fact that you have multiple possible solutions for a given input is a hint that there isn't a really robust algorithm (other than a look up table) to achieve what you want. So your table might look like this:
int A[] = {0,0,-1,0,0,1,0,1,1,-1,1,1};
int B[] = {0,0,-1,1,1,0,2,1,1,-1,2,2};
int C[] = {0,1,-1,0,2,1,0,1,1,-1,0,1};
Then look up A, B, C. If A < 0, there is no solution.
I've read every answer here, Wikipedia and WikiHow, the indian guy's lecture, and other sources, and I'm pretty sure I understand what they're saying and have implemented it that way. But I'm confused about a statement that all of these explanations make that is clearly false.
They all say to cover the zeros in the matrix with a minimum number of lines, and if that is equal to N (that is, there's a zero in every row and every column), then there's a zero solution and we're done. But then I found this:
a b c d e
A 0 7 0 0 0
B 0 8 0 0 6
C 5 0 7 3 4
D 5 0 5 9 3
E 0 4 0 0 9
There's a zero in every row and column, and no way to cover the zeros with fewer than five lines, but there's clearly no zero solution. Row C has only the zero in column b, but that leaves no zero for row D.
Do I misunderstand something here? Do I need a better test for whether or not there's a zero assignment possible? Are all these sources leaving out something essential?
You can cover the zeros in the matrix in your example with only four lines: column b, row A, row B, row E.
Here is a step-by-step walkthrough of the algorithm as it is presented in the Wikipedia article as of June 25 applied to your example:
a b c d e
A 0 7 0 0 0
B 0 8 0 0 6
C 5 0 7 3 4
D 5 0 5 9 3
E 0 4 0 0 9
Step 1: The minimum in each row is zero, so the subtraction has no effect. We try to assign tasks such that every task is performed at zero cost, but this turns out to be impossible. Proceed to next step.
Step 2: The minimum in each column is also zero, so this step also has no effect. Proceed to next step.
Step 3: We locate a minimal number of lines to cover up all the zeros. We find [b,A,B,E].
a b c d e
A ---|---------
B ---|---------
C 5 | 7 3 4
D 5 | 5 9 3
E ---|---------
Step 4: We locate the minimal uncovered element. This is 3, at (C,d) and (D,e). We subtract 3 from every unmarked element and add 3 to every element covered by two lines:
a b c d e
A 0 10 0 0 0
B 0 11 0 0 6
C 2 0 4 0 1
D 2 0 2 6 0
E 0 7 0 0 9
Immediately the minimum number of lines to cover up all the zeros becomes 5. This is easy to verify as there is a zero in every row and a zero in every column. The algorithm asserts that an assignment like the one we were looking for in step 1 should now be possible on the new matrix.
We try to assign tasks such that every task is performed at zero cost (according to the new matrix). This is now possible. We find the solution [(A,e),(B,c),(C,d),(D,b),(E,a)].
We can now go back and verify that the solution that we found actually is optimal. We see that every assigned job has zero cost, except (C,d), which has cost 3. Since 3 is actually the lowest nonzero element in the matrix, and we have seen that there is no zero-cost solution, it is clear that this is an optimal solution.