Question : Given an integer(n) denoting the no. of particles initially
Given an array of sizes of these particles
These particles can go into any number of simulations (possibly none)
In one simualtion two particles combines to give another particle with size as the difference between the size of them (possibly 0).
Find the smallest particle that can be formed.
constraints
n<=1000
size<=1e9
Example 1
3
30 10 8
Output
2
Explaination- 10 - 8 is the smallest we can achive
Example 2
4
1 2 4 8
output
1
explanation
We cannot make another 1 so as to get 0 so smallest without any simulation is 1
example 3
5
30 27 26 10 6
output
0
30-26=4
10-6 =4
4-4 =0
My thinking: I can only think of the brute force solution which will obviously time out. Can anyone help me out here with just the approach? I think it's related to dynamic programming
I think this can be solved in O(n^2log(n))
Consider your third example: 30 27 26 10 6
Sort the input to make it : 6 10 26 27 30
Build a list of differences for each (i,j) combination.
For:
i = 1 -> 4 20 21 24
i = 2 -> 16, 17, 20
i = 3 -> 1, 4
i = 4 -> 3
There is no list for i = 5 why? because it is already considered for combination with other particles before.
Now consider the below cases:
Case 1
The particle i is not combined with any other particle yet. This means some other particle should have been combined with a particle other than i.
This suggests us that we need to search for A[i] in the lists j = 1 to N except for j = i.
Get the nearest value. This can be done using binary search. Because your difference lists are sorted! Then your result for now is |A[i] - NearestValueFound|
Case 2
The particle i is combined with some other particle.
Take example i = 1 above and lets consider that its combined with particle 2. The result is 4.
So search for 4 in all the lists except list 2 - because we consider that particle 2 is already combined with particle 1 and we shouldn't search list 2.
Do we have a best match? It seems we have a match 4 found in the list 3. It needn't be 0 - in this case it is 0 so just return 0.
Repeat Case 1, 2 for all particles. Time complexity is O(n^2log(n)), because you are doing a binary search on all lists for each i except the list i.
import itertools as it
N = int(input())
nums = list()
for i in range(N):
nums.append(int(input()))
_min = min(nums)
def go(li):
global _min
if len(li)>1:
for i in it.combinations(li, 2):
temp = abs(i[0] - i[1])
if _min > temp:
_min = temp
k = li.copy()
k.remove(i[0])
k.remove(i[1])
k.append(temp)
go(k)
go(nums)
print(_min)
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.
OK, this is a question I got for my advance algorithm class. I already turned in my solution once but got rejected by my instructor due to efficiency issue, in other words, I already made the efforts on my part but could not get it even after his hint, so please be gentle. I will give his hint below
Given an array of intervals with both start point and end point, find the number of other intervals fall within it for each interval. number of intervals is less than 10^9 and their ids are distinct. start and end are less than 10^18, the input files don't contain duplicate number for start and end. All the numbers above are integers
the hint is: considering a data structure with buckets. The algorithm should be faster than O(n^2)
sample input and output
input:
5 %% number of intervals
2 100 200 %% id, start,end. all lines below follows this
3 110 190
4 105 145
1 90 150
5 102 198
output:
3 0
4 0
1 1
5 2
2 3
The numbers are pretty big so O(N log N) might be a little to much but here's an idea.
First things first normalize the values, that means turning them smaller while keepinging the same ordering. In your example the normalize would be
90 100 102 105 110 145 150 190 198 200
1 2 3 4 5 6 7 8 9 10
So you're new intervals are:
5
2 2 10
3 5 8
4 4 6
1 1 7
5 3 9
Now the edges of the intervals are in the range of [1, 2N].
Now sort the intervals by their end:
5
4 4 6
1 1 7
3 5 8
5 3 9
2 2 10
When you reach an interval you can say that all the intervals that start before it and have not been encountered yet should have their answer increased by one. This can be done with a SegmentTree.
What you do when you get an interval [x, y] you increase all values in the range [1, x - 1] by 1 and then compute its answer as the value at x in the segment tree. That's just addition on an interval and query on a point, a common segment tree problem.
I don't really think you can solve this problem with less than O(N log N) time and O(N) memory, so this solution should be the asymptotically best solution in both time and space.