Given four numbers NAA, NAB, NBA, NBB each of them are less or equal to 10^5.
It is needed to the find number of different strings that contain only AB characters and contain:
exactly NAA occurences of "AA";
exactly NAB occurence of "AB";
exactly NBA occurences of "BA";
exactly NBB occurence of "BB".
I know only brute force solution.
Could you please give me some hint or advice how to do it faster?
You can use two 5-dimensional DP table with state of cell defined as :
tableA[N][AA][BB][AB][BA] = no. of strings of length N ending at 'A' having AA times "AA", BB times "BB", AB times "AB" and BA times "BA" and
tableB[N][AA][BB][AB][BA] = no. of strings of length N ending at 'B' having AA times "AA", BB times "BB", AB times "AB" and BA times "BA"
Now you can use following state-transition rules to fill the table:
tableA[N][AA][BB][AB][BA] = tableA[N-1][AA-1][BB][AB][BA] + tableB[N-1][AA][BB][AB][BA-1] and
tableB[N][AA][BB][AB][BA] = tableA[N-1][AA][BB][AB-1][BA] + tableB[N-1][AA][BB-1][AB][BA]
Use appropriate base cases and finally retrieve answer:
answer = tableA[AB][NAA][NBB][NAB][NBA] + tableB[AB][NAA][NBB][NAB][NBA]
So, time for the answer.
Given a string you can do the following transformations without changing NAB and NBA
AA -> A
A -> AA
BB -> B
B -> BB
So, we can arrive at any valid string by starting from a string that contains neither AA nor BB by "inflation".
Inflating
Lets assume we have a string without AA and BB substrings that has the correct number of ABs and BAs.
Let NA the number of As and NB the number of Bs in that string (calculation of these are shown below).
Than we need to distribute NAA As to NA places. There are binom(NAA+NA-1, NA-1) ways to do that (look up integer compositions). There are also binom(NBB+NB-1, NB-1) possibilities to place the additional Bs. Because insertion of As and Bs is independant we can simply multiply the numbers to get all possibilities.
Calculating the actual result
There are four cases to consider:
NAB = NBA+1: The string starts with A and ends with B. NA=NB=NAB
So, N = binom(NAA+NAB-1, NAB-1) * binom(NBB+NAB-1, NAB-1)
NAB+1 = NBA: Same as above just excchange A and B everywhere
NAB = NBA:
The string either begins with A and ends with A or begins with B and ends with B. So there are two possibilities that need to be added.
String starts with A gives NA=NAB+1 and NB=NAB
String starts with B gives NA=NAB and NB=NAB+1
So, N = binom(NAA+NAB, NAB)*binom(NBB+NAB-1, NAB-1)
+ binom(NAA+NAB-1, NAB-1)*binom(NBB+NAB, NAB)
For all other pairs of NAB and NBA: N=0.
I hope, I have got all the "-1"s right. And sorry for the bad layout.
Related
I want to find the common elements in multiple (>=2) cell arrays of strings.
A related question is here, and the answer proposes to use the function intersect(), however it works for only 2 inputs.
In my case, I have more than two cells, and I want to obtain a single common subset. Here is an example of what I want to achieve:
c1 = {'a','b','c','d'}
c2 = {'b','c','d'}
c3 = {'c','d'}
c_common = my_fun({c1,c2,c3});
in the end, I want c_common={'c','d'}, since only these two strings occur in all the inputs.
How can I do this with MATLAB?
Thanks in advance,
P.S. I also need the indices from each input, but I can probably do that myself using the output c_common, so not necessary in the answer. But if anyone wants to tackle that too, my actual output will be like this:
[c_common, indices] = my_fun({c1,c2,c3});
where indices = {[3,4], [2,3], [1,2]} for this case.
Thanks,
Listed in this post is a vectorized approach to give us the common strings and indices using unique and accumarray. This would work even when the strings are not sorted within each cell array to give us indices corresponding to their positions within it, but they have to be unique. Please have a look at the sample input, output section* to see such a case run. Here's the implementation -
C = {c1,c2,c3}; % Add more cell arrays here
% Get unique strings and ID each of the strings based on their uniqueness
[unqC,~,unqID] = unique([C{:}]);
% Get count of each ID and the IDs that have counts equal to the number of
% cells arrays in C indicate that they are present in all cell arrays and
% thus are the ones to be finally selected
match_ID = find(accumarray(unqID(:),1)==numel(C));
common_str = unqC(match_ID)
% ------------ Additional work to get indices ----------------
N_str = numel(common_str);
% Store matches as a logical array to be used at later stages
matches = ismember(unqID,match_ID);
% Use ismember to find all those indices in unqID and subtract group
% lengths from them to give us the indices within each cell array
clens = [0 cumsum(cellfun('length',C(1:end-1)))];
match_index = reshape(find(matches),N_str,[]);
% Sort match_index along each column based on the respective unqID elements
[m,n] = size(match_index);
[~,sidx] = sort(reshape(unqID(matches),N_str,[]),1);
sorted_match_index = match_index(bsxfun(#plus,sidx,(0:n-1)*m));
% Subtract cumulative group lens to give us indices corres. to each cell array
common_idx = bsxfun(#minus,sorted_match_index,clens).'
Please note that at the step that calculates match_ID : accumarray(unqID(:),1) could be replaced by histc(unqID,1:max(unqID)). Also, histcounts be another alternative there.
*Sample input, output -
c1 =
'a' 'b' 'c' 'd'
c2 =
'b' 'c' 'a' 'd'
c3 =
'c' 'd' 'a'
common_str =
'a' 'c' 'd'
common_idx =
1 3 4
3 2 4
3 1 2
As noted in the comments to this question, there is a file in File Exchange called "MINTERSECT -- Multiple set intersection." at http://www.mathworks.com/matlabcentral/fileexchange/6144-mintersect-multiple-set-intersection that contains simple code to generalize intersect to multiple sets. In a nutshell, the code gets the output from performing intersect on the first pair of cells and then perform intersect on this output with the next cell. This process continues until all cells have been compared. Note that the author points out that the code is not particularly efficient but it may be sufficient for your use case.
If I have strings that I know have no more than 2 distinct characters,
example set:
aab
abbbbabb
bbbaa
aaaaaaa
aaaa
abab
a
aa
aaaaa
aaabba
aabbbab
What's the most efficient way to put them into alphabetical order?
the resulting sorted set:
a
aa
aaaa
aaaaa
aaaaaaa
aaabba
aab
aabbbab
abab
abbbbabb
bbbaa
edit:
I know I could just use a normal sorting algorithm (quick sort, merge sort), but the question is: Does the fact that there are not more than 2 distinct characters make something else more efficient?
If the maximum length of the string matters, I would like to know the answer for 2 different scenarios:
maximum length of the string is the same as the number of strings (n strings being sorted, n maximum length of the string)
maximum length of the string is log n, with n as the number of strings being sorted
I can also assume that all of the strings are distinct.
The String compareTo or compareToIgnoresCase method will return a negative integer, 0, or a polsitive integer depending on the alphabetical ordering of the two Strings being compared. Try that.
General sorting algorithm based on comparisons only asymptotically can't achieve results better than O(nlogn). In your case there is an additional information (2 distinct chars) which has a potential of improving this result. A simple approach that will yield a O(n) result:
Check the first character (let's mark it x).
Scan the string till the end
whenever x is encountered increase a counter.
when encountered the non-x character (let's mark it y) for the first time store it in a dedicated variable
Compare x and y.
if x < y fill the string from the beginning with x's according to the counter and the rest with y
if x > y fill the string from the beginning with y's string length-num of x's slots and the rest with x's.
Disclaimer: This is a problem lifted from HackerRank, but their editorial answer wasn't sufficient so I hoped to get better answers. If it's against any policy, please let me know and I'll take this down.
Problem:
You are given two integers, N and M. Count the number of strings of length N under the alphabet set of size M that doesn't contain any palindromic string of the length greater than 1 as a consecutive substring.
N=2,M=2 -> 2 :: AA, AB, BA, BB
N=2,M=3 -> 6 :: AA, AB, AC, BA, BB, BC, CA, CB, CC
ABCDE counts as it does not contain any palindromic substrings.
ABCCC does not count as it does contain "CCC", a palindrome of length >1.
Editorial
Here is the provided answer which I think is wrong:
For N>=3, there are (M−2) ways to choose any next symbol (after the first two) - basically it should not coincide with the previous and pred-previous symbols, that aren't equal.
If N=1, return M
If N=2, return M * (M-1)
If N>=3, return M * (M-1) * (M-2)^(N-2)
counterexample: N=4, M=3, "ABCC"
My Solution Try
When I was working on this problem, I tried to find all the strings that contained palindromic substrings and subtracting that from the total, M^N. I ran into a lot of problems with over counting. For example, "ABABA" has "ABA","BAB","ABA" of n=3, and "ABABA" of n=5.
Thanks for any help in elucidating this problem. I really hope for a good answer to figure this out!
Suppose you build up palindrome-free strings one letter at a time. For the first letter, you have M choices, and for the second, you have M-1, since you can't use the first letter. This much is obvious.
For every letter after the first two, you can't use the previous letter, and you can't use the letter before that, so that's two choices eliminated. What about the other letters? Well, if using one of those creates a palindrome, it would have to be a palindrome of length at least 4 - but if adding a letter creates a palindrome of length K+2 for K>=2, the string must already have had a palindrome of length K for the new palindrome to build off of. (For K<2, this is okay.) Since the string didn't have any palindromes of length >=2, we can conclude that adding any letter other than the previous two letters is fine.
Thus, we have M choices for the first letter, M-1 choices for the second, and M-2 for every letter after that.
Deterministic automata to find number of subsequences in string ?
How can I construct a DFA to find number of occurence string as a subsequence in another string?
eg. In "ssstttrrriiinnngggg" we have 3 subsequences which form string "string" ?
also both string to be found and to be searched only contain characters from specific character Set .
I have some idea about storing characters in stack poping them accordingly till we match , if dont match push again .
Please tell DFA solution ?
OVERLAPPING MATCHES
If you wish to count the number of overlapping sequences then you simply construct a DFA that matches the string, e.g.
1 -(if see s)-> 2 -(if see t)-> 3 -(if see r)-> 4 -(if see i)-> 5 -(if see n)-> 6 -(if see g)-> 7
and then compute the number of ways of being in each state after seeing each character using dynamic programming. See the answers to this question for more details.
DP[a][b] = number of ways of being in state b after seeing the first a characters
= DP[a-1][b] + DP[a-1][b-1] if character at position a is the one needed to take state b-1 to b
= DP[a-1][b] otherwise
Start with DP[0][b]=0 for b>1 and DP[0][1]=1.
Then the total number of overlapping strings is DP[len(string)][7]
NON-OVERLAPPING MATCHES
If you are counting the number of non-overlapping sequences, then if we assume that the characters in the pattern to be matched are distinct, we can use a slight modification:
DP[a][b] = number of strings being in state b after seeing the first a characters
= DP[a-1][b] + 1 if character at position a is the one needed to take state b-1 to b and DP[a-1][b-1]>0
= DP[a-1][b] - 1 if character at position a is the one needed to take state b to b+1 and DP[a-1][b]>0
= DP[a-1][b] otherwise
Start with DP[0][b]=0 for b>1 and DP[0][1]=infinity.
Then the total number of non-overlapping strings is DP[len(string)][7]
This approach will not necessarily give the correct answer if the pattern to be matched contains repeated characters (e.g. 'strings').
Given are 6 strings of any length. The words are to be arranged in the pattern shown below. They can be arranged either vertically or horizontally.
--------
| |
| |
| |
---------------
| |
| |
| |
--------
The pattern need not to be symmetric and there need to be two empty areas as shown.
For example:
Given strings
PQF
DCC
ACTF
CKTYCA
PGYVQP
DWTP
The pattern can be
DCC...
W.K...
T.T...
PGYVQP
..C..Q
..ACTF
where dot represent empty areas.
The other example is
RVE
LAPAHFUIK
BIRRE
KZGLPFQR
LLHU
UUZZSQHILWB
Pattern is
LLHU....
A..U....
P..Z....
A..Z....
H..S....
F..Q....
U..H....
I..I....
KZGLPFQR
...W...V
...BIRRE
If multiple patterns are possible then pattern with lexicographically smallest first line, then second line and so on is to be formed. What algorithm can be used to solve this?
Find strings which suits to this constraint:
strlen(a) + strlen(b) - 1 = strlen(c)
strlen(d) + strlen(e) - 1 = strlen(f)
After that try every possible situation if they are valid. For example;
aaa.....
d.f.....
d.f.....
d.f.....
cccccccc
..f....e
..f....e
..bbbbbb
There will be 2*2*2 = 8 different situation.
There are a number of heuristics that you can apply, but before that, let's go over some properties of the puzzle.
+aa+
c f
+ee+eee+
f d
+bbb+
Let us call the length of the string with the same character as appeared in the diagram above. We have:
a + b - 1 = e
c + d - 1 = f
I will refer to the 2 strings for the cross in the middle as middle strings.
We also infer that the length of the string cannot be less than 2. Therefore, we can infer:
e > a, e > b
f > c, f > d
From this, we know that the 2 shortest strings cannot be middle strings, due to the inequality above.
The 3 largest strings cannot be equal also, since after choosing any of 3 string as middle string, we are left with 2 largest strings that are equal, and it is impossible according to the inequality above.
The puzzle is only tricky when the lengths are regular. When the lengths are irregular, you can do direct mapping from length to position.
If we have the 2 largest strings being equal, due to the inequality above, they are the 2 middle strings. The worst case for this one is a "regular" puzzle, where the length a, b, c, d are equal.
If the 2 largest strings are unequal, the largest string's position can be determined immediately (since its length is unique in the puzzle) - as one of the middle string. In worst case, there can be 3 candidates for the other middle string - just brute force and check all of them.
Algorithm:
Try to map unique length string to the position.
Brute force the 2 strings in the middle (taken into consideration what I mentioned above), and brute force to fill in the rest.
Even with stupid brute force, there are only 6! = 720 cases, if the string can only go from left to right, up to down (no reverse). There will be 46080 cases (* 2^6) if the string is allowed to be in any direction.