I need to find the minimal number of insertions needed to convert a string into a palindrome. Note: the insertions can happen at any place, at the end, or within. If it was only at the end, we have a question here.
So I found out that this can be done in O(N**2) time by this simple trick:
Let the string be s1. Reverse it. Let it be s2. Say the length is l.
Now find the longest common subsequence of s1 and s2. Let its length be x.
The answer is l-x.
For example, suppose s1 = abcda. Therefore s2 = adcba. Length is 5. Longest common subsequence is aba of length 3. So the minimal number of insertions is 5-3 = 2, which is the actual answer, with the resulting string - adcbcda.
However, I cannot understand the logic behind it. Can anyone explain it to me why it works?
And, is there any O(N) solution possible for this?
I don't know whether there is a O(N) solution but by comparing with the reverse, you find a subsequence which is a palindrome. Then you have l-x letters that are not paired. (You can consider a letter's pair as its reflection if you have a mirror right at the middle of the word. e.g. ab|ba) Later, by insertions you just complete the picture.
Now,firstly, how do we find a (maximum)subsequence that is common to two strings? There is a polynomial algorithm for finding it see it here
https://en.wikipedia.org/wiki/Longest_common_subsequence_problem
When we try to find the longest common subsequence(lcs) between s1 and s2(reverse of s1) we actually find lcs between the first half of s1 and first half s2, also second half of s1 and second half of s2.
Assume
s1 = abcddzac
so s2 = cazddcba. Here we can see it as comparison of abcd with cazd(first half) plus comparison of dzac with dcba(second half). We can see that both of comparisons are the same except they are reverse of each other so their concatenation has to be palindrome, so lcs of s1 and s2 has to be palindrome.
Once we have the lcs(ad|da) which is of length 4, we have 4 more letters that break the symmetry(b,c,z,c). Then we insert one letter for each of them to make a symmetry, i.e. a palindrome. We set our middle point as the middle point of the lcs and consider that we break s1 into two from that middle point so we have
s1 = a bc d|d z a c and we break it like a stick into two from d|d and we end up with:
dzac
dcba
now we simply fill between the letters of lcs so that they are the same. In our case steps are as follows:
dzac
dcba
dzac
dzcba
dzcac
dzcba
dzcbac
dzcba
dzcbac
dzcbac
Now we unbreak it from the same point and we have
cabczddzcbac which is a palindrome.
Note: cddc is also an ldc but that doesn't change the number of steps.
Related
I found below problem in one website.
A wonderful string is a string where at most one letter appears an odd number of times.
For example, "ccjjc" and "abab" are wonderful, but "ab" is not.
Given a string word that consists of the first ten lowercase English letters ('a' through 'j'), return the number of wonderful non-empty substrings in word. If the same substring appears multiple times in word, then count each occurrence separately.
A substring is a contiguous sequence of characters in a string.
Example 1 :
Input: word = "aba"
Output: 4
Explanation: The four wonderful substrings are a , b , a(last character) , aba.
I tried to solve it. I implemented a O(n^2) solution (n is input string length). But expected time complexity is O(n). I could not solve it in O(n). I found below solution but could not understood it. Can you please help me to understand below O(n) solution for this problem or come up with an O(n) solution?
My O(N^2) approach - for every substring check whether it has at most one odd count char. This check can be done in O(1) time using an 10 character array.
class Solution {
public:
long long wonderfulSubstrings(string str) {
long long ans=0;
int idx=0; long long xorsum=0;
unordered_map<long long,long long>mp;
mp[xorsum]++;
while(idx<str.length()){
xorsum=xorsum^(1<<(str[idx]-'a'));
// if xor is repeating it means it is having even ouccrences of all elements
// after the previos ouccerence of xor.
if(mp.find(xorsum)!=mp.end())
ans+=mp[xorsum];
mp[xorsum]++;
// if xor will have at most 1 odd character than check by xoring with (a to j)
// check correspondingly in the map
for(int i=0;i<10;i++){
long long temp=xorsum;
temp=temp^(1<<i);
if(mp.find(temp)!=mp.end())
ans+=mp[temp];
}
idx++;
}
return ans;
}
};
There's two main algorithmic tricks in the code, bitmasks and prefix-sums, which can be confusing if you've never seen them before. Let's look at how the problem is solved conceptually first.
For any substring of our string S, we want to count the number of appearances for each of the 10 possible letters, and ask if each number is even or odd.
For example, with a substring s = accjjc, we can summarize it as: odd# a, even# b, odd# c, even# d, even# e, even# f, even# g, even# h, even# i, even# j. This is kind of long, so we can summarize it using a bitmask: for each letter a-j, put a 1 if the count is odd, or 0 if the count is even. This gives us a 10-digit binary string, which is 1010000000 for our example.
You can treat this as a normal integer (or long long, depending on how ints are represented). When we see another character, the count flips whether it was even or odd. On bitmasks, this is the same as flipping a single bit, or an XOR operation. If we add another 'a', we can update the bitmask to start with 'even# a' by XORing it with the number 1000000000.
We want to count the number of substrings where at most one character count is odd. This is the same as counting the number of substrings whose bitmask has at most one 1. There are 11 of these bitmasks: the ten-zero string, and each string with exactly one 1 for each of the ten possible spots. If you interpret these as integers, the last ten strings are the first ten powers of 2: 1<<0, 1<<1, 1<<2, ... 1<<9.
Now, we want to count the bitmasks for all substrings in O(n) time. First, solve a simpler problem: count the bitmasks for just all prefixes, and store these counts in a hashmap. We can do this by keeping a running bitmask from the start, and performing updates by an XOR of the bit corresponding to that letter: xorsum=xorsum^(1<<(str[idx]-'a')). This can clearly be done in a single, O(n) time pass through the string.
How do we get counts of arbitrary substrings? The answer is prefix-sums: the count of letters in any substring can be expressed as a different of two prefix-counts. For example, with s = accjjc, suppose we want the bitmask corresponding to the substring 'jj'. This substring can be written as the difference of two prefixes: 'jj' = 'accjj' - 'acc'.
In the same way, we want to subtract the counts for the two prefix strings. However, we only have the bitmasks telling us whether each letter has an even or odd frequency. In the arithmetic of bitmasks, we treat each position mod 2, so coordinate-wise subtraction becomes XOR.
This means counts(jj) = counts(accjj) - counts(acc) becomes
bitmask(jj) = bitmask(accjj) ^ bitmask(acc).
There's still a problem: the algorithm I've described is still quadratic. If, at every prefix, we iterate over all previous prefix-bitmasks and check if our mask XOR the old mask is one of the 11 goal-bitmasks, we still have a quadratic runtime. Instead, you can use the fact that XOR is its own inverse: if a ^ b = c, then b = a ^ c. So, instead of doing XORs with old prefix masks, you XOR with the 11 goal masks and add the number of times we've seen that mask: ans+=mp[xorsum] counts the substrings ending at our current index whose bitmask is xorsum ^ 0000000000 = xorsum. The loop after that counts substrings whose bitmask is one of the ten goal bitmasks.
Lastly, you just have to add your current prefix-mask to update the counts: mp[xorsum]++.
I have a collection S, typically containing 10-50 long strings. For illustrative purposes, suppose the length of each string ranges between 1000 and 10000 characters.
I would like to find strings of specified length k (typically in the range of 5 to 20) that are substrings of every string in S. This can obviously be done using a naive approach - enumerating every k-length substring in S[0] and checking if they exist in every other element of S.
Are there more efficient ways of approaching the problem? As far as I can tell, there are some similarities between this and the longest common subsequence problem, but my understanding of LCS is limited and I'm not sure how it could be adapted to the situation where we bound the desired common substring length to k, or if subsequence techniques can be applied to finding substrings.
Here's one fairly simple algorithm, which should be reasonably fast.
Using a rolling hash as in the Rabin-Karp string search algorithm, construct a hash table H0 of all the |S0|-k+1 length k substrings of S0. That's roughly O(|S0|) since each hash is computed in O(1) from the previous hash, but it will take longer if there are collisions or duplicate substrings. Using a better hash will help you with collisions but if there are a lot of k-length duplicate substrings in S0 then you could end up using O(k|S0|).
Now use the same rolling hash on S1. This time, look each substring up in H0 and if you find it, remove it from H0 and insert it into a new table H1. Again, this should be around O(|S1|) unless you have some pathological case, like both S0 and S1 are just long repetitions of the same character. (It's also going to be suboptimal if S0 and S0 are the same string, or have lots of overlapping pieces.)
Repeat step 2 for each Si, each time creating a new hash table. (At the end of each iteration of step 2, you can delete the hash table from the previous step.)
At the end, the last hash table will contain all the common k-length substrings.
The total run time should be about O(Σ|Si|) but in the worst case it could be O(kΣ|Si|). Even so, with the problem size as described, it should run in acceptable time.
Some thoughts (N is number of strings, M is average length, K is needed substring size):
Approach 1:
Walk through all strings, computing rolling hash for k-length strings and storing these hashes in the map (store tuple {key: hash; string_num; position})
time O(NxM), space O(NxM)
Extract groups with equal hash, check step-by-step:
1) that size of group >= number of strings
2) all strings are represented in this group 3
3) thorough checking of real substrings for equality (sometimes hashes of distinct substrings might coincide)
Approach 2:
Build suffix array for every string
time O(N x MlogM) space O(N x M)
Find intersection of suffix arrays for the first string pair, using merge-like approach (suffixes are sorted), considering only part of suffixes of length k, then continue with the next string and so on
I would treat each long string as a collection of overlapped short strings, so ABCDEFGHI becomes ABCDE, BCDEF, CDEFG, DEFGH, EFGHI. You can represent each short string as a pair of indexes, one specifying the long string and one the starting offset in that string (if this strikes you as naive, skip to the end).
I would then sort each collection into ascending order.
Now you can find the short strings common to the first two collection by merging the sorted lists of indexes, keeping only those from the first collection which are also present in the second collection. Check the survivors of this against the third collection, and so on and the survivors at the end correspond to those short strings which are present in all long strings.
(Alternatively you could maintain a set of pointers into each sorted list and repeatedly look to see if every pointer points at short strings with the same text, then advancing the pointer which points at the smallest short string).
Time is O(n log n) for the initial sort, which dominates. In the worst case - e.g. when every string is AAAAAAAA..AA - there is a factor of k on top of this, because all string compares check all characters and take time k. Hopefully, there is a clever way round this with https://en.wikipedia.org/wiki/Suffix_array which allows you to sort in time O(n) rather than O(nk log n) and the https://en.wikipedia.org/wiki/LCP_array, which should allow you to skip some characters when comparing substrings from different suffix arrays.
Thinking about this again, I think the usual suffix array trick of concatenating all of the strings in question, separated by a character not found in any of them, works here. If you look at the LCP of the resulting suffix array you can split it into sections, splitting at points where where the difference between suffixes occurs less than k characters in. Now each offset in any particular section starts with the same k characters. Now look at the offsets in each section and check to see if there is at least one offset from every possible starting string. If so, this k-character sequence occurs in all starting strings, but not otherwise. (There are suffix array constructions which work with arbitrarily large alphabets so you can always expand your alphabet to produce a character not in any string, if necessary).
I would try a simple method using HashSets:
Build a HashSet for each long string in S with all its k-strings.
Sort the sets by number of elements.
Scan the first set.
Lookup the term in the other sets.
The first step takes care of repetitions in each long string.
The second ensures the minimum number of comparisons.
let getHashSet k (lstr:string) =
let strs = System.Collections.Generic.HashSet<string>()
for i in 0..lstr.Length - k do
strs.Add lstr.[i..i + k - 1] |> ignore
strs
let getCommons k lstrs =
let strss = lstrs |> Seq.map (getHashSet k) |> Seq.sortBy (fun strs -> strs.Count)
match strss |> Seq.tryHead with
| None -> [||]
| Some h ->
let rest = Seq.tail strss |> Seq.toArray
[| for s in h do
if rest |> Array.forall (fun strs -> strs.Contains s) then yield s
|]
Test:
let random = System.Random System.DateTime.Now.Millisecond
let generateString n =
[| for i in 1..n do
yield random.Next 20 |> (+) 65 |> System.Convert.ToByte
|] |> System.Text.Encoding.ASCII.GetString
[ for i in 1..3 do yield generateString 10000 ]
|> getCommons 4
|> fun l -> printfn "found %d\n %A" l.Length l
result:
found 40
[|"PPTD"; "KLNN"; "FTSR"; "CNBM"; "SSHG"; "SHGO"; "LEHS"; "BBPD"; "LKQP"; "PFPH";
"AMMS"; "BEPC"; "HIPL"; "PGBJ"; "DDMJ"; "MQNO"; "SOBJ"; "GLAG"; "GBOC"; "NSDI";
"JDDL"; "OOJO"; "NETT"; "TAQN"; "DHME"; "AHDR"; "QHTS"; "TRQO"; "DHPM"; "HIMD";
"NHGH"; "EARK"; "ELNF"; "ADKE"; "DQCC"; "GKJA"; "ASME"; "KFGM"; "AMKE"; "JJLJ"|]
Here it is in fiddle: https://dotnetfiddle.net/ZK8DCT
Suppose we are given two strings s1 and s2(both lowercase). We have two find the minimal lexographic string that can be formed by merging two strings.
At the beginning , it looks prettty simple as merge of the mergesort algorithm. But let us see what can go wrong.
s1: zyy
s2: zy
Now if we perform merge on these two we must decide which z to pick as they are equal, clearly if we pick z of s2 first then the string formed will be:
zyzyy
If we pick z of s1 first, the string formed will be:
zyyzy which is correct.
As we can see the merge of mergesort can lead to wrong answer.
Here's another example:
s1:zyy
s2:zyb
Now the correct answer will be zybzyy which will be got only if pick z of s2 first.
There are plenty of other cases in which the simple merge will fail. My question is Is there any standard algorithm out there used to perform merge for such output.
You could use dynamic programming. In f[x][y] store the minimal lexicographical string such that you've taken x charecters from the first string s1 and y characters from the second s2. You can calculate f in bottom-top manner using the update:
f[x][y] = min(f[x-1][y] + s1[x], f[x][y-1] + s2[y]) \\ the '+' here represents
\\ the concatenation of a
\\ string and a character
You start with f[0][0] = "" (empty string).
For efficiency you can store the strings in f as references. That is, you can store in f the objects
class StringRef {
StringRef prev;
char c;
}
To extract what string you have at certain f[x][y] you just follow the references. To udapate you point back to either f[x-1][y] or f[x][y-1] depending on what your update step says.
It seems that the solution can be almost the same as you described (the "mergesort"-like approach), except that with special handling of equality. So long as the first characters of both strings are equal, you look ahead at the second character, 3rd, etc. If the end is reached for some string, consider the first character of the other string as the next character in the string for which the end is reached, etc. for the 2nd character, etc. If the ends for both strings are reached, then it doesn't matter from which string to take the first character. Note that this algorithm is O(N) because after a look-ahead on equal prefixes you know the whole look-ahead sequence (i.e. string prefix) to include, not just one first character.
EDIT: you look ahead so long as the current i-th characters from both strings are equal and alphabetically not larger than the first character in the current prefix.
I have given n strings . I have to find a string S so that, given n strings are sub-sequence of S.
For example, I have given the following 5 strings:
AATT
CGTT
CAGT
ACGT
ATGC
Then the string is "ACAGTGCT" . . Because, ACAGTGCT contains all given strings as super-sequence.
To solve this problem I have to know the algorithm . But I have no idea how to solve this . Guys, can you help me by telling technique to solve this problem ?
This is a NP-complete problem called multiple sequence alignment.
The wiki page describes solution methods such as dynamic programming which will work for small n, but becomes prohibitively expensive for larger n.
The basic idea is to construct an array f[a,b,c,...] representing the length of the shortest string S that generates "a" characters of the first string, "b" characters of the second, and "c" characters of the third.
My Approach: using Trie
Building a Trie from the given words.
create empty string (S)
create empty string (prev)
for each layer in the trie
create empty string (curr)
for each character used in the current layer
if the character not used in the previous layer (not in prev)
add the character to S
add the character to curr
prev = curr
Hope this helps :)
1 Definitions
A sequence of length n is a concatenation of n symbols taken from an alphabet .
If S is a sequence of length n and T is a sequence of length m and n m then S is a subsequence of T if S can be obtained by deleting m-n symbols from T. The symbols need not be contiguous.
A sequence T of length m is a supersequence of S of length n if T can be obtained by inserting m-n symbols. That is, T is a supersequence of S if and only if S is a subsequence of T.
A sequence T is a common supersequence of the sequences S1 and S2 of T is a supersequence of both S1 and S2.
2 The problem
The problem is to find a shortest common supersequence (SCS), which is a common supersequence of minimal length. There could be more than one SCS for a given problem.
2.1 Example
S= {a, b, c}
S1 = bcb
S2 = baab
S3 = babc
One shortest common supersequence is babcab (babacb, baabcb, bcaabc, bacabc, baacbc).
3 Techniques
Dynamic programming Requires too much memory unless the number of input-sequences are very small.
Branch and bound Requires too much time unless the alphabet is very small.
Majority merge The best known heuristic when the number of sequences is large compared to the alphabet size. [1]
Greedy (take two sequences and replace them by their optimal shortest common supersequence until a single string is left) Worse than majority merge. [1]
Genetic algorithms Indications that it might be better than majority merge. [1]
4 Implemented heuristics
4.1 The trivial solution
The trivial solution is at most || times the optimal solution length and is obtained by concatenating the concatenation of all characters in sigma as many times as the longest sequence. That is, if = {a, b, c} and the longest input sequence is of length 4 we get abcabcabcabc.
4.2 Majority merge heuristic
The Majority merge heuristic builds up a supersequence from the empty sequence (S) in the following way:
WHILE there are non-empty input sequences
s <- The most frequent symbol at the start of non-empty input-sequences.
Add s to the end of S.
Remove s from the beginning of each input sequence that starts with s.
END WHILE
Majority merge performs very well when the number of sequences is large compared to the alphabet size.
5 My approach - Local search
My approach was to apply a local search heuristic to the SCS problem and compare it to the Majority merge heuristic to see if it might do better in the case when the alphabet size is larger than the number of sequences.
Since the length of a valid supersequence may vary and any change to the supersequence may give an invalid string a direct representation of a supersequence as a feasible solution is not an option.
I chose to view a feasible solution (S) as a sequence of mappings x1...xSl where Sl is the sum of the lengths of all sequences and xi is a mapping to a sequencenumber and an index.
That means, if L={{s1,1...s1,m1}, {s2,1...s2,m2} ...{sn,1...s3,mn}} is the set of input sequences and L(i) is the ith sequence the mappings are represented like this:
xi {k, l}, where k L and l L(k)
To be sure that any solution is valid we need to introduce the following constraints:
Every symbol in every sequence may only have one xi mapped to it.
If xi ss,k and xj ss,l and k < l then i < j.
If xi ss,k and xj ss,l and k > l then i > j.
The second constraint enforces that the order of each sequence is preserved but not its position in S. If we have two mappings xi and xj then we may only exchange mappings between them if they map to different sequences.
5.1 The initial solution
There are many ways to choose an initial solution. As long as the order of the sequences are preserved it is valid. I chose not to in some way randomize a solution but try two very different solution-types and compare them.
The first one is to create an initial solution by simply concatenating all the sequences.
The second one is to interleave the sequences one symbol at a time. That is to start with the first symbol of every sequence then, in the same order, take the second symbol of every sequence and so on.
5.2 Local change and the neighbourhood
A local change is done by exchanging two mappings in the solution.
One way of doing the iteration is to go from i to Sl and do the best exchange for each mapping.
Another way is to try to exchange the mappings in the order they are defined by the sequences. That is, first exchange s1,1, then s2,1. That is what we do.
There are two variants I have tried.
In the first one, if a single mapping exchange does not yield a better value I return otherwise I go on.
In the second one, I seperately for each sequence do as many exchanges as there are sequences so a symbol in each sequence will have a possibility of moving. The exchange that gives the best value I keep and if that value is worse than the value of the last step in the algorithm I return otherwise I go on.
A symbol may move any number of position to the left or to the right as long as the exchange does not change the order of the original sequences.
The neighbourhood in the first variant is the number of valid exchanges that can be made for the symbol. In the second variant it is the sum of valid exchanges of each symbol after the previous symbol has been exchanged.
5.3 Evaluation
Since the length of the solution is always constant it has to be compressed before the real length of the solution may be obtained.
The solution S, which consists of mappings is converted to a string by using the symbols each mapping points to. A new, initialy empty, solution T is created. Then this algorithm is performed:
T = {}
FOR i = 0 TO Sl
found = FALSE
FOR j = 0 TO |L|
IF first symbol in L(j) = the symbol xi maps to THEN
Remove first symbol from L(j)
found = TRUE
END IF
END FOR
IF found = TRUE THEN
Add the symbol xi maps to to the end of T
END IF
END FOR
Sl is as before the sum of the lengths of all sequences. L is the set of all sequences and L(j) is sequence number j.
The value of the solution S is obtained as |T|.
With Many Many Thanks to : Andreas Westling
Given n strings S1, S2, ..., Sn, and an alphabet set A={a_1,a_2,....,a_m}. Assume that the alphabets in each string are all distinct. Now I want to create an inverted-index for each a_i (i=1,2...,m). My inverted-index has also something special: The alphabets in A are in some sequential order, if in the inverted-index a_i has included one string (say S_2), then a_j (j=i+1,i+2,...,m) don't need to include S_2 any more. In short, every string just appears in the inverted list only once. My question is how to build such list in a fast and efficient way? Any time complexity is bounded?
For example, A={a,b,e,g}, S1={abg}, S2={bg}, S3={gae}, S4={g}. Then my inverted-list should be:
a: S1,S3
b: S2 (since S1 has appeared previously, so we don't need to include it here)
e:
g: S4
If I understand your question correctly, a straightforward solution is:
for each string in n strings
find the "smallest" character in the string
put the string in the list for the character
The complexity is proportional to the total length of the strings, multiplying by a constant for the order testing.
If there is a simple way for testing, (e.g. the characters are in alphabetical order and all lower-case, a < will be enough), simply compare them; otherwise, I suggest using a hash table, each pair of which is a character and its order, later simply compare them.