I would like to generate a Search tree for a permutation problem. My requirement is as follows: I want to use a Divide and Conquer strategy for doing so
I am giving an example tree length 3 Permutation.
Given a set of n numbers, divide the problem into n subproblems, each having one of the numbers from the set as the first number and the chosen number removed from the set. For each subproblem, repeat the process. If set is empty, stop.
Related
I encountered the following problem for which I couldn't quite find the appropriate solution.
The problem says for a given string having a specific hash value, find the lowest string (which is not the same as the given one) of the
same length and same hash value (if one exists). E.g. For the
following value mapping of alphabets: {a:0, b:1, c:2,...,z:25}
If the given string is: ady with hash value - 27. The
lexicographically smallest one (from all possible ones excluding the
given one) would be: acz
Solution approach I could think of:
I reduced the problem to Coin-Change problem and resorted to finding all possible combinations for the given sum. Out of all the obtained solutions, I sort them up and find the lowest (or the next smallest if the given string is smallest).
The problem however lies with finding all possible solutions (even in a DP approach) which might be inefficient for larger inputs.
My doubt is:
What solution strategy (possibly even Greedy) could give a better time complexity than above?
I cannot guarantee that this will give you a lower complexity, but a couple of things:1 you don't need to check all the space, just the space of lexicographic value less than or equal to the given string. 2: you can formulate it as an integer programming problem:
Assuming your character space is the letters, and each letter is given its number index[0-25] so a corresponds to 0, b to 1 and so forth. let x_i be the number of letters in your string corresponding to index i. You can formulate your problem as:
min sum_i(wi*xi)
st xi*ai = M
xi>=0,
sum_i(xi)=n
sum_i(wi*xi)<= N
xi integer
Where wi= 26^i, ai is equal to hash(letter(i)), n is the number of letters of the original string, N is the hash value of the original string. This is an integer programming problem so you can try plugging it to a solver. The original problem is very similar to subset sum problem with fixed subset size (where the hash values are the elements you are summing over, and the subset size is the length of the string) so you might also want to take a look at that, although as you will see from the answer it is a complicated problem.
How would you efficiently solve this problem
Suppose we were given a list of words [“apple”, “banana”, “mango”]
If we are given a word in the list that is one typo away,
“Dpple”
“Adple”
“Appld”
We output true
If there is more than one typo, we output false.
For optimizations, I’ve tried storing the list in a hashtable containing the number of letters of each word and looking for the same number of letters upon the given input to reduce the size in which we look for our input. Is there a faster optimization we can make to this problem?
One possible optimisation would be to generate all one-typo words for the given list and put them in a map (or some better string lookup structure). Then lookup the given words - if found output true, else false. The total number of one-typo words is: 25*L, where L is the total number of letters in the input list (assuming case does not matter).
The interleaving rule is to form a new word by inserting one word into another, in a letter by letter fashion, like showing below:
a p p l e
o l d
=
aoplpdle
It does not matter which word goes first. (oalpdple is also valid)
The problem is given a vector of strings {old, apple, talk, aoplpdle, otladlk}, find all the words that are valid interleavings of two word from the vector.
The simplest solution asks for at least O(n^2) time complexity, taking every two word and form a interleaving word, check if it is in the vector.
Is there better solutions?
Sort by length. You only need to check combinations where the sum of lengths of 2 entries (words...) is equal to the length of existing entry(ies).
This will reduce your average complexity. I didn't take the time to compute the worst complexity, but it's probably lower then O(n^2) as well.
You can also optimize the "inner loop" by rejecting matches early - you don't really need to construct the entire interleaved word to reject a match - iterate the candidate word alongside the 2 input words till you find a mismatch. This won't reduce your worst complexity, but will have a positive effect on overall performance.
Suppose we have a string A of length n. And we have k <= n.
Now I want to know all distinct strings generated by cyclic shift of any segment of that string of length k, any number of times.
Ex: A = "asdfgh" and k=3.
Then possible permutations are "dasfgh" when segment "asd" is chosen for shifting. Now "dasfgh" can give another permutation "dfasgh" when segment "asf" is chosen.
I want to know if a specific given permutation can be formed or not by such shifts.
Can someone help me by providing some good algorithms or literature or link telling about best approach for solving such questions. I know backtracking can be used but it won't be efficient as n can be as large as 100000.
I am on an interview ride here. One more interview question I had difficulties with.
“A rose is a rose is a rose” Write an
algorithm that prints the number of
times a character/word occurs. E.g.
A – 3 Rose – 3 Is – 2
Also ensure that when you are printing
the results, they are in order of
what was present in the original
sentence. All this in order n.
I did get solution to count number of occurrences of each word in sentence in the order as present in the original sentence. I used Dictionary<string,int> to do it. However I did not understand what is meant by order of n. That is something I need an explanation from you guys.
There are 26 characters, So you can use counting sort to sort them, in your counting sort you can have an index which determines when specific character visited first time to save order of occurrence. [They can be sorted by their count and their occurrence with sort like radix sort].
Edit: by words first thing every one can think about it, is using Hash table and insert words in hash, and in this way count them, and They can be sorted in O(n), because all numbers are within 1..n steel you can sort them by counting sort in O(n), also for their occurrence you can traverse string and change position of same values.
Order of n means you traverse the string only once or some lesser multiple of n ,where n is number of characters in the string.
So your solution to store the String and number of its occurences is O(n) , order of n, as you loop through the complete string only once.
However it uses extra space in form of the list you created.
Order N refers to the Big O computational complexity analysis where you get a good upper bound on algorithms. It is a theory we cover early in a Data Structures class, so we can torment, I mean help the student gain facility with it as we traverse in a balanced way, heaps of different trees of knowledge, all different. In your case they want your algorithm to grow in compute time proportional to the size of the text as it grows.
It's a reference to Big O notation. Basically the interviewer means that you have to complete the task with an O(N) algorithm.
"Order n" is referring to Big O notation. Big O is a way for mathematicians and computer scientists to describe the behavior of a function. When someone specifies searching a string "in order n", that means that the time it takes for the function to execute grows linearly as the length of that string increases. In other words, if you plotted time of execution vs length of input, you would see a straight line.
Saying that your function must be of Order n does not mean that your function must equal O(n), a function with a Big O less than O(n) would also be considered acceptable. In your problems case, this would not be possible (because in order to count a letter, you must "touch" that letter, thus there must be some operation dependent on the input size).
One possible method is to traverse the string linearly. Then create a hash and list. The idea is to use the word as the hash key and increment the value for each occurance. If the value is non-existent in the hash, add the word to the end of the list. After traversing the string, go through the list in order using the hash values as the count.
The order of the algorithm is O(n). The hash lookup and list add operations are O(1) (or very close to it).