Problem statement:
Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences.
Note:
The same word in the dictionary may be reused multiple times in the segmentation.
You may assume the dictionary does not contain duplicate words.
Sample test case:
Input:
s = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
Output:
[
"cats and dog",
"cat sand dog"
]
My Solution:
class Solution {
unordered_set<string> words;
unordered_map<string, vector<string> > memo;
public:
vector<string> getAllSentences(string s) {
if(s.size()==0){
return {""};
}
if(memo.count(s)) {
return memo[s];
}
string curWord = ""; vector<string> result;
for(int i = 0; i < s.size(); i++ ) {
curWord+=s[i];
if(words.count(curWord)) {
auto sentences = getAllSentences(s.substr(i+1));
for(string s : sentences) {
string sentence = curWord + ((int)s.size()>0? ((" ") + s) : "");
result.push_back(sentence);
}
}
}
return memo[s] = result;
}
vector<string> wordBreak(string s, vector<string>& wordDict) {
for(auto word : wordDict) {
words.insert(word);
}
return getAllSentences(s);
}
};
I am not sure about the time and space complexity. I think it should be 2^n where n is the length of given string s. Can anyone please help me to prove time and space complexity?
I have also some following questions:
If I don't use memo in the getAllSentences function what will be the
time complexity in this case?
Is there any better solution than this?
Let's try to go through the algorithm step by step but for specific wordDict to simplify the things.
So let wordDict be all the characters from a to z,
wordDict = ["a",..., "z"]
In this case if(words.count(curWord)) would be true every time when i = 0 and false otherwise.
Also, let's skip using memo cache (we'll add it later).
In the case above, we just got though string s recursively until we reach the end without any additional memory except result vector which gives the following:
time complexity is O(n!)
space complexity is O(1) - just 1 solution exists
where n - lenght of s
Now let's examine how using memo cache changes the situation in our case. Cache would contain n items - size of our string s which changes space complexity to O(n). Our time is the same since every there will be no hits by using memo cache.
This is the basis for us to move forward.
Now let's try to find how the things are changed if wordDict contains all the pairs of letters (and length of s is 2*something, so we could reach the end).
So, wordDict = ['aa','ab',...,'zz']
In this case we move forward with for 2 letters instead of 1 and everything else is the same, which gives us the following complexity withoug using memo cache:
time complexity is O((n/2)!)
space complexity is O(1) - just 1 solution exists
Memo cache would contain (n/2) items, giving a complexity of O(n) which also changes space complexity to O(n) but all the checks there are of different length.
Let's now imagine that wordDict contains both dictionaries we mentioned before ('a'...'z','aa'...'zz').
In this case we have the following complexity without using memo cache
time complexity is O((n)!) as we need to check the case for i=0 and i=1 which roughly doubles the number of checks we need to do for each step but on the other size it reduces the number of checks we have to do later since we move forward by 2 letters instead of one (this is the trickiest part for me).
Space complexity is ~O(2^n) since every additional char doubles the number of results.
Now let's think of the memo cache we have. It would be usefull for every 3 letters, because for example '...ab c...' gives the same as '...a bc...', so it reduces the number of calculations by 2 at every step, so our complexity would be the following
time complexity is roughly O((n/2)!) and we need O(2*n)=O(n) memory to store the memo. Let's also remember that in n/2 expression 2 reflects the cache effectiveness.
space complexity is O(2^n) - 2 here is a charateristic of the wordDict we've constructed
These were 3 cases for us to understand how the complexity is changing depending of the curcumstances. Now let's try to generalize it to the generic case:
time complexity is O((n/(l*e))!) where l = min length of words in wordDict, e - cache effectiveness (I would assume it 1 in general case but there might bt situations where it's different as we saw in the case above
space complexity is O(a^n) where a is a similarity of words in our wordDict, could be very very roughly estimated as P(h/l)=(h/l)! where h is max word length in a dictionary and l is min word length as (for example, if wordDict contains all combinations of up 3 letters, this gives us 3! combinations for every 6 letters)
This is how I see your approach and it's complexity.
As for improving the solution itself, I don't see any simple way to improve it. There might be an alternative way to divide the string in 3 parts and then processing each part separately but it would definitely work if we could get rid of searching the results and just count the number of results without displaying them.
I hope it helps.
I have written this recursive function to find the palindrome.
def palindrome(string):
print("palindrome called with:"+string)
if(len(string)<=3):
return string[0]==string[-1]
else:
res=palindrome(string[1:-1])
print("palindrome returned:"+str(res))
return res
I have yo find the time complexity of this algorithm of it now.
My questions
is my base case right? which is len<=3?
I'm unable to relate this to classic examples of fibonacci and factorial algorithm which are everywhere on internet.
Yes, the fact is that only your base case is correct.
So what you should be doing here is, check if the first and last character are same, then check if the remaining string is also a palindrome.
But at no point you are checking that.
So, with minimal changes in your code, the following solution will work, this will fail if an empty string is a passed as an argument.
def palindrome(string):
print("palindrome called with:"+string)
if(len(string)<=3):
return string[0]==string[-1]
else:
if string[0] == string[-1]:
res=palindrome(string[1:-1])
print("palindrome returned:"+str(res))
return res
else:
return False
Definitely, there are better ways of writing this.
def palindrome(s):
return s == '' or (s[0]==s[-1] and palindrome(s[1:-1]))
All I have done is, further reduced your base case, by letting it make two more recursive calls.
Now, coming to the time complexity, which is same for both the codes.
In one function call, we are doing an O(1) operation of comparing the first and last character. And this recursive call is being done for at most n/2 times. n/2 because, in a string of length n, we are removing 2 characters in each call. Thus, the overall complexity will be O(n).(Just mind that, this ignores the string copying/slicing in every recursive call.)
Finally, you should avoid this recursively, as we are making a new string(at the time of slicing) before every recursive call.
def palindrome(s):
def _palindrome(string, i, j):
if i >= j:
return True
return string[i] == string[j] and _palindrome(string, i + 1, j - 1)
return _palindrome(s, 0, len(s) - 1)
This will not make copy at every call. Thus, is definitely an O(n) solution.
I would like an algorithm/function that, given a number N, generates random numbers from 0 to N - 1 in constant time. After the Nth call, the function may do as it pleases. Also, it is important that the algorithm generates the numbers when requested rather than using shuffling, because I may (and in the average case do) not need the full list of numbers. What is the best approach to take?
(optional to read) I thought of having a hash set of numbers, and pulling the numbers out one at a time, but this would require inserting all elements (which I often do not need) into the hash set first... this will not work... Argh
Thanks for any help in advance.
Modify the Fisher–Yates shuffle by replacing the array with a map that stores only the elements that have been moved. In Python:
import random
class Shuffle:
def __init__(self, n):
self.d = {}
self.n = n
def generate(self):
i = random.randrange(self.n)
self.n -= 1
di = self.d[i] if i in self.d else i # idiomatically, self.d.get(i, i)
dn = self.d[self.n] if self.n in self.d else self.n
self.d[i] = dn
self.d[self.n] = di
return di
The space usage and amortized expected running time is O(1) words per element actually generated. Up to log factors, this is optimal.
Say I have an array of arrays in Ruby,
[[100,300],
[400,500]]
that I'm building by adding successive lines of CSV data.
What's the best way, when adding a new subarray, to test if the range covered by the two numbers in the subarray is covered by any previous subarrays?
In other words, each subarray comprises a linear range (100-300 and 400-500) in the example above. If I want an exception to be thrown if I tried to add [499,501], for example, to the array because there would be overlap, how could I best test for this?
Since your subarrays are supposed to represent ranges, it might be a good idea to actually use an array of ranges instead of an array of array.
So your array becomes [100..300, 400..500].
For two ranges, we can easily define a method which checks whether two ranges overlap:
def overlap?(r1, r2)
r1.include?(r2.begin) || r2.include?(r1.begin)
end
Now to check whether a range r overlaps with any range in your array of ranges, you just need to check whether overlap?(r, r2) is true for any r2 in the array of ranges:
def any_overlap?(r, ranges)
ranges.any? do |r2|
overlap?(r, r2)
end
end
Which can be used like this:
any_overlap?(499..501, [100..300, 400..500])
#=> true
any_overlap?(599..601, [100..300, 400..500])
#=> false
Here any_overlap? takes O(n) time. So if you use any_overlap? every time you add a range to the array, the whole thing will be O(n**2).
However there's a way to do what you want without checking each range:
You add all the ranges to the array without checking for overlap. Then you check whether any range in the array overlaps with any other. You can do this efficiently in O(n log n) time by sorting the array on the beginning of each range and then testing whether two adjacent ranges overlap:
def any_overlap?(ranges)
ranges.sort_by(&:begin).each_cons(2).any? do |r1,r2|
overlap?(r1, r2)
end
end
Use multi_range and call overlaps? method to check whether there is any overlaps:
MultiRange.new([100..300, 400..500]).overlaps?(499..501)
Note that you may need to transform your input from array of arrays to array of ranges before using it. Possible working example is:
arrays = [[100, 300], [400, 500]]
subarray = [499, 501]
ranges = arrays.map{|a, b| a..b }
MultiRange.new(ranges).overlaps?(subarray[0].. subarray[1])
# => true
I would like to randomly iterate through a range. Each value will be visited only once and all values will eventually be visited. For example:
class Array
def shuffle
ret = dup
j = length
i = 0
while j > 1
r = i + rand(j)
ret[i], ret[r] = ret[r], ret[i]
i += 1
j -= 1
end
ret
end
end
(0..9).to_a.shuffle.each{|x| f(x)}
where f(x) is some function that operates on each value. A Fisher-Yates shuffle is used to efficiently provide random ordering.
My problem is that shuffle needs to operate on an array, which is not cool because I am working with astronomically large numbers. Ruby will quickly consume a large amount of RAM trying to create a monstrous array. Imagine replacing (0..9) with (0..99**99). This is also why the following code will not work:
tried = {} # store previous attempts
bigint = 99**99
bigint.times {
x = rand(bigint)
redo if tried[x]
tried[x] = true
f(x) # some function
}
This code is very naive and quickly runs out of memory as tried obtains more entries.
What sort of algorithm can accomplish what I am trying to do?
[Edit1]: Why do I want to do this? I'm trying to exhaust the search space of a hash algorithm for a N-length input string looking for partial collisions. Each number I generate is equivalent to a unique input string, entropy and all. Basically, I'm "counting" using a custom alphabet.
[Edit2]: This means that f(x) in the above examples is a method that generates a hash and compares it to a constant, target hash for partial collisions. I do not need to store the value of x after I call f(x) so memory should remain constant over time.
[Edit3/4/5/6]: Further clarification/fixes.
[Solution]: The following code is based on #bta's solution. For the sake of conciseness, next_prime is not shown. It produces acceptable randomness and only visits each number once. See the actual post for more details.
N = size_of_range
Q = ( 2 * N / (1 + Math.sqrt(5)) ).to_i.next_prime
START = rand(N)
x = START
nil until f( x = (x + Q) % N ) == START # assuming f(x) returns x
I just remembered a similar problem from a class I took years ago; that is, iterating (relatively) randomly through a set (completely exhausting it) given extremely tight memory constraints. If I'm remembering this correctly, our solution algorithm was something like this:
Define the range to be from 0 to
some number N
Generate a random starting point x[0] inside N
Generate an iterator Q less than N
Generate successive points x[n] by adding Q to
the previous point and wrapping around if needed. That
is, x[n+1] = (x[n] + Q) % N
Repeat until you generate a new point equal to the starting point.
The trick is to find an iterator that will let you traverse the entire range without generating the same value twice. If I'm remembering correctly, any relatively prime N and Q will work (the closer the number to the bounds of the range the less 'random' the input). In that case, a prime number that is not a factor of N should work. You can also swap bytes/nibbles in the resulting number to change the pattern with which the generated points "jump around" in N.
This algorithm only requires the starting point (x[0]), the current point (x[n]), the iterator value (Q), and the range limit (N) to be stored.
Perhaps someone else remembers this algorithm and can verify if I'm remembering it correctly?
As #Turtle answered, you problem doesn't have a solution. #KandadaBoggu and #bta solution gives you random numbers is some ranges which are or are not random. You get clusters of numbers.
But I don't know why you care about double occurence of the same number. If (0..99**99) is your range, then if you could generate 10^10 random numbers per second (if you have a 3 GHz processor and about 4 cores on which you generate one random number per CPU cycle - which is imposible, and ruby will even slow it down a lot), then it would take about 10^180 years to exhaust all the numbers. You have also probability about 10^-180 that two identical numbers will be generated during a whole year. Our universe has probably about 10^9 years, so if your computer could start calculation when the time began, then you would have probability about 10^-170 that two identical numbers were generated. In the other words - practicaly it is imposible and you don't have to care about it.
Even if you would use Jaguar (top 1 from www.top500.org supercomputers) with only this one task, you still need 10^174 years to get all numbers.
If you don't belive me, try
tried = {} # store previous attempts
bigint = 99**99
bigint.times {
x = rand(bigint)
puts "Oh, no!" if tried[x]
tried[x] = true
}
I'll buy you a beer if you will even once see "Oh, no!" on your screen during your life time :)
I could be wrong, but I don't think this is doable without storing some state. At the very least, you're going to need some state.
Even if you only use one bit per value (has this value been tried yes or no) then you will need X/8 bytes of memory to store the result (where X is the largest number). Assuming that you have 2GB of free memory, this would leave you with more than 16 million numbers.
Break the range in to manageable batches as shown below:
def range_walker range, batch_size = 100
size = (range.end - range.begin) + 1
n = size/batch_size
n.times do |i|
x = i * batch_size + range.begin
y = x + batch_size
(x...y).sort_by{rand}.each{|z| p z}
end
d = (range.end - size%batch_size + 1)
(d..range.end).sort_by{rand}.each{|z| p z }
end
You can further randomize solution by randomly choosing the batch for processing.
PS: This is a good problem for map-reduce. Each batch can be worked by independent nodes.
Reference:
Map-reduce in Ruby
you can randomly iterate an array with shuffle method
a = [1,2,3,4,5,6,7,8,9]
a.shuffle!
=> [5, 2, 8, 7, 3, 1, 6, 4, 9]
You want what's called a "full cycle iterator"...
Here is psudocode for the simplest version which is perfect for most uses...
function fullCycleStep(sample_size, last_value, random_seed = 31337, prime_number = 32452843) {
if last_value = null then last_value = random_seed % sample_size
return (last_value + prime_number) % sample_size
}
If you call this like so:
sample = 10
For i = 1 to sample
last_value = fullCycleStep(sample, last_value)
print last_value
next
It would generate random numbers, looping through all 10, never repeating If you change random_seed, which can be anything, or prime_number, which must be greater than, and not be evenly divisible by sample_size, you will get a new random order, but you will still never get a duplicate.
Database systems and other large-scale systems do this by writing the intermediate results of recursive sorts to a temp database file. That way, they can sort massive numbers of records while only keeping limited numbers of records in memory at any one time. This tends to be complicated in practice.
How "random" does your order have to be? If you don't need a specific input distribution, you could try a recursive scheme like this to minimize memory usage:
def gen_random_indices
# Assume your input range is (0..(10**3))
(0..3).sort_by{rand}.each do |a|
(0..3).sort_by{rand}.each do |b|
(0..3).sort_by{rand}.each do |c|
yield "#{a}#{b}#{c}".to_i
end
end
end
end
gen_random_indices do |idx|
run_test_with_index(idx)
end
Essentially, you are constructing the index by randomly generating one digit at a time. In the worst-case scenario, this will require enough memory to store 10 * (number of digits). You will encounter every number in the range (0..(10**3)) exactly once, but the order is only pseudo-random. That is, if the first loop sets a=1, then you will encounter all three-digit numbers of the form 1xx before you see the hundreds digit change.
The other downside is the need to manually construct the function to a specified depth. In your (0..(99**99)) case, this would likely be a problem (although I suppose you could write a script to generate the code for you). I'm sure there's probably a way to re-write this in a state-ful, recursive manner, but I can't think of it off the top of my head (ideas, anyone?).
[Edit]: Taking into account #klew and #Turtle's answers, the best I can hope for is batches of random (or close to random) numbers.
This is a recursive implementation of something similar to KandadaBoggu's solution. Basically, the search space (as a range) is partitioned into an array containing N equal-sized ranges. Each range is fed back in a random order as a new search space. This continues until the size of the range hits a lower bound. At this point the range is small enough to be converted into an array, shuffled, and checked.
Even though it is recursive, I haven't blown the stack yet. Instead, it errors out when attempting to partition a search space larger than about 10^19 keys. I has to do with the numbers being too large to convert to a long. It can probably be fixed:
# partition a range into an array of N equal-sized ranges
def partition(range, n)
ranges = []
first = range.first
last = range.last
length = last - first + 1
step = length / n # integer division
((first + step - 1)..last).step(step) { |i|
ranges << (first..i)
first = i + 1
}
# append any extra onto the last element
ranges[-1] = (ranges[-1].first)..last if last > step * ranges.length
ranges
end
I hope the code comments help shed some light on my original question.
pastebin: full source
Note: PW_LEN under # options can be changed to a lower number in order to get quicker results.
For a prohibitively large space, like
space = -10..1000000000000000000000
You can add this method to Range.
class Range
M127 = 170_141_183_460_469_231_731_687_303_715_884_105_727
def each_random(seed = 0)
return to_enum(__method__) { size } unless block_given?
unless first.kind_of? Integer
raise TypeError, "can't randomly iterate from #{first.class}"
end
sample_size = self.end - first + 1
sample_size -= 1 if exclude_end?
j = coprime sample_size
v = seed % sample_size
each do
v = (v + j) % sample_size
yield first + v
end
end
protected
def gcd(a,b)
b == 0 ? a : gcd(b, a % b)
end
def coprime(a, z = M127)
gcd(a, z) == 1 ? z : coprime(a, z + 1)
end
end
You could then
space.each_random { |i| puts i }
729815750697818944176
459631501395637888351
189447252093456832526
919263002791275776712
649078753489094720887
378894504186913665062
108710254884732609237
838526005582551553423
568341756280370497598
298157506978189441773
27973257676008385948
757789008373827330134
487604759071646274309
217420509769465218484
947236260467284162670
677052011165103106845
406867761862922051020
136683512560740995195
866499263258559939381
596315013956378883556
326130764654197827731
55946515352016771906
785762266049835716092
515578016747654660267
...
With a good amount of randomness so long as your space is a few orders smaller than M127.
Credit to #nick-steele and #bta for the approach.
This isn't really a Ruby-specific answer but I hope it's permitted. Andrew Kensler gives a C++ "permute()" function that does exactly this in his "Correlated Multi-Jittered Sampling" report.
As I understand it, the exact function he provides really only works if your "array" is up to size 2^27, but the general idea could be used for arrays of any size.
I'll do my best to sort of explain it. The first part is you need a hash that is reversible "for any power-of-two sized domain". Consider x = i + 1. No matter what x is, even if your integer overflows, you can determine what i was. More specifically, you can always determine the bottom n-bits of i from the bottom n-bits of x. Addition is a reversible hash operation, as is multiplication by an odd number, as is doing a bitwise xor by a constant. If you know a specific power-of-two domain, you can scramble bits in that domain. E.g. x ^= (x & 0xFF) >> 5) is valid for the 16-bit domain. You can specify that domain with a mask, e.g. mask = 0xFF, and your hash function becomes x = hash(i, mask). Of course you can add a "seed" value into that hash function to get different randomizations. Kensler lays out more valid operations in the paper.
So you have a reversible function x = hash(i, mask, seed). The problem is that if you hash your index, you might end up with a value that is larger than your array size, i.e. your "domain". You can't just modulo this or you'll get collisions.
The reversible hash is the key to using a technique called "cycle walking", introduced in "Ciphers with Arbitrary Finite Domains". Because the hash is reversible (i.e. 1-to-1), you can just repeatedly apply the same hash until your hashed value is smaller than your array! Because you're applying the same hash, and the mapping is one-to-one, whatever value you end up on will map back to exactly one index, so you don't have collisions. So your function could look something like this for 32-bit integers (pseudocode):
fun permute(i, length, seed) {
i = hash(i, 0xFFFF, seed)
while(i >= length): i = hash(i, 0xFFFF, seed)
return i
}
It could take a lot of hashes to get to your domain, so Kensler does a simple trick: he keeps the hash within the domain of the next power of two, which makes it require very few iterations (~2 on average), by masking out the unnecessary bits. The final algorithm looks like this:
fun next_pow_2(length) {
# This implementation is for clarity.
# See Kensler's paper for one way to do it fast.
p = 1
while (p < length): p *= 2
return p
}
permute(i, length, seed) {
mask = next_pow_2(length)-1
i = hash(i, mask, seed) & mask
while(i >= length): i = hash(i, mask, seed) & mask
return i
}
And that's it! Obviously the important thing here is choosing a good hash function, which Kensler provides in the paper but I wanted to break down the explanation. If you want to have different random permutations each time, you can add a "seed" value to the permute function which then gets passed to the hash function.