Given two sorted array A and B length N. Each elements may contain natural number less than M. Determine all possible distances for all combinations elements A and B. In this case, if A[i] - B[j] < 0, then the distance is M + (A[i] - B[j]).
Example :
A = {0,2,3}
B = {1,2}
M = 5
Distances = {0,1,2,3,4}
Note: I know O(N^2) solution, but I need faster solution than O(N^2) and O(N x M).
Edit: Array A, B, and Distances contain distinct elements.
You can get a O(MlogM) complexity solution in the following way.
Prepare an array Ax of length M with Ax[i] = 1 if i belongs to A (and 0 otherwise)
Prepare an array Bx of length M with Bx[M-1-i] = 1 if i belongs to B (and 0 otherwise)
Use the Fast Fourier Transform to convolve these 2 sequences together
Inspect the output array, non-zero values correspond to possible distances
Note that the FFT is normally done with floating point numbers, so in step 4 you probably want to test if the output is greater than 0.5 to avoid potential rounding noise issues.
I possible done with optimized N*N.
If convert A to 0 and 1 array where 1 on positions which present in A (in range [0..M].
After convert this array into bitmasks, size of A array will be decreased into 64 times.
This will allow insert results by blocks of size 64.
Complexity still will be N*N but working time will be greatly decreased. As limitation mentioned by author 50000 for A and B sizes and M.
Expected operations count will be N*N/64 ~= 4*10^7. It will passed in 1 sec.
You can use bitvectors to accomplish this. Bitvector operations on large bitvectors is linear in the size of the bitvector, but is fast, easy to implement, and may work well given your 50k size limit.
Initialize two bitvectors of length M. Call these vectA and vectAnswer. Set the bits of vectA that correspond to the elements in A. Leave vectAnswer with all zeroes.
Define a method to rotate a bitvector by k elements (rotate down). I'll call this rotate(vect,k).
Then, for every element b of B, vectAnswer = vectAnswer | rotate(vectA,b).
Related
I have a mathematical/algorithmic problem here.
Given an array of numbers, find a way to separate it to 5 subarrays, so that sum of each subarrays is less than or equal to a given number. All numbers from the initial array, must go to one of the subarrays, and be part of one sum.
So the input to the algorithm would be:
d - representing the number that each subarrays sum has to be less or equal
A - representing the array of numbers that will be separated to different subarrays, and will be part of one sum
Algorithm complexity must be polynomial.
Thank you.
If by "subarray" you mean "subset" as opposed to "contiguous slice", it is impossible to find a polynomial time algorithm for this problem (unless P = NP). The Partition Problem is to partition a list of numbers into to sets such that the sum of both sets are equal. It is known to be NP-complete. The partition problem can be reduced to your problem as follows:
Suppose that x1, ..., x_n are positive numbers that you want to partition into 2 sets such that their sums are equal. Let d be this common sum (which would be the sum of the xi divided by 2). extend x_i to an array, A, of size n+3 by adding three copies of d. Clearly the only way to partition A into 5 subarrays so that the sum of each is less than or equal to d is if the sum of each actually equals d. This would in turn require 3 of the subarrays to have length 1, each consisting of the number d. The remaining 2 subarrays would be exactly a partition of the original n numbers.
On the other hand, if there are additional constraints on what the numbers are and/or the subarrays need to be, there might be a polynomial solution. But, if so, you should clearly spell out what there constraints are.
Set up of the problem:
d : the upper bound for the subarray
A : the initial array
Assuming A is not sorted.
(Heuristic)
Algorithm:
1.Sort A in ascending order using standard sorting algorithm->O(nlogn)
2.Check if the largest element of A is greater than d ->(constant)
if yes, no solution
if no, continue
3.Sum up all the element in A, denote S. Check if S/5 > d ->O(n)
if yes, no solution
if no, continue
4.Using greedy approach, create a new subarray Asi, add next biggest element aj in the sorted A to Asi so that the sum of Asi does not exceed d. Remove aj from sorted A ->O(n)
repeat step4 until either of the condition satisfied:
I.At creating subarray Asi, there are only 5-i element left
In this case, split the remaining element to individual subarray, done
II. i = 5. There are 5 subarray created.
The algorithm described above is bounded by O(nlogn) therefore in polynomial time.
We have an array consisting of each entry as a tuple of two integers. Let the array be A = [(a1, b1), (a2, b2), .... , (an, bn)]. Now we have multiple queries where we are given an integer x, we need to find the maximum value of ai + |x - bi| for 1 <= i <= n.
I understand this can be easily achieved in O(n) time complexity for each query but I am looking for something faster than that, probably O(log n) for each query. I can preprocess the array in O(n) time, but the queries should be done faster than O(n).
Any kind of help would be appreciated.
It seems to be way too easy to over-think this.
For n = 1, the function is v-shaped with a minimum of a1 at b1, with slopes of -1 and 1, respectively - let's call these values ac and bc (for combined).
For an additional pair (ai, bi), one of the pairs may dominate the other (|bc - bi| ≤ |ac - ai), which may then be ignored.
Otherwise, the falling slope of the combination will be from the pair with the larger b, the rising slope from the other.
The minimum will be between the individual b, closer to the b of the pair with the larger a, the distance being half the difference between the (absolute value of the) "coordinate" differences, the minimum value that amount higher.
The main catch is that neither needs to be an integer - the only alternative being exactly in the middle between two integers.
(Ending up with the falling slope from max ai + bi, and the rising slope of max ai - bi.)
I have a question in algorithm design about arrays, which should be implement in C language.
Suppose that we have an array which has n elements. For simplicity n is power of '2' like 1, 2, 4, 8, 16 , etc. I want to separate this to 2 parts with (n/2) elements. Condition of separating is lowest absolute difference between sum of all elements in two arrays for example if I have this array (9,2,5,3,6,1,4,7) it will be separate to these arrays (9,5,1,3) and (6,7,4,2) . summation of first array's elements is 18 and the summation of second array's elements is 19 and the difference is 1 and these two arrays are the answer but two arrays like (9,5,4,2) and (7,6,3,1) isn't the answer because the difference of element summation is 4 and we have found 1 . so 4 isn't the minimum difference. How to solve this?
Thank you.
This is the Partition Problem, which is unfortunately NP-Hard.
However, since your numbers are integers, if they are relatively low, there is a pseudo polynomial O(W*n^2) solution using Dynamic Programming (where W is sum of all elements).
The idea is to create the DP matrix of size (W/2+1)*(n+1)*(n/2+1), based on the following recursive formula:
D(0,i,0) = true
D(0,i,k) = false k != 0
D(x,i,k) = false x < 0
D(x,0,k) = false x > 0
D(x,i,0) = false x > 0
D(x,i,k) = D(x,i-1,k) OR D(x-arr[i], i-1,k-1)
The above gives a 3d matrix, where each entry D(x,i,k) says if there is a subset containing exactly k elements, that sums to x, and uses the first i elements as candidates.
Once you have this matrix, you just need to find the highest x (that is smaller than SUM/2) such that D(x,n,n/2) = true
Later, you can get the relevant subset by going back on the table and "retracing" your choices at each step. This thread deals with how it is done on a very similar problem.
For small sets, there is also the alternative of a naive brute force solution, which basically splits the array to all possible halves ((2n)!/(n!*n!) of those), and picks the best one out of them.
there is an array of numbers an this array is irregular and we should find a maximum number (n) that at least n number is bigger than it (this number may be in array and may not be in array )
for example if we give 2 5 7 6 9 number 4 is maximum number that at least 4 number (or more than it ) is bigger than 4 (5 6 7 9 are bigger)
i solve this problem but i think it gives time limit in big array of numbers so i want to resolve this problem in another way
so i use merge sort for sorting that because it take nlog(n) and then i use a counter an it counts from 1 to k if we have k number more than k we count again for example we count from 1 to 4 then in 5 we don't have 5 number more than 5 so we give k-1 = 4 and this is our n .
it's good or it maybe gives time limit ? does anybody have another idea ?
thanks
In c++ there is a function called std::nth_element and it can find the nth element of an array in linear time. Using this function you should find the N - n- th element (where N is the total number of elements in the array) and subtract 1 from it.
As you seek a solution in C you can not make use of this function, but you can implement your solution similarly. nth_element performs something quite similar to qsort, but it only performs partition on the part of the array where the n-th element is.
Now let's assume you have nth_element implemented. We will perform something like combination of binary search and nth_element. First we assume that the answer of the question is the middle element of the array (i.e. the N/2-th element). We use nth_element and we find the N/2th element. If it is more than N/2 we know the answer to your problem is at least N/2, otherwise it will be less. Either way in order to find the answer we will only continue with one of the two partitions created by the N/2th element. If this partition is the right one(elements bigger than N/2) we continue solving the same problem, otherwise we start searching for the max element M on the left of the N/2th element that has at least x bigger elements such that x + N/2 > M. The two subproblems will have the same complexity. You continue performing this operation until the interval you are interested in is of length 1.
Now let's prove the complexity of the above algorithm is linear. First nth_element is linear performing operations in the order of N, second nth_element that only considers one half of the array will perform operations in the order of N/2 the third - in the order of N/4 and so on. All in all you will perform operations in the order of N + N/2 + N/4 + ... + 1. This sum is less than 2 * N thus your complexity is still linear.
Your solution is asymptotically slower than what I propose above as it has a complexity O(n*log(n)), while my solution has complexity of O(n).
I would use a modified variant of a sorting algorithm that uses pivot values.
The reason is that you want to sort as few elements as possible.
So I would use qsort as my base algorithm and let the pivot element control which partition to sort (you will only need to sort one).
Description
Given an Array of size (n*k+b) where n elements occur k times and one element occurs b times, in other words there are n+1 distinct Elements. Given that 0 < b < k find the element occurring b times.
My Attempted solutions
Obvious solution will be using hashing but it will not work if the numbers are very large. Complexity is O(n)
Using map to store the frequencies of each element and then traversing map to find the element occurring b times.As Map's are implemented as height balanced trees Complexity will be O(nlogn).
Both of my solution were accepted but the interviewer wanted a linear solution without using hashing and hint he gave was make the height of tree constant in tree in which you are storing frequencies, but I am not able to figure out the correct solution yet.
I want to know how to solve this problem in linear time without hashing?
EDIT:
Sample:
Input: n=2 b=2 k=3
Aarray: 2 2 2 3 3 3 1 1
Output: 1
I assume:
The elements of the array are comparable.
We know the values of n and k beforehand.
A solution O(n*k+b) is good enough.
Let the number occuring only b times be S. We are trying to find the S in an array of n*k+b size.
Recursive Step: Find the median element of the current array slice as in Quick Sort in lineer time. Let the median element be M.
After the recursive step you have an array where all elements smaller than M occur on the left of the first occurence of M. All M elements are next to each other and all element larger than M are on the right of all occurences of M.
Look at the index of the leftmost M and calculate whether S<M or S>=M. Recurse either on the left slice or the right slice.
So you are doing a quick sort but delving only one part of the divisions at any time. You will recurse O(logN) times but each time with 1/2, 1/4, 1/8, .. sizes of the original array, so the total time will still be O(n).
Clarification: Let's say n=20 and k = 10. Then, there are 21 distinct elements in the array, 20 of which occur 10 times and the last occur let's say 7 times. I find the medium element, let's say it is 1111. If the S<1111 than the index of the leftmost occurence of 1111 will be less than 11*10. If S>=1111 then the index will be equal to 11*10.
Full example: n = 4. k = 3. Array = {1,2,3,4,5,1,2,3,4,5,1,2,3,5}
After the first recursive step I find the median element is 3 and the array is something like: {1,2,1,2,1,2,3,3,3,5,4,5,5,4} There are 6 elements on the left of 3. 6 is a multiple of k=3. So each element must be occuring 3 times there. So S>=3. Recurse on the right side. And so on.
An idea using cyclic groups.
To guess i-th bit of answer, follow this procedure:
Count how many numbers in array has i-th bit set, store as cnt
If cnt % k is non-zero, then i-th bit of answer is set. Otherwise it is clear.
To guess whole number, repeat the above for every bit.
This solution is technically O((n*k+b)*log max N), where max N is maximal value in the table, but because number of bits is usually constant, this solution is linear in array size.
No hashing, memory usage is O(log k * log max N).
Example implementation:
from random import randint, shuffle
def generate_test_data(n, k, b):
k_rep = [randint(0, 1000) for i in xrange(n)]
b_rep = [randint(0, 1000)]
numbers = k_rep*k + b_rep*b
shuffle(numbers)
print "k_rep: ", k_rep
print "b_rep: ", b_rep
return numbers
def solve(data, k):
cnts = [0]*10
for number in data:
bits = [number >> b & 1 for b in xrange(10)]
cnts = [cnts[i] + bits[i] for i in xrange(10)]
return reduce(lambda a,b:2*a+(b%k>0), reversed(cnts), 0)
print "Answer: ", solve(generate_test_data(10, 15, 13), 3)
In order to have a constant height B-tree containing n distinct elements, with height h constant, you need z=n^(1/h) children per nodes: h=log_z(n), thus h=log(n)/log(z), thus log(z)=log(n)/h, thus z=e^(log(n)/h), thus z=n^(1/h).
Example, with n=1000000, h=10, z=3.98, that is z=4.
The time to reach a node in that case is O(h.log(z)). Assuming h and z to be "constant" (since N=n.k, then log(z)=log(n^(1/h))=log(N/k^(1/h))=ct by properly choosing h based on k, you can then say that O(h.log(z))=O(1)... This is a bit far-fetched, but maybe that was the kind of thing the interviewer wanted to hear?
UPDATE: this one use hashing, so it's not a good answer :(
in python this would be linear time (set will remove the duplicates):
result = (sum(set(arr))*k - sum(arr)) / (k - b)
If 'k' is even and 'b' is odd, then XOR will do. :)