For example, given an array
a = [1, 2, 3, 7, 8, 9]
and an integer
i = 2. Find maximal subarrays where the distance between the largest and the smallest elements is at most i. The output for the example above would be:
[1,2,3] [7,8,9]
The subarrays are maximal in the sense given two subarrays A and B. There exists no element b in B such that A + b satisfies the condition given. Does there exist a non-polynomial time algorithm for said problem ?
This problem might be solved in linear time using method of two pointers and two deques storing indices, the first deque keeps minimum, another keeps maximum in sliding window.
Deque for minimum (similar for maximum):
current_minimum = a[minq.front]
Adding i-th element of array: //at the right index
while (!minq.empty and a[minq.back] > a[i]):
//last element has no chance to become a minimum because newer one is better
minq.pop_back
minq.push_back(i)
Extracting j-th element: //at the left index
if (!minq.empty and minq.front == j)
minq.pop_front
So min-deque always contains non-decreasing sequence.
Now set left and right indices in 0, insert index 0 into deques, and start to move right. At every step add index in order into deques, and check than left..right interval range is good. When range becomes too wide (min-max distance is exceeded), stop moving right index, check length of the last good interval, compare with the best length.
Now move left index, removing elements from deques. When max-min becomes good, stop left and start with right again. Repeat until array end.
I have an array A of size N. All elements are positive integers. In one step, I can add two adjacent elements and replace them with their sum. That said, the array size reduces by 1. Now I need to make all the elements same by performing minimum number of steps.
For example: A = [1,2,3,2,1,3].
Step 1: Merge index 0 and 1 ==> A = [3,3,2,1,3]
Step 2: Merge index 2 and 3 (of new array) ==> [3,3,3,3]
Hence number of steps are 2.
I couldn't think of a straight solution, so tried a recursive approach by merging all indices one by one and returning the min level I can get when either array size is 1 or all elements are equal.
Below is the code I tried:
# Checks if all the elements are same or not
def check(A):
if len(set(A)) == 1:
return True
return False
# Recursive function to find min steps
def min_steps(N,A,level):
# If all are equal return the level
if N == 1 or check(A):
return level
# Initialize min variable
mn = float('+inf')
# Try merging all one by one and recur
for i in range(N-1):
temp = A[:]
temp[i]+=temp[i+1]
temp.pop(i+1)
mn = min(mn, min_steps(N-1,temp, level+1))
return mn
This solution has complexity of O(N^N). I want to reduce it to polynomial time near to O(N^2) or O(N^3). Can anyone help me modify this solution or tell me if I am missing something?
Combining any k adjacent pairs of elements (even if they include elements formed from previous combining steps) leaves exactly n-k elements in total, each of which we can map back to the contiguous subarray of the original problem that constitutes the elements that were added together to form it. So, this problem is equivalent to partitioning the array into the largest possible number of contiguous subarrays such that all subarrays have the same sum: Any adjacent pair of elements within the same subarray can be combined into a single element, and this process repeated within the subarray with adjacent pairs chosen in any order, until all elements have been combined into a single element.
So, if there are n elements and they sum to T, then a simple O(nT) algorithm is:
For i from 0 to T:
Try partitioning the elements into subarrays each having sum i. This amounts to scanning along the array, greedily adding the current element to the current subarray if the sum of elements in the current subarray is strictly < i. When we reach a total of exactly i, the current subarray ends and a new subarray (initially having sum 0) begins. If adding the current element takes us above the target of i, or if we run out of elements before reaching the target, stop this scan and try the next outer loop iteration (value of i). OTOH if we get to the end, having formed k subarrays in the process, stop and report n-k as the optimal (minimum possible) number of combining moves.
A small speedup would be to only try target i values that evenly divide T.
EDIT: To improve the time complexity from O(nT) to O(n^2), it suffices to only try target i values corresponding to sums of prefixes of the array (since there must be a subarray containing the first element, and this subarray can only have such a sum).
I have 2 sorted integer lists or arrays a and b, both having same number of elements. I want to pair an element in a with an element in b such that when I take smaller element in all pairs, their sum is minimum.
For example,
a=[1,7,14,18]
b=[8,9,10,12]
I would be pairing [(1,12),(7,10),(14,9),(18,8)] and then taking smaller element in each pair, namely, [1,7,9,8], I will get minimum sum. This is just one possibility I took. I want to know if this method of pairing elements of first list from the first element and moving forward with elements of the second list starting from end and going backwards will give me the minimum sum.
Yes, that method of pairing the largest with the smallest will work:
If the largest element in the second array is smaller than the smallest in the first array, any pairing method will work and so pair the remaining elements using your method.
If not, pairing the first pair with your method will ensure the smallest from the first (which should be counted) will be counted and the largest from the second (which should not be counted) will not be counted
Repeat steps 1 and 2 with the remaining elements from both arrays until you run out of elements
As you can see, the smallest remaining elements from the arrays will always be counted at each step along the way, and so the sum of the smallest from the resulting pairs will be minimized as desired.
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.
The problem is a variant of subarray counting. Given an array of numbers, let's say, 1,2,2,3,2,1,2,2,2,2 I look for subarrays and count the frequency of each. I start with looking from some K length subarrays (example K = 3).
Count of subarray 1,2,2 is C1:2.
Count of subarray 2,2,3 is 1.
Count of subarray 2,3,2 is 1.
and so on.
Now, I look for subarrays of length 2.
Count of subarray 1,2 is C2: 2. But (1,2) is a subset of the subarray 1,2,2. So, I calculate its count by subtracting C1 from C2 which gives count of 1,2 as 0. Similarly, count of 2,2 is 1.
My problem is in handling cases where more than one parent subset exists. I don't consider the sub-arrays in my result set whose frequency comes out to be 1. Example:
1,2,3,1,2,3,1,2,2,3
Here, Count of 1,2,3 is 2.
Count of 2,3,1 is 2.
Now, when I look for count of 2,3, it should be 1 as all the greater length parents have covered the occurrences. How shall I handle these cases?
The approach I thought was to mark all the pattern occurrences of the parent. In the above case, mark all the occurrences of 1,2,3 and 2,3,1. Array looks like this:
1,2,3,1,2,3,1,2,2,3
X,X,X,X,X,X,X,2,2,3
where X denotes the marked position. Now, frequency of 2,3 we see is 1 as per the positions which are unmarked. So, basically, I mark all the pattern occurrences I find at the current step. For the next step, I start looking for patterns from the unmarked locations only to get the correct count.
I am dealing with the large data on which this seems a bit not-so-good thing to do. Also, I'm not sure if it's correct or not. Any other approaches or ideas can be of big help?
Build suffix array for given array.
To count all repeating subarrays with given length - walk through this suffix array, comparing neighbor suffixes by needed prefix length.
For your first example
source array
1,2,2,3,2,1,2,2,2,2
suffix array is
5,0,9,4,8,7,6,1,2,3:
1,2,2,2,2 (5)
1,2,2,3,2,1,2,2,2,2 (0)
2 (9)
2,1,2,2,2,2 (4)
2,2 (8)
2,2,2 (7)
2,2,2,2 (6)
2,2,3,2,1,2,2,2,2 (1)
2,3,2,1,2,2,2,2 (2)
3,2,1,2,2,2,2 (3)
With length 2 we can count two subarrays 1,2 and four subarrays 2,2
If you want to count any given subarray - for example, all suffixes beginning with (1,2), just use binary search to get the first and the last indexes (like std:upperbound and std:lowerbound operations in C++ STL).
For the same example indexes of the first and last occurrences of (1,2) in suffix array are 0 and 1, so count is last-first+1=2