This is an algorithm question:
Input is an array with non-duplicate positive integers. Find a continuous subarray(size > 1) which has the maximum median value.
Example: input: [100, 1, 99, 2, 1000], output should be the result of (1000 + 2) / 2 = 501
I can come up the brute force solution: try all lengths from 2 -> array size to find the maximum median. But it seems too slow. I also tried to use two pointer on this question but not sure when to move left and right pointer.
Anyone has a better idea to solve this question?
tl;dr - We can show that the answer must be of length 2 or 3, after which it's linear time to check all the possibilities.
Let's say the input is A and the smallest subarray with the biggest median is a. The biggest median is either a single element or the average of a pair of elements from a. Notice that every element in a bigger than the largest element of the median can only be next to elements less than the smallest element of the median (otherwise such a pair could be chosen as a subarray to form a bigger median).
If either end of a had a pair of elements that didn't include an element of the median, it could be eliminated from a without affecting the median, a contradiction.
If either end of a was smaller than the smallest element of the median, eliminating it would increase the median, a contradiction.
Thus each end of a is either an element of the median or larger than the largest element of the median (because it's larger than the smallest elt of the median and not equal to the largest elt of the median).
Thus each end of a is an element of the median because otherwise, we'd have an element larger than an element of the median adjacent to an elt of the median, forming a larger median.
If a is odd then it must be of length three, since any larger odd length could have 2 removed from the end of a farthest from the median without changing the median.
If a is even then it must be of length 2 because any larger even length bookended by the elements of the median with interior elements alternating between smaller and larger than the median must have one of the median elements adjacent to a larger element than the other elt of the median, forming a larger median.
This proof outline could use some editing, but regardless, the conclusion is that the smallest array containing the largest median must be of length 2 or 3.
Given that, check every such subarray in linear time. O(n).
This is a Python implementation of an algorithm that solves the problem in O(n):
import random
import statistics
n = 50
numbers = random.sample(range(n),n)
max_m = 0;
max_a = [];
for i in range(2,3):
for j in range(0,n-i+1):
a = numbers[j:j+i]
m = statistics.median(a)
if m > max_m:
max_m = m
max_a = a
print(numbers)
print(max_m)
print(max_a)
This is a variation of the brute force algorithm (O(n^3)) that performs only the search for sub-arrays of length 2 or 3. The reason is that for every array of size n, there exists a sub-array that has the same or improved median. Applying this reasoning recursively, we can reduce the size of the sub-array to 2 or 3. Thus, by looking only at sub-arrays of size 2 or 3, we are guaranteed to obtain the sub-array with the maximum median.
The operation is the following: If, for a contiguous sub-array (at the beginning or at the end), at least half of the elements are lower than the median (or lower than both values forming the median, if this is the case), remove them to improve or at least preserve the median.
If in all sub-arrays there is always at least one more element above or equal to the median(s) than below, there will come a point where the size of the sub-array will be that of the median. In that case, it means that the complement will have more elements below the median, and thus, we can simply remove the complement and improve (or preserve) the median. Thus, we can always perform the operation. For n=3, it can happen that you need to remove 2 or 3 elements to perform the operation, which is not allowed. In this case, the result is the list itself.
Related
I'm having a problem to solve this one. The task is to make a program that will for a given array of N numbers ( N <= 10^5 ), print a new array that's made by joining any two adjacent elements into their sum, (the sum is replacing these two adjacent elements and the size of the array is smaller by 1), until array's size is K. I need to print a solution where GCD of new elements is maximized. (and also print GCD after printing the array).
Note: Sum of all elements in the given array is not higher than 10^6.
I've realized that I could use prefix sum somehow because the sum of all elements isn't higher than 10^6, but that didn't help me that much.
What is an optimal solution to this problem?
Your GCD will be a divisor of the sum of all elements in the array. Your sum is not greater then 10^6, so the number of divisors is not greater than 240, so you can just check all of this GCDs, and it will be fast enough. You can check if asked gcd is possible in linear time: just go through array while the current sum is not the divisor of wanted gcd. When it is just set the current sum to 0. If you have found at least k blocks, it is possible to get current gcd (you can join any 2 blocks, and gcd will be the same).
I've been asked to write a program to find the kth order statistic of a data set consisting of character and their occurrences. For example, I have a data set consisting of
B,A,C,A,B,C,A,D
Here I have A with 3 occurrences, B with 2 occurrences C with 2 occurrences and D with on occurrence. They can be grouped in pairs (characters, number of occurrences), so, for example, we could represent the above sequence as
(A,3), (B,2), (C,2) and (D,1).
Assuming than k is the number of these pairs, I am asked to find the kth of the data set in O(n) where n is the number of pairs.
I thought could sort the element based their number of occurrence and find their kth smallest elements, but that won't work in the time bounds. Can I please have some help on the algorithm for this problem?
Assuming that you have access to a linear-time selection algorithm, here's a simple divide-and-conquer algorithm for solving the problem. I'm going to let k denote the total number of pairs and m be the index you're looking for.
If there's just one pair, return the key in that pair.
Otherwise:
Using a linear-time selection algorithm, find the median element. Let medFreq be its frequency.
Sum up the frequencies of the elements less than the median. Call this less. Note that the number of elements less than or equal to the median is less + medFreq.
If less < m < less + medFreq, return the key in the median element.
Otherwise, if m ≤ less, recursively search for the mth element in the first half of the array.
Otherwise (m > less + medFreq), recursively search for the (m - less - medFreq)th element in the second half of the array.
The key insight here is that each iteration of this algorithm tosses out half of the pairs, so each recursive call is on an array half as large as the original array. This gives us the following recurrence relation:
T(k) = T(k / 2) + O(k)
Using the Master Theorem, this solves to O(k).
I have to divide the elements of an array into 3 groups. This needs to be done without sorting the array. Consider the example
we have 120 unsorted values thus the smallest 40 values need to be in the first group and next 40 in the second and the largest 40 in the third group
I was thinking of the median of median approach but not able to apply it to my problem, kindly suggest an algorithm.
You can call quickselect twice on your array to do this in-place and in average case linear time. The worst case runtime can also be improved to O(n) by using the linear time median of medians algorithm to choose an optimal pivot for quickselect.
For both calls to quickselect, use k = n / 3. On your first call, use quickselect on the entire array, and on your second call, use it from arr[k..n-1] (for a 0-indexed array).
Wikipedia explanation of quickselect:
Quickselect uses the same overall approach as quicksort, choosing one
element as a pivot and partitioning the data in two based on the
pivot, accordingly as less than or greater than the pivot. However,
instead of recursing into both sides, as in quicksort, quickselect
only recurses into one side – the side with the element it is
searching for. This reduces the average complexity from O(n log n) (in
quicksort) to O(n) (in quickselect).
As with quicksort, quickselect is generally implemented as an in-place
algorithm, and beyond selecting the kth element, it also partially
sorts the data. See selection algorithm for further discussion of the
connection with sorting.
To divide the elements of the array into 3 groups, use the following algorithm written in Python in combination with quickselect:
k = n / 3
# First group smallest elements in array
quickselect(L, 0, n - 1, k) # Call quickselect on your entire array
# Then group middle elements in array
quickselect(L, k, n - 1, k) # Call quickselect on subarray
# Largest elements in array are already grouped so
# there is no need to call quickselect again
The key point of all this is that quickselect uses a subroutine called partition. Partition arranges an array into two parts, those greater than a given element and those less than a given element. Thus it partially sorts an array around this element and returns its new sorted position. Thus by using quickselect, you actually partially sort the array around the kth element (note that this is different from actually sorting the entire array) in-place and in average-case linear time.
Time Complexity for quickselect:
Worst case performance O(n2)
Best case performance O(n)
Average case performance O(n)
The runtime of quickselect is almost always linear and not quadratic, but this depends on the fact that for most arrays, simply choosing a random pivot point will almost always yield linear runtime. However, if you want to improve the worst case performance for your quickselect, you can choose to use the median of medians algorithm before each call to approximate an optimal pivot to be used for quickselect. In doing so, you will improve the worst case performance of your quickselect algorithm to O(n). This overhead probably isn't necessary but if you are dealing with large lists of randomized integers it can prevent some abnormal quadratic runtimes of your algorithm.
Here is a complete example in Python which implements quickselect and applies it twice to a reverse-sorted list of 120 integers and prints out the three resulting sublists.
from random import randint
def partition(L, left, right, pivotIndex):
'''partition L so it's ordered around L[pivotIndex]
also return its new sorted position in array'''
pivotValue = L[pivotIndex]
L[pivotIndex], L[right] = L[right], L[pivotIndex]
storeIndex = left
for i in xrange(left, right):
if L[i] < pivotValue:
L[storeIndex], L[i] = L[i], L[storeIndex]
storeIndex = storeIndex + 1
L[right], L[storeIndex] = L[storeIndex], L[right]
return storeIndex
def quickselect(L, left, right, k):
'''retrieve kth smallest element of L[left..right] inclusive
additionally partition L so that it's ordered around kth
smallest element'''
if left == right:
return L[left]
# Randomly choose pivot and partition around it
pivotIndex = randint(left, right)
pivotNewIndex = partition(L, left, right, pivotIndex)
pivotDist = pivotNewIndex - left + 1
if pivotDist == k:
return L[pivotNewIndex]
elif k < pivotDist:
return quickselect(L, left, pivotNewIndex - 1, k)
else:
return quickselect(L, pivotNewIndex + 1, right, k - pivotDist)
def main():
# Setup array of 120 elements [120..1]
n = 120
L = range(n, 0, -1)
k = n / 3
# First group smallest elements in array
quickselect(L, 0, n - 1, k) # Call quickselect on your entire array
# Then group middle elements in array
quickselect(L, k, n - 1, k) # Call quickselect on subarray
# Largest elements in array are already grouped so
# there is no need to call quickselect again
print L[:k], '\n'
print L[k:k*2], '\n'
print L[k*2:]
if __name__ == '__main__':
main()
I would take a look at order statistics. The kth order statistic of a statistical sample is equal to its kth-smallest value. The problem of computing the kth smallest (or largest) element of a list is called the selection problem and is solved by a selection algorithm.
It is right to think the median of the medians way. However, instead of finding the median, you might want to find both 20th and 40th smallest elements from the array. Just like finding the median, it takes only linear time to find both of them using a selection algorithm. Finally you go over the array and partition the elements according to these two elements, which is linear time as well.
PS. If this is your exercise in an algorithm class, this might help you :)
Allocate an array of the same size of the input array
scan the input array once and keep track of the min and max values of the array.
and at the same time set to 1 all the values of the second array.
compute delta = (max - min) / 3.
Scan the array again and set the second array to two if the number is > min+delta and < max-delta; otherwise if >= max-delta, set it to 3;
As a result you will have an array that tells in which group the number is.
I am assuming that all the numbers are different from each other.
Complexity: O(2n)
First let me phrase the proper question:
Q: There is a file containing more than a million points (x,y) each of which represents a star. There is a planet earth at (a,b). Now, the task is to build an algorithm that would return the 100 closest stars to earth. What would be the time and space complexities of your algorithm.
This question has been asked many times in various interviews. I tried looking up the answers but could not find a satisfactory one.
One way to do it which I thought might be using a max heap of size 100. Calculate distances for each star and check if the distance is lesser than the root of the max heap. If yes, replace it with the root and call heapify.
Any other better/faster answers?
P.S: This is not a homework question.
You can actually do this in time O(n) and space O(k), where k is the number of closest points that you want, by using a very clever trick.
The selection problem is as follows: Given an array of elements and some index i, rearrange the elements of the array such that the ith element is in the right place, all elements smaller than the ith element are to the left, and all elements greater than the ith element are to the right. For example, given the array
40 10 00 30 20
If I tried to select based on index 2 (zero-indexed), one result might be
10 00 20 40 30
Since the element at index 2 (20) is in the right place, the elements to the left are smaller than 20, and the elements to the right are greater than 20.
It turns out that since this is a less strict requirement than actually sorting the array, it's possible to do this in time O(n), where n is the number of elements of the array. Doing so requires some complex algorithms like the median-of-medians algorithm, but is indeed O(n) time.
So how do you use this here? One option is to load all n elements from the file into an array, then use the selection algorithm to select the top k in O(n) time and O(n) space (here, k = 100).
But you can actually do better than this! For any constant k that you'd like, maintain a buffer of 2k elements. Load 2k elements from the file into the array, then use the selection algorithm to rearrange it so that the smallest k elements are in the left half of the array and the largest are in the right, then discard the largest k elements (they can't be any of the k closest points). Now, load k more elements from the file into the buffer and do this selection again, and repeat this until you've processed every line of the file. Each time you do a selection you discard the largest k elements in the buffer and retain the k closest points you've seen so far. Consequently, at the very end, you can select the top k elements one last time and find the top k.
What's the complexity of the new approach? Well, you're using O(k) memory for the buffer and the selection algorithm. You end up calling select on a buffer of size O(k) a total of O(n / k) times, since you call select after reading k new elements. Since select on a buffer of size O(k) takes time O(k), the total runtime here is O(n + k). If k = O(n) (a reasonable assumption), this takes time O(n), space O(k).
Hope this helps!
To elaborate on the MaxHeap solution you would build a max-heap with the first k elements from the file ( k = 100 in this case ).
The key for the max-heap would be its distance from Earth (a,b). Distance between 2 points on a 2d plane can be calculated using:
dist = (x1,y1) to (x2,y2) = square_root((x2 - x1)^2 + (y2 - y1)^2);
This would take O(k) time to construct. For every subsequent element from k to n. ie (n - k) elements you need to fetch its distance from earth and compare it with the top of max-heap. If the new element to be inserted is closer to earth than the top of the max-heap, replace the top of the max-heap and call heapify on the new root of the heap.
This would take O((n-k)logk) time to complete.
Finally we would be left with just the k elements in the max-heap. You can call heapify k times to return all these k elements. This is another O(klogk).
Overall time complexity would be O(k + (n-k)logk + klogk).
It's a famous question and there has been lot's of solution for that:
http://en.wikipedia.org/wiki/K-nearest_neighbor_algorithm
if you did not find it useful, there are some other resources such as Rurk's computational geometry book.
Your algorithm is correct. Just remember that time complexity of your program is O(n . log 100 ) = O(n), unless number of closest points to find can vary.
import sys,os,csv
iFile=open('./file_copd.out','rU')
earth = [0,0]
##getDistance return distance given two stars
def getDistance(star1,star2):
return sqrt((star1[0]-star2[0])**2 +(star1[1]-star2[1])**2 )
##diction dict_galaxy looks like this {key,distance} key is the seq assign to each star, value is a list [distance,its cordinance]
##{1,[distance1,[x,y]];2,[distance2,[x,y]]}
dict_galaxy={}
#list_galaxy=[]
count = 0
sour=iFile.readlines()
for line in sour:
star=line.split(',') ##Star is a list [x,y]
dict_galaxy[count]=[getDistance(earth,star),star]
count++
###Now sort this dictionary based on their distance, and return you a list of keys.
list_sorted_key = sorted(dict_galaxy,key=lambda x:dict_galaxy[x][0])
print 'is this what you want %s'%(list_sorted_key[:100].to_s)
iFile.close()
I can use the median of medians selection algorithm to find the median in O(n). Also, I know that after the algorithm is done, all the elements to the left of the median are less that the median and all the elements to the right are greater than the median. But how do I find the k nearest neighbors to the median in O(n) time?
If the median is n, the numbers to the left are less than n and the numbers to the right are greater than n.
However, the array is not sorted in the left or the right sides. The numbers are any set of distinct numbers given by the user.
The problem is from Introduction to Algorithms by Cormen, problem 9.3-7
No one seems to quite have this. Here's how to do it. First, find the median as described above. This is O(n). Now park the median at the end of the array, and subtract the median from every other element. Now find element k of the array (not including the last element), using the quick select algorithm again. This not only finds element k (in order), it also leaves the array so that the lowest k numbers are at the beginning of the array. These are the k closest to the median, once you add the median back in.
The median-of-medians probably doesn't help much in finding the nearest neighbours, at least for large n. True, you have each column of 5 partitioned around it's median, but this isn't enough ordering information to solve the problem.
I'd just treat the median as an intermediate result, and treat the nearest neighbours as a priority queue problem...
Once you have the median from the median-of-medians, keep a note of it's value.
Run the heapify algorithm on all your data - see Wikipedia - Binary Heap. In comparisons, base the result on the difference relative to that saved median value. The highest priority items are those with the lowest ABS(value - median). This takes O(n).
The first item in the array is now the median (or a duplicate of it), and the array has heap structure. Use the heap extract algorithm to pull out as many nearest-neighbours as you need. This is O(k log n) for k nearest neighbours.
So long as k is a constant, you get O(n) median of medians, O(n) heapify and O(log n) extracting, giving O(n) overall.
med=Select(A,1,n,n/2) //finds the median
for i=1 to n
B[i]=mod(A[i]-med)
q=Select(B,1,n,k) //get the kth smallest difference
j=0
for i=1 to n
if B[i]<=q
C[j]=A[i] //A[i], the real value should be assigned instead of B[i] which is only the difference between A[i] and median.
j++
return C
You can solve your problem like that:
You can find the median in O(n), w.g. using the O(n) nth_element algorithm.
You loop through all elements substutiting each with a pair:
the absolute difference to the median, element's value.
Once more you do nth_element with n = k. after applying this algorithm you are guaranteed to have the k smallest elements in absolute difference first in the new array. You take their indices and DONE!
Four Steps:
Use Median of medians to locate the median of the array - O(n)
Determine the absolute difference between the median and each element in the array and store them in a new array - O(n)
Use Quickselect or Introselect to pick k smallest elements out of the new array - O(k*n)
Retrieve the k nearest neighbours by indexing the original array - O(k)
When k is small enough, the overall time complexity becomes O(n).
Find the median in O(n). 2. create a new array, each element is the absolute value of the original value subtract the median 3. Find the kth smallest number in O(n) 4. The desired values are the elements whose absolute difference with the median is less than or equal to the kth smallest number in the new array.
You could use a non-comparison sort, such as a radix sort, on the list of numbers L, then find the k closest neighbors by considering windows of k elements and examining the window endpoints. Another way of stating "find the window" is find i that minimizes abs(L[(n-k)/2+i] - L[n/2]) + abs(L[(n+k)/2+i] - L[n/2]) (if k is odd) or abs(L[(n-k)/2+i] - L[n/2]) + abs(L[(n+k)/2+i+1] - L[n/2]) (if k is even). Combining the cases, abs(L[(n-k)/2+i] - L[n/2]) + abs(L[(n+k)/2+i+!(k&1)] - L[n/2]). A simple, O(k) way of finding the minimum is to start with i=0, then slide to the left or right, but you should be able to find the minimum in O(log(k)).
The expression you minimize comes from transforming L into another list, M, by taking the difference of each element from the median.
m=L[n/2]
M=abs(L-m)
i minimizes M[n/2-k/2+i] + M[n/2+k/2+i].
You already know how to find the median in O(n)
if the order does not matter, selection of k smallest can be done in O(n)
apply for k smallest to the rhs of the median and k largest to the lhs of the median
from wikipedia
function findFirstK(list, left, right, k)
if right > left
select pivotIndex between left and right
pivotNewIndex := partition(list, left, right, pivotIndex)
if pivotNewIndex > k // new condition
findFirstK(list, left, pivotNewIndex-1, k)
if pivotNewIndex < k
findFirstK(list, pivotNewIndex+1, right, k)
don't forget the special case where k==n return the original list
Actually, the answer is pretty simple. All we need to do is to select k elements with the smallest absolute differences from the median moving from m-1 to 0 and m+1 to n-1 when the median is at index m. We select the elements using the same idea we use in merging 2 sorted arrays.
If you know the index of the median, which should just be ceil(array.length/2) maybe, then it just should be a process of listing out n(x-k), n(x-k+1), ... , n(x), n(x+1), n(x+2), ... n(x+k)
where n is the array, x is the index of the median, and k is the number of neighbours you need.(maybe k/2, if you want total k, not k each side)
First select the median in O(n) time, using a standard algorithm of that complexity.
Then run through the list again, selecting the elements that are nearest to the median (by storing the best known candidates and comparing new values against these candidates, just like one would search for a maximum element).
In each step of this additional run through the list O(k) steps are needed, and since k is constant this is O(1). So the total for time needed for the additional run is O(n), as is the total runtime of the full algorithm.
Since all the elements are distinct, there can be atmost 2 elements with the same difference from the mean. I think it is easier for me to have 2 arrays A[k] and B[k] the index representing the absolute value of the difference from the mean. Now the task is to just fill up the arrays and choose k elements by reading the first k non empty values of the arrays reading A[i] and B[i] before A[i+1] and B[i+1]. This can be done in O(n) time.
All the answers suggesting to subtract the median from the array would produce incorrect results. This method will find the elements closest in value, not closest in position.
For example, if the array is 1,2,3,4,5,10,20,30,40. For k=2, the value returned would be (3,4); which is incorrect. The correct output should be (4,10) as they are the nearest neighbor.
The correct way to find the result would be using the selection algorithm to find upper and lower bound elements. Then by direct comparison find the remaining elements from the list.