I have a sorted array, and a number, how many maximum comparisons should I do to find if the number is contained in the array ?
Suppose we have a million of numbers in the array.
To complete the answer from #aponeme, the maximum number of comparisons is equal to
2*upper(log2(n))
The reason is that the size of the array you examine is equal to
n, n/2, n/4, ...n/(2^steps).
Then the maximum number of steps is such that
n/(2^nsteps) = 1, i.e. nsteps = log2(n)
With divide and Conquer or Binary search (pseudo - code):
Split the array in half
If number is bigger than max of first half work with the second half else work with first half
Repeat step 1 - 2 with the remaining half
Worst case scenario: Need to divide into two arrays each of length 1 => O(logN)
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.
Some sorting algorithms, like Insertion Sort, have a Θ(n) asymptotic runtime for some subset of the n! possible permutations of n elements, which means that for those permutations, the number of comparisons that Insertion Sort does is kn for some constant k. For a given constant k, what is the maximum number of permutations for which any given comparison sort could terminate within kn comparisons?
Number of operations in insertion sort depends on the number of inversions. So we need to evaluate number of permutations of n values (1..n for simplicity), containing exactly k inversions.
We can see that Inv(n, 0) = 1 - sorted array
Also Inv(0, k) = 0 - empty array
We can get array with n elements and k inversions:
-adding value n to the end of array with n-1 items and k inversions (so number of inversions remains the same)
-inserting value n before the end of array with n-1 items and k-1 inversions (so adding one inversion)
-inserting value n before two elements in the end of array with n-1 items and k-2 inversions (so adding two inversions)
-and so on
Using this approach, we can just fill a table Inv[n][k] row-by-row and cell-by-cell
Inv[n][k] = Sum(Inv[n-1][i]) where j=0..k
Every comparison at most doubles the never of input permutations you can distinguish. Thus, with kn comparisons you can sort at most 2^(kn) permutations.
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 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.