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Given an array A with all elements appearing twice except one element which appears only once. How do we find the element which appears only once in O(logn) time? Let's discuss two cases.
Array is always sorted and elements are in sequential order. Let's assume A = [1, 1, 2, 2, 3, 4, 4, 5, 5, 6, 6], we want to find 3 in log n time because it appears only once.
When the array is not sorted and the elements are not in sequential order.
I can only come up with a solution of using the XOR operator on the binary representation of the integers as explained Here, and at the end, the binary string will represent the element which appears only once because duplicates will cancel out. But it takes O(n) time. How can we do better than that?
using Haroon S' comment this is the solution which I think is correct, given the constraints for time.
class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
low = 0
high = len(nums)-1
while(low<high):
mid = (low+high)//2
if(mid%2==0):
mid+=1
if(nums[mid]==nums[mid+1]):
# answer in second half
high = mid-1
elif(nums[mid]==nums[mid-1]):
# answer in first half
low = mid+1
return nums[low]
If the elements are sorted (i.e., the first case you mentioned) then I believe a strategy not unlike binary search could work in O(logN) time.
Starting from the left endpoint in a sorted array, until we encounter the unique element, all the index pairs (2i, 2i + 1) we encounter along the way will have the same value. (i.e., due to the array being sorted) However, as we go towards the right endpoint of the array, as soon as we consider an array that includes the unique element, that structure of "same values within (2i, 2i+1) index pairs" will be invalid.
Using that information, a search algorithm similar to binary search can find out in which half of the array the unique element is. Basically, you can deduce that, "in the left half of the array, if the values in the rightmost index pair (2i, 2i+1) are the same, then the unique value is in the right half". (i.e., with the exception of the last index on the left half-array being even; but you can overcome that case with various O(1) time operations)
The overall complexity then becomes O(logN), due to the halving of the array size at each step.
For the demonstration of the index notion I mentioned above, see your own example. In the left of the unique element(i.e. 3) all index pairs (2i, 2i+1) have the same values. And all subarrays starting from index 0 and ending with an index that is to the right of the unique element, all index pairs (2i, 2i+1) have a correspond to cells that contain different values.
Unless the array is sorted, though, since you'd have to investigate each and every element, I believe any algorithm you may come up with would take at least O(n) time. This is what I think will happen in the second case you mention in your question.
In the general case this is impossible, as to make sure an element doesn't repeat you need to check every other element.
From your example, it seems the array might be a sorted sequence of integers with no "gaps" (or some other clearly defined sequence, like all even numbers, etc). In this case it is possible with a modified binary search.
You have the array [1,1,2,2,3,4,4,5,5,6,6].
You check the middle element and the element following it and see 3 and 4. Now you know there are only 5 elements from the set {1, 2, 3}, while there are 6 elements from the set {4, 5, 6}. Which means, the missing elements is in {1, 2, 3}.
Then you recurse on [1,1,2,2,3]. You see 2,2. Now you know there are 2 "1" elements and 1 "3" element, so 3 is the answer.
The reason you check 2 elements in each step is that if you see just "3", you don't know whether you hit the first 3 in "3,3" or the second one. But if you read 2 elements you always find a "boundary" between 2 different elements.
The condition for this to be viable is that, given the value of an element, you need to be able to calculate in O(1) how many different elements come before this element. In your case this is trivial, but it is also possible for any arithmetic series, geometric series (with fixed size numbers)...
This is not a O(log n) solution. I have no idea how to solve it in logarithmic time without the constraints that the array is sorted and we have a known difference between consecutive numbers so we can recognise when we are to the left or right of the singleton. The other solutions already deal with that special case and I couldn’t do better there either.
I have a suggestion that might solve the general case in O(n), rather than O(n log n) when you first sort the array. It’s not as fast as the xor solution, but it will also work for non-integers. The elements must have an order, so it is not completely general, but it will work anywhere you can sort the elements.
The idea is the same as the k’th order element algorithm based on Quicksort. You partition and recurse on one half of the array. The time recurrence is T(n) = T(n/2) + O(n) = O(n).
Given array x and indices i,j, representing sub-array x[i:j], partition with quicksort’s partitioning method. You want a variant that partitions x[i:j] into three segments, x[i:k] x[k:l], x[l:j] where all elements in the first part are smaller than the pivot (whatever it is) all elements in x[k:l] are equal to the pivot, and all elements in the last segment are greater than the pivot.
(you might be able to use a version that only partitions in two, or explicitly count the number of pivots, but with this version is easier to work with here)
Now, if the middle segment has length one, you have your singleton. It is the pivot.
If not, the length of the segment that has the singleton is odd while the other is even. So recurse on the segment with the odd length.
It doesn’t give you worst case linear time, for the same reason that Quicksort isn’t worst case log-linear, but you get an expected linear time algorithm and likely a fast one at that.
Not, of course, as fast as those solutions based on binary search, but here the elements do not need to be sorted and we can handle elements with arbitrary gaps between them. We are also not restricted to data where we can easily manipulate their bit-patterns. So it is more general. If you can compare the elements, this approach will find the singleton in O(n).
This solution will find the element in the array that appeared only once but there should not be more than one element of that type and the array should be sorted. This is Binary Search and will return the element in O(log n) time.
var singleNonDuplicate = function(nums) {
let s=0,e= nums.length-1
while(s < e){
let mid = Math.trunc(s+(e-s)/2)
if((mid%2 == 0&& nums[mid] ==nums[mid+1])||(mid%2==1 && nums[mid] == nums[mid-1]) ){
s= mid+1
}
else{
e = mid
}
}
return nums[s] // can return nums[e] also
};
I don't believe there is a O(log n) solution for that. The reason is that in order to find which element is appearing only once, you at least need to iterate over the elements of that array once.
I have a list of numbers and I have a sum value. For instance,
list = [1, 2, 3, 5, 7, 11, 10, 23, 24, 54, 79 ]
sum = 20
I would like to generate a sequence of numbers taken from that list, such that the sequence sums up to that target. In order to help achieve this, the sequence can be of any length and repetition is allowed.
result = [2, 3, 5, 10] ,or result = [1, 1, 2, 3, 3, 5, 5] ,or result = [10, 10]
I've been doing a lot of research into this problem and have found the subset sum problem to be of interest. My problem is, in a few ways, similar to the subset sum problem in that I would like to find a subset of numbers that produces the targeted sum.
However, unlike the subset sum problem which finds all sets of numbers that sum up to the target (and so runs in exponential time if brute forcing), I only want to find one set of numbers. I want to find the first set that gives me the sum. So, in a certain sense, speed is a factor.
Additionally, I would like there to be some degree of randomness (or pseudo-randomness) to the algorithm. That is, should I run the algorithm using the same list and sum multiple times, I should get a different set of numbers each time.
What would be the best algorithm to achieve this?
Additional Notes:
What I've achieved so far is using a naive method where I cycle through the list adding it to every combination of values. This obviously takes a long time and I'm currently not feeling too happy about it. I'm hoping there is a better way to do this!
If there is no sequence that gives me the exact sum, I'm satisfied with a sequence that gives me a sum that is as close as possible to the targeted sum.
As others said, this is a NP-problem.
However, this doesn't mean small improvements aren't possible:
Is 1 in the list? [1,1,1,1...] is the solution. O(1) in a sorted list
Remove list element bigger than the target sum. O(n)
Is there any list element x with (x%sum)==0 ? Again, easy solution. O(n)
Are there any list elements x,y with (x%y)==0 ? Remove x. O(n^2)
(maybe even: Are there any list elements x,y,z with (x%y)==z or (x+y)==z ? Remove x. O(n^3))
Before using the full recursion, try if you can get the sum
just with the smallest even and smallest odd number.
...
Subset Sum problem isn't about finding all subsets, but rather about determining if there is some subset. It is a decision problem. All problems in NP are like this. And even this simpler problem is NP-complete.
This means that if you want an exact answer (the subset must sum exactly some value) you won't be able to do much better than the any subset sum algorithm (it is exponential unless P=NP).
I would attempt to reduce the problem to a brute-force search of a smaller set.
Sort the list smallest to largest.
Keep a sum and result list.
Repeat {
Draw randomly from the subset of list less than target - sum.
Increment sum by drawn value, add drawn value to result list.
} until list[0] > sum or sum == 0
If sum != 0, brute force search for small combinations from list that match the difference between sum and small combinations of result.
This approach may fail to find valid solutions, even if they exist. It can, however, quickly find a solution or quickly fail before having to resort to a slower brute force approach using the entire set at a greater depth.
This is a greedy approach to the problem:
Without 'randomness':
Obtain the single largest number in the set that is smaller than your desired sum- we'll name it X. Given it's ordered, at best it's O(1), and O(N) at worst if the sum is 2.
As you can repeat the value- say c times, do so as many times until you get closest to the sum, but be careful! Create a range of values- essentially now you'll be finding another sum! You'll now be find numbers that add up to R = (sum - X * c). So find the largest number smaller than R. Check if R - (number you just found) = 0 or if any [R - (number you just found)] % (smaller #s) == 0.
If it becomes R > 0, make partial sums of the smaller numbers less than R (this will not be more than 5 ~ 10 computations because of the nature of this algorithm). See if these would then satisfy it.
If that step makes R < 0, remove one X and start the process again.
With 'randomness':
Just get X randomly! :-)
Note: This would work best if you have a few single digit numbers.
I have a large array with increasing values - like this:
array = [0, 1, 6, 6, 12, 13, 22, ..., 92939, 92940]
and I want to use interpolation-search algorithm on it. Size of the array is variable, new elements are added to the end of the array.
I need to find index of some element, let's call it X.
Y = find(X in array)
Y must be an index of the element from the array such that array[Y] >= X
find can be implemented with binary search but for some complicated reasons I want to implemented it using interpolation search. Interpolation search tries to guess correct position of the X by looking at the bounds of the array. If first array value is 0 and last is 100 and I want to find position of the value 25 than, if array length is 1000, I need to look at value at index 250 first. This works as a charm if values of the array is evenly distributed. But if they not evenly distributed, interpolation search can work slower than binary search (there is some optimizations possible).
I'm trying to speed up search in such cases using Count-Min Sketch data structure. When I appending new element to the array I just add some data to count-min sketch data-structure.
Z = 1005000
elements_before_Z = len(array)
array.append(Z)
count_min_sketch.add(Z, elements_before_Z)
# Z is the key and elenents_before_Z - is count
using this approach I can guess position of the searched element X approximately. This can result in search speedup if guess is correct, but I've ran into some problems.
I don't know if X is in array and my count_min_sketch have seen this value. If this is the case I can get correct value out of count_min_sketch datastructure. If it's not - I will get 0 or some other value (worst case scenario).
Collisions. If value X have been seen by my count_min_sketch object - than I get back correct value or larger value. If count min sketch used for something like counting word occurrence in document - this is not a problem because collisions is rare and error is less or equal than number of collisions (it usually used like this: count_min_sketch.add(Z, 1)). In my case, every collision can result in large error, because I usually add large numbers for every key.
Is it possible to use count-min sketch in such way (adding large number of entries every time)?
Suppose you are given a range and a few numbers in the range (exceptions). Now you need to generate a random number in the range except the given exceptions.
For example, if range = [1..5] and exceptions = {1, 3, 5} you should generate either 2 or 4 with equal probability.
What logic should I use to solve this problem?
If you have no constraints at all, i guess this is the easiest way: create an array containing the valid values, a[0]...a[m] . Return a[rand(0,...,m)].
If you don't want to create an auxiliary array, but you can count the number of exceptions e and of elements n in the original range, you can simply generate a random number r=rand(0 ... n-e), and then find the valid element with a counter that doesn't tick on exceptions, and stops when it's equal to r.
Depends on the specifics of the case. For your specific example, I'd return a 2 if a Uniform(0,1) was below 1/2, 4 otherwise. Similarly, if I saw a pattern such as "the exceptions are odd numbers", I'd generate values for half the range and double. In general, though, I'd generate numbers in the range, check if they're in the exception set, and reject and re-try if they were - a technique known as acceptance/rejection for obvious reasons. There are a variety of techniques to make the exception-list check efficient, depending on how big it is and what patterns it may have.
Let's assume, to keep things simple, that arrays are indexed starting at 1, and your range runs from 1 to k. Of course, you can always shift the result by a constant if this is not the case. We'll call the array of exceptions ex_array, and let's say we have c exceptions. These need to be sorted, which shall turn out to be pretty important in a while.
Now, you only have k-e useful numbers to work with, so it'll be meaningful to find a random number in the range 1 to k-e. Say we end up with the number r. Now, we just need to find the r-th valid number in your array. Simple? Not so much. Remember, you can never simply walk over any of your arrays in a linear fashion, because that can really slow down your implementation when you have a lot of numbers. You have do some sort of binary search, say, to come up with a fast enough algorithm.
So let's try something better. The r-th number would nominally have lied at index r in your original array had you had no exceptions. The number at index r is r, of course, since your range and your array indices start from 1. But, you have a bunch of invalid numbers between 1 and r, and you want to somehow get to the r-th valid number. So, lets do a binary search on the array of exceptions, ex_array, to find how many invalid numbers are equal to or less than r, because we have these many invalid numbers lying between 1 and r. If this number is 0, we're all done, but if it isn't, we have a bit more work to do.
Assume you found there were n invalid numbers between 1 and r after the binary search. Let's advance n indices in your array to the index r+n, and find the number of invalid numbers lying between 1 and r+n, using a binary search to find how many elements in ex_array are less than or equal to r+n. If this number is exactly n, no more invalid numbers were encountered, and you've hit upon your r-th valid number. Otherwise, repeat again, this time for the index r+n', where n' is the number of random numbers that lay between 1 and r+n.
Repeat till you get to a stage where no excess exceptions are found. The important thing here is that you never once have to walk over any of the arrays in a linear fashion. You should optimize the binary searches so they don't always start at index 0. Say if you know there are n random numbers between 1 and r. Instead of starting your next binary search from 1, you could start it from one index after the index corresponding to n in ex_array.
In the worst case, you'll be doing binary searches for each element in ex_array, which means you'll do c binary searches, the first starting from index 1, the next from index 2, and so on, which gives you a time complexity of O(log(n!)). Now, Stirling's approximation tells us that O(ln(x!)) = O(xln(x)), so using the algorithm above only makes sense if c is small enough that O(cln(c)) < O(k), since you can achieve O(k) complexity using the trivial method of extracting valid elements from your array first.
In Python the solution is very simple (given your example):
import random
rng = set(range(1, 6))
ex = {1, 3, 5}
random.choice(list(rng-ex))
To optimize the solution, one needs to know how long is the range and how many exceptions there are. If the number of exceptions is very low, it's possible to generate a number from the range and just check if it's not an exception. If the number of exceptions is dominant, it probably makes sense to gather the remaining numbers into an array and generate random index for fetching non-exception.
In this answer I assume that it is known how to get an integer random number from a range.
Here's another approach...just keep on generating random numbers until you get one that isn't excluded.
Suppose your desired range was [0,100) excluding 25,50, and 75.
Put the excluded values in a hashtable or bitarray for fast lookup.
int randNum = rand(0,100);
while( excludedValues.contains(randNum) )
{
randNum = rand(0,100);
}
The complexity analysis is more difficult, since potentially rand(0,100) could return 25, 50, or 75 every time. However that is quite unlikely (assuming a random number generator), even if half of the range is excluded.
In the above case, we re-generate a random value for only 3/100 of the original values.
So 3% of the time you regenerate once. Of those 3%, only 3% will need to be regenerated, etc.
Suppose the initial range is [1,n] and and exclusion set's size is x. First generate a map from [1, n-x] to the numbers [1,n] excluding the numbers in the exclusion set. This mapping with 1-1 since there are equal numbers on both sides. In the example given in the question the mapping with be as follows - {1->2,2->4}.
Another example suppose the list is [1,10] and the exclusion list is [2,5,8,9] then the mapping is {1->1, 2->3, 3->4, 4->6, 5->7, 6->10}. This map can be created in a worst case time complexity of O(nlogn).
Now generate a random number between [1, n-x] and map it to the corresponding number using the mapping. Map looks can be done in O(logn).
You can do it in a versatile way if you have enumerators or set operations. For example using Linq:
void Main()
{
var exceptions = new[] { 1,3,5 };
RandomSequence(1,5).Where(n=>!exceptions.Contains(n))
.Take(10)
.Select(Console.WriteLine);
}
static Random r = new Random();
IEnumerable<int> RandomSequence(int min, int max)
{
yield return r.Next(min, max+1);
}
I would like to acknowledge some comments that are now deleted:
It's possible that this program never ends (only theoretically) because there could be a sequence that never contains valid values. Fair point. I think this is something that could be explained to the interviewer, however I believe my example is good enough for the context.
The distribution is fair because each of the elements has the same chance of coming up.
The advantage of answering this way is that you show understanding of modern "functional-style" programming, which may be interesting to the interviewer.
The other answers are also correct. This is a different take on the problem.
The binary search is highly efficient for uniform distributions. Each member of your list has equal 'hit' probability. That's why you try the center each time.
Is there an efficient algorithm for no uniform distributions ? e.g. a distribution following a 1/x distribution.
There's a deep connection between binary search and binary trees - binary tree is basically a "precalculated" binary search where the cutting points are decided by the structure of the tree, rather than being chosen as the search runs. And as it turns out, dealing with probability "weights" for each key is sometimes done with binary trees.
One reason is because it's a fairly normal binary search tree but known in advance, complete with knowledge of the query probabilities.
Niklaus Wirth covered this in his book "Algorithms and Data Structures", in a few variants (one for Pascal, one for Modula 2, one for Oberon), at least one of which is available for download from his web site.
Binary trees aren't always binary search trees, though, and one use of a binary tree is to derive a Huffman compression code.
Either way, the binary tree is constructed by starting with the leaves separate and, at each step, joining the two least likely subtrees into a larger subtree until there's only one subtree left. To efficiently pick the two least likely subtrees at each step, a priority queue data structure is used - perhaps a binary heap.
A binary tree that's built once then never modified can have a number of uses, but one that can be efficiently updated is even more useful. There are some weight-balanced binary tree data structures out there, but I'm not familiar with them. Beware - the term "weight balanced" is commonly used where each node always has weight 1, but subtree weights are approximately balanced. Some of these may be adaptable for varied node weights, but I don't know for certain.
Anyway, for a binary search in an array, the problem is that it's possible to use an arbitrary probability distribution, but inefficient. For example, you could have a running-total-of-weights array. For each iteration of your binary search, you want to determine the half-way-through-the-probability distribution point, so you determine the value for that then search the running-total-of-weights array. You get the perfectly weight-balanced next choice for your main binary search, but you had to do a complete binary search into your running total array to do it.
The principle works, however, if you can determine that weighted mid-point without searching for a known probability distribution. The principle is the same - you need the integral of your probability distribution (replacing the running total array) and when you need a mid-point, you choose it to get an exact centre value for the integral. That's more an algebra issue than a programming issue.
One problem with a weighted binary search like this is that the worst-case performance is worse - usually by constant factors but, if the distribution is skewed enough, you may end up with effectively a linear search. If your assumed distribution is correct, the average-case performance is improved despite the occasional slow search, but if your assumed distribution is wrong you could pay for that when many searches are for items that are meant to be unlikely according to that distribution. In the binary tree form, the "unlikely" nodes are further from the root than they would be in a simply balanced (flat probability distribution assumed) binary tree.
A flat probability distribution assumption works very well even when it's completely wrong - the worst case is good, and the best and average cases must be at least that good by definition. The further you move from a flat distribution, the worse things can be if actual query probabilities turn out to be very different from your assumptions.
Let me make it precise. What you want for binary search is:
Given array A which is sorted, but have non-uniform distribution
Given left & right index L & R of search range
Want to search for a value X in A
To apply binary search, we want to find the index M in [L,R]
as the next position to look at.
Where the value X should have equal chances to be in either range [L,M-1] or [M+1,R]
In general, you of course want to pick M where you think X value should be in A.
Because even if you miss, half the total 'chance' would be eliminated.
So it seems to me you have some expectation about distribution.
If you could tell us what exactly do you mean by '1/x distribution', then
maybe someone here can help build on my suggestion for you.
Let me give a worked example.
I'll use similar interpretation of '1/x distribution' as #Leonid Volnitsky
Here is a Python code that generate the input array A
from random import uniform
# Generating input
a,b = 10,20
A = [ 1.0/uniform(a,b) for i in range(10) ]
A.sort()
# example input (rounded)
# A = [0.0513, 0.0552, 0.0562, 0.0574, 0.0576, 0.0602, 0.0616, 0.0721, 0.0728, 0.0880]
Let assume the value to search for is:
X = 0.0553
Then the estimated index of X is:
= total number of items * cummulative probability distribution up to X
= length(A) * P(x <= X)
So how to calculate P(x <= X) ?
It this case it is simple.
We reverse X back to the value between [a,b] which we will call
X' = 1/X ~ 18
Hence
P(x <= X) = (b-X')/(b-a)
= (20-18)/(20-10)
= 2/10
So the expected position of X is:
10*(2/10) = 2
Well, and that's pretty damn accurate!
To repeat the process on predicting where X is in each given section of A require some more work. But I hope this sufficiently illustrate my idea.
I know this might not seems like a binary search anymore
if you can get that close to the answer in just one step.
But admit it, this is what you can do if you know the distribution of input array.
The purpose of a binary search is that, for an array that is sorted, every time you half the array you are minimizing the worst case, e.g. the worst possible number of checks you can do is log2(entries). If you do some kind of an 'uneven' binary search, where you divide the array into a smaller and larger half, if the element is always in the larger half you can have worse worst case behaviour. So, I think binary search would still be the best algorithm to use regardless of expected distribution, just because it has the best worse case behaviour.
You have a vector of entries, say [x1, x2, ..., xN], and you're aware of the fact that the distribution of the queries is given with probability 1/x, on the vector you have. This means your queries will take place with that distribution, i.e., on each consult, you'll take element xN with higher probability.
This causes your binary search tree to be balanced considering your labels, but not enforcing any policy on the search. A possible change on this policy would be to relax the constraint of a balanced binary search tree -- smaller to the left of the parent node, greater to the right --, and actually choosing the parent nodes as the ones with higher probabilities, and their child nodes as the two most probable elements.
Notice this is not a binary search tree, as you are not dividing your search space by two in every step, but rather a rebalanced tree, with respect to your search pattern distribution. This means you're worst case of search may reach O(N). For example, having v = [10, 20, 30, 40, 50, 60]:
30
/ \
20 50
/ / \
10 40 60
Which can be reordered, or, rebalanced, using your function f(x) = 1 / x:
f([10, 20, 30, 40, 50, 60]) = [0.100, 0.050, 0.033, 0.025, 0.020, 0.016]
sort(v, f(v)) = [10, 20, 30, 40, 50, 60]
Into a new search tree, that looks like:
10 -------------> the most probable of being taken
/ \ leaving v = [[20, 30], [40, 50, 60]]
20 30 ---------> the most probable of being taken
/ \ leaving v = [[40, 50], [60]]
40 50 -------> the most probable of being taken
/ leaving v = [[60]]
60
If you search for 10, you only need one comparison, but if you're looking for 60, you'll perform O(N) comparisons, which does not qualifies this as a binary search. As pointed by #Steve314, the farthest you go from a fully balanced tree, the worse will be your worst case of search.
I will assume from your description:
X is uniformly distributed
Y=1/X is your data which you want to search and it is stored in sorted table
given value y, you need to binary search it in the above table
Binary search usually uses value in center of range (median). For uniform distribution it is possible to to speed up search by knowing approximately where in the table to we need to look for searched value.
For example if we have uniformly distributed values in [0,1] range and query is for 0.25, it is best to look not in center of range but in 1st quarter of the range.
To use the same technique for 1/X data, store in table not Y but inverse 1/Y. Search not for y but for inverse value 1/y.
Unweighted binary search isn't even optimal for uniformly distributed keys in expected terms, but it is in worst case terms.
The proportionally weighted binary search (which I have been using for decades) does what you want for uniform data, and by applying an implicit or explicit transform for other distributions. The sorted hash table is closely related (and I've known about this for decades but never bothered to try it).
In this discussion I will assume that the data is uniformly selected from 1..N and in an array of size N indexed by 1..N. If it has a different solution, e.g. a Zipfian distribution where the value is proportional to 1/index, you can apply an inverse function to flatten the distribution, or the Fisher Transform will often help (see Wikipedia).
Initially you have 1..N as the bounds, but in fact you may know the actual Min..Max. In any case we will assume we always have a closed interval [Min,Max] for the index range [L..R] we are currently searching, and initially this is O(N).
We are looking for key K and want index I so that
[I-R]/[K-Max]=[L-I]/[Min-K]=[L-R]/[Min-Max] e.g. I = [R-L]/[Max-Min]*[Max-K] + L.
Round so that the smaller partition gets larger rather than smaller (to help worst case). The expected absolute and root mean square error is <√[R-L] (based on a Poisson/Skellam or a Random Walk model - see Wikipedia). The expected number of steps is thus O(loglogN).
The worst case can be constrained to be O(logN) in several ways. First we can decide what constant we regard as acceptable, perhaps requiring steps 1. Proceeding for loglogN steps as above, and then using halving will achieve this for any such c.
Alternatively we can modify the standard base b=B=2 of the logarithm so b>2. Suppose we take b=8, then effectively c~b/B. we can then modify the rounding above so that at step k the largest partition must be at most N*b^-k. Viz keep track of the size expected if we eliminate 1/b from consideration each step which leads to worst case b/2 lgN. This will however bring our expected case back to O(log N) as we are only allowed to reduce the small partition by 1/b each time. We can restore the O(loglog N) expectation by using simple uprounding of the small partition for loglogN steps before applying the restricted rounding. This is appropriate because within a burst expected to be local to a particular value, the distribution is approximately uniform (that is for any smooth distribution function, e.g. in this case Skellam, any sufficiently small segment is approximately linear with slope given by its derivative at the centre of the segment).
As for the sorted hash, I thought I read about this in Knuth decades ago, but can't find the reference. The technique involves pushing rather than probing - (possibly weighted binary) search to find the right place or a gap then pushing aside to make room as needed, and the hash function must respect the ordering. This pushing can wrap around and so a second pass through the table is needed to pick them all up - it is useful to track Min and Max and their indexes (to get forward or reverse ordered listing start at one and track cyclically to the other; they can then also be used instead of 1 and N as initial brackets for the search as above; otherwise 1 and N can be used as surrogates).
If the load factor alpha is close to 1, then insertion is expected O(√N) for expected O(√N) items, which still amortizes to O(1) on average. This cost is expected to decrease exponentially with alpha - I believe (under Poisson assumptions) that μ ~ σ ~ √[Nexp(α)].
The above proportionally weighted binary search can used to improve on the initial probe.