First of all, this is not a homework or something like that, this is a suggestive problem from my last question Candidate in finding majority element in an array.
There are n kinds of objects O1, O2, ..., On, and there is an array F[1...n], F[i] is the number of Oi(i.e. there are F[i] Ois, array F[1...n] is given), and every F[i] > 0 .
Now use the following rule to pair objects:
if i != j, Oi can be paired with Oj,
else if i == j, Oi can not be paired with Oj.
i.e., only two different kinds of objects can be paired with each other.
Whether or not a valid pairing method may exist for the input array F[1...n]? Give out an algorithm with best time complexity to tell true or false and proves its correctness.
If there exist, output one valid paring sequence. Give out your algorithm and time/space complexity analysis.
For example,
input F[] = {10, 2, 4, 4};
Then there exists at least one valid pairing method:
2 O1s and 2 O2s paired, 4 O1s and 4 O3s paired, 4 O1s and 4 O4s paired.
One valid pairing sequence is:
(1,2) (1,2) (1,3) (1,3) (1,3) (1,3) (1,4) (1,4) (1,4) (1,4)
Checking if a solution exists in O(n)
Let s be the sum of F.
If s is odd there is no solution (intuitive)
If there exists an i such that F[i] > s/2 there is no solution (intuitive)
Otherwise, a solution exists (proof by construction follows)
Finding a solution in O(n)
# Find s
s = 0
for i in 1..n:
s += F[i]
# Find m such that F[m] is maximal
m = 1
for i in 1..n:
if F[i] > F[m]:
m = i
if s % 2 != 0 or F[m] > s/2:
fail
a = 1
b = 1
# Pair off arbitrary objects (except those of type m) until F[m] = s/2
while s/2 > F[m]:
# Find a type with a non-zero frequency
until F[a] > 0 and a != m:
a = a + 1
# Find another type with a non-zero frequency
until F[b] > 0 and b != m and b != a:
b = b + 1
count = min(F[a], F[b], s/2 - F[m])
pair(a, b, count)
# Pair off objects of type m with objects of different types
while F[m] > 0:
# Find a type with a non-zero frequency
until F[a] > 0 and a != m:
a = a + 1
pair(a, m, F[a])
end of algorithm
def pair(a, b, count):
# Pairs off 'count' pairs of types a and b
F[a] = F[a] - count
F[b] = F[b] - count
s = s - (2 * count)
output "Pairing off $a and $b ($count times)"
The two while loops are both linear. The first while loop increments either a or b by at least one at each iteration, because after matching up count pairs either F[a] is zero, or F[b] is zero, or s/2 = F[m] and the loop terminates. a and b can be incremented at most n times each before all the elements have been visited. The second while loop is also linear since it increments a at by at least one during each iteration.
The key invariants are
(1) F[m] is the largest element of F
(2) F[m] <= s/2
I think both are fairly obvious upon inspection.
With the inner loop, as long as s/2 > F[m] there must be at least two other object types with non-zero frequencies. If there was only one, say a, then
F[a] + F[m] = s
F[a] = s - F[m] > s - s/2 (from the loop condition)
F[a] > s/2
F[a] > F[m]
which is a contradiction of invariant (1).
Since there are at least two types with non-zero frequencies (besides m) the loop will be able to find types a and b and pair off objects until s/2 = F[m].
The second loop is trivial. Since exactly half the objects are of type m, each object of type m can be paired with an object of a different type.
Here's one suggestion. I'm not sure if it succeeds for every possible case, though, or if it's the most efficient possible algorithm.
Let n be the total number of indices. Construct a highest-value priority queue of the numbers of object types, where each object type is its index i. In other words, make a priority queue where the sorting values in the queue are the values of F. Associate each node with the list of all objects of that type. This will take O(n log(n)) time.
For each object type, starting with the type that has the most duplicates and proceeding to the type with the fewest, pair one object from that "class" of objects with one object for each of the other classes that still have objects remaining in them, and remove that object from that node in the queue. Since every queue item except the top one will have one fewer object in it, most of the queue will still be in priority-queue order; the top node, however, will have n-1 fewer items (or it will be empty), so heapify down to maintain the queue. Also, remove nodes with no objects left. Repeat this process with the new highest queue value until all nodes are paired. This will take O(n log(n) + k) time, where k is the total number of items. Assuming that k is significantly larger than n, the total time complexity is O(k).
Again, I'm not quite sure that this will always find a solution if one is possible. My intuition is that if you re-heapify (if necessary) after every pairing, you'll ensure that if a full pairing is possible it will be found, but (1) this would be much less efficient, and (2) I'm not sure what cases it would succeed at that the original algorithm wouldn't, and (3) I'm not entirely certain even that would work every time.
As for what values of F have no solution, obviously if there exists one class of objects that has more elements than all the other classes combined, no pairing is possible. Beyond that...I'm not really sure. It would be interesting to investigate whether my "improved" algorithm correctly evaluates every case or not.
Related
I have this problem, that I feel I am vastly overcomplicating. I feel like this should be incredibly basic, but I am stumbling on a mental block.
The question reads as follows:
Given an array of integers A[1..n], such that A[1] ≤ A[n] and for all
i, 1 ≤ i < n, we have |A[i] − A[i+ 1]| ≤ 1. Devise an semi-efficient
algorithm (better in the worst case then the native case of looking at
every cell in the array) to find any j such that A[j] = z for a given
value of z, A[1] ≤ z ≤ A[n].
My understanding of the given array is as follows: You have an array that is 1-indexed where the first element of the array is smaller than or equal to the last element of the array. Each element of the array is with in 1 of the previous one (So A[2] could be -1, 0, or +1 of A[1]'s value).
I have had several solutions to this question all of which have had there issues, here is an example of one to show my thought process.
i = 2
while i <= n {
if (A[i] == x) then
break // This can be changed into a less messy case where
// I don't use break, but this is a rough concept
else if (abs(A[i] - j) <= 1) then
i--
else
i += 2
}
This however fails when most of the values inside the array are repeating.
An array of [1 1 1 1 1 1 1 1 1 1 2] where searching for 2 for example, it would run forever.
Most of my attempted algorithms follow a similar concept of incrementing by 2, as that seems like the most logical approach when dealing with with an array that is increasing by a maximum of 1, however, I am struggling to find any that would work in a case such as [1 1 1 1 1 1 1 1 1 1 2] as they all either fail, or match the native worst case of n.
I am unsure if I am struggling because I don't understand what the question is asking, or if I am simply struggling to to put together an algorithm.
What would an algorithm look like that fits the requirements?
This can be solved via a form of modified binary search. The most important premises:
the input array always contains the element
distance between adjacent elements is always 1
there's always an increasingly ordered subarray containing the searched value
Taking it from there we can apply two strategies:
divide and conquer: we can reduce the range searched by half, since we always know which subarray will definitely contain the specified value as a part of an increasing sequence.
limiting the search-range: suppose the searched value is 3 and the limiting value on the right half of the range is 6, we can then shift the right limit to the left by 3 cells.
As code (pythonesque, but untested):
def search_semi_binary(arr, val):
low, up = 0, len(arr) - 1
while low != up:
# reduce search space
low += abs(val - arr[low])
up -= abs(val - arr[up])
# binary search
mid = (low + up) // 2
if arr[mid] == val:
return mid
elif val < arr[mid]:
# value is definitely in the lower part of the array
up = mid - 1
else:
# value is definitely in the upper part of the array
low = mid + 1
return low
The basic idea consists of two parts:
First we can reduce the search space. This uses the fact that adjacent cells of the array may only differ by one. I.e. if the lower bound of our search space has an absolute difference of 3 to val, we can shift the lower bound to the right by at least three without shifting the value out of the search window. Same applies to the upper bound.
The next step follows the basic principle of binary search using the following loop-invariant:
At the start of each iteration there exists an array-element in arr[low:up + 1] that is equal to val and arr[low] <= val <= arr[up]. This is also guaranteed after applying the search-space reduction. Depending on how mid is chosen, one of three cases can happen:
arr[mid] == val: in this case, the searched index is found
arr[mid] < val: In this case arr[mid] < val <= arr[up] must hold due to the assumption of an initial valid state
arr[mid] > val: analogous for arr[mid] > val >= arr[low]
For the latter two cases, we can pick low = mid + 1 (or up = mid - 1 respectively) and start the next iteration.
In the worst case, you'll have to look at all array elements.
Assume all elements are zero, except that a[k] = 1 for one single k, 1 ≤ k ≤ n. k isn't known, obviously. And you look for the value 1. Until you visit a[k], whatever you visit has a value of 0. Any element that you haven't visited could be equal to 1.
Let's say we are looking for a number 5. If they array starts with A[1]=1, the best case scenario is having the 5 in A[5] as it needs to be incremented at least 4 times. If A[5] = 3, then let's check A[7] as it's the closest possible solution. How do we decide it's A[7]? From the number we are looking for, let's call it R for result, we subtract what we currently have, let's call it C for current, and add the result to i as in A[i+(R-C)]
Unfortunately the above solution would apply to every scenario but the worst case scenario (when we iterate through the whole array).
Input: n arrays of integers of length p.
Output: An array of p integers built by copying contiguous subarrays of the input arrays into matching indices of the output, satisfying the following conditions.
At most one subarray is used from each input array.
Every index of the output array is filled from exactly one subarray.
The output array has the minimum possible sum.
Suppose I have 2 arrays:
[1,7,2]
[2,1,8]
So if I choose a subarray [1,7] from array 1 and subarray [8] from array 2. since these 2 subarrays are not overlapping for any index and are contiguous. We are also not taking any subarray twice from an array from which we have already chosen a subarray.
We have the number of elements in the arrays inside the collection = 2 + 1 = 3, which is the same as the length of the individual array (i.e. len(array 1) which is equal to 3). So, this collection is valid.
The sum here for [1,7] and [8] is 1 + 7 + 8 = 16
We have to find a collection of such subarrays such that the total sum of the elements of subarrays is minimum.
A solution to the above 2 arrays would be a collection [2,1] from array 1 and [2] from array 2.
This is a valid collection and the sum is 2 + 1 + 2 = 5 which is the minimum sum for any such collection in this case.
I cannot think of any optimal or correct approach, so I need help.
Some Ideas:
I tried a greedy approach by choosing minimum elements from all array for a particular index since the index is always increasing (non-overlapping) after a valid choice, I don't have to bother about storing minimum value indices for every array. But this approach is clearly not correct since it will visit the same array twice.
Another method I thought was to start from the 0th index for all arrays and start storing their sum up to k elements for every array since the no. of arrays are finite, I can store the sum upto k elements in an array. Now I tried to take a minimum across these sums and for a "minimum sum", the corresponding subarray giving this sum (i.e. k such elements in that array) can be a candidate for a valid subarray of size k, thus if we take this subarray, we can add a k + 1-th element corresponding to every array into their corresponding sum and if the original minimum still holds, then we can keep on repeating this step. When the minima fail, we can consider the subarray up to the index for which minima holds and this will be a valid starting subarray. However, this approach will also clearly fail because there could exist another subarray of size < k giving minima along with remaining index elements from our subarray of size k.
Sorting is not possible either, since if we sort then we are breaking consecutive condition.
Of course, there is a brute force method too.
I am thinking, working through a greedy approach might give a progress in the approach.
I have searched on other Stackoverflow posts, but couldn't find anything which could help my problem.
To get you started, here's a recursive branch-&-bound backtracking - and potentially exhaustive - search. Ordering heuristics can have a huge effect on how efficient these are, but without mounds of "real life" data to test against there's scant basis for picking one over another. This incorporates what may be the single most obvious ordering rule.
Because it's a work in progress, it prints stuff as it goes along: all solutions found, whenever they meet or beat the current best; and the index at which a search is cut off early, when that happens (because it becomes obvious that the partial solution at that point can't be extended to meet or beat the best full solution known so far).
For example,
>>> crunch([[5, 6, 7], [8, 0, 3], [2, 8, 7], [8, 2, 3]])
displays
new best
L2[0:1] = [2] 2
L1[1:2] = [0] 2
L3[2:3] = [3] 5
sum 5
cut at 2
L2[0:1] = [2] 2
L1[1:3] = [0, 3] 5
sum 5
cut at 2
cut at 2
cut at 2
cut at 1
cut at 1
cut at 2
cut at 2
cut at 2
cut at 1
cut at 1
cut at 1
cut at 0
cut at 0
So it found two ways to get a minimal sum 5, and the simple ordering heuristic was effective enough that all other paths to full solutions were cut off early.
def disp(lists, ixs):
from itertools import groupby
total = 0
i = 0
for k, g in groupby(ixs):
j = i + len(list(g))
chunk = lists[k][i:j]
total += sum(chunk)
print(f"L{k}[{i}:{j}] = {chunk} {total}")
i = j
def crunch(lists):
n = len(lists[0])
assert all(len(L) == n for L in lists)
# Start with a sum we know can be beat.
smallest_sum = sum(lists[0]) + 1
smallest_ixs = [None] * n
ixsofar = [None] * n
def inner(i, sumsofar, freelists):
nonlocal smallest_sum
assert sumsofar <= smallest_sum
if i == n:
print()
if sumsofar < smallest_sum:
smallest_sum = sumsofar
smallest_ixs[:] = ixsofar
print("new best")
disp(lists, ixsofar)
print("sum", sumsofar)
return
# Simple greedy heuristic: try available lists in the order
# of smallest-to-largest at index i.
for lix in sorted(freelists, key=lambda lix: lists[lix][i]):
L = lists[lix]
newsum = sumsofar
freelists.remove(lix)
# Try all slices in L starting at i.
for j in range(i, n):
newsum += L[j]
# ">" to find all smallest answers;
# ">=" to find just one (potentially faster)
if newsum > smallest_sum:
print("cut at", j)
break
ixsofar[j] = lix
inner(j + 1, newsum, freelists)
freelists.add(lix)
inner(0, 0, set(range(len(lists))))
How bad is brute force?
Bad. A brute force way to compute it: say there are n lists each with p elements. The code's ixsofar vector contains p integers each in range(n). The only constraint is that all occurrences of any integer that appears in it must be consecutive. So a brute force way to compute the total number of such vectors is to generate all p-tuples and count the number that meet the constraints. This is woefully inefficient, taking O(n**p) time, but is really easy, so hard to get wrong:
def countb(n, p):
from itertools import product, groupby
result = 0
seen = set()
for t in product(range(n), repeat=p):
seen.clear()
for k, g in groupby(t):
if k in seen:
break
seen.add(k)
else:
#print(t)
result += 1
return result
For small arguments, we can use that as a sanity check on the next function, which is efficient. This builds on common "stars and bars" combinatorial arguments to deduce the result:
def count(n, p):
# n lists of length p
# for r regions, r from 1 through min(p, n)
# number of ways to split up: comb((p - r) + r - 1, r - 1)
# for each, ff(n, r) ways to spray in list indices = comb(n, r) * r!
from math import comb, prod
total = 0
for r in range(1, min(n, p) + 1):
total += comb(p-1, r-1) * prod(range(n, n-r, -1))
return total
Faster
Following is the best code I have for this so far. It builds in more "smarts" to the code I posted before. In one sense, it's very effective. For example, for randomized p = n = 20 inputs it usually finishes within a second. That's nothing to sneeze at, since:
>>> count(20, 20)
1399496554158060983080
>>> _.bit_length()
71
That is, trying every possible way would effectively take forever. The number of cases to try doesn't even fit in a 64-bit int.
On the other hand, boost n (the number of lists) to 30, and it can take an hour. At 50, I haven't seen a non-contrived case finish yet, even if left to run overnight. The combinatorial explosion eventually becomes overwhelming.
OTOH, I'm looking for the smallest sum, period. If you needed to solve problems like this in real life, you'd either need a much smarter approach, or settle for iterative approximation algorithms.
Note: this is still a work in progress, so isn't polished, and prints some stuff as it goes along. Mostly that's been reduced to running a "watchdog" thread that wakes up every 10 minutes to show the current state of the ixsofar vector.
def crunch(lists):
import datetime
now = datetime.datetime.now
start = now()
n = len(lists[0])
assert all(len(L) == n for L in lists)
# Start with a sum we know can be beat.
smallest_sum = min(map(sum, lists)) + 1
smallest_ixs = [None] * n
ixsofar = [None] * n
import threading
def watcher(stop):
if stop.wait(60):
return
lix = ixsofar[:]
while not stop.wait(timeout=600):
print("watch", now() - start, smallest_sum)
nlix = ixsofar[:]
for i, (a, b) in enumerate(zip(lix, nlix)):
if a != b:
nlix.insert(i,"--- " + str(i) + " -->")
print(nlix)
del nlix[i]
break
lix = nlix
stop = threading.Event()
w = threading.Thread(target=watcher, args=[stop])
w.start()
def inner(i, sumsofar, freelists):
nonlocal smallest_sum
assert sumsofar <= smallest_sum
if i == n:
print()
if sumsofar < smallest_sum:
smallest_sum = sumsofar
smallest_ixs[:] = ixsofar
print("new best")
disp(lists, ixsofar)
print("sum", sumsofar, now() - start)
return
# If only one input list is still free, we have to take all
# of its tail. This code block isn't necessary, but gives a
# minor speedup (skips layers of do-nothing calls),
# especially when the length of the lists is greater than
# the number of lists.
if len(freelists) == 1:
lix = freelists.pop()
L = lists[lix]
for j in range(i, n):
ixsofar[j] = lix
sumsofar += L[j]
if sumsofar >= smallest_sum:
break
else:
inner(n, sumsofar, freelists)
freelists.add(lix)
return
# Peek ahead. The smallest completion we could possibly get
# would come from picking the smallest element in each
# remaining column (restricted to the lists - rows - still
# available). This probably isn't achievable, but is an
# absolute lower bound on what's possible, so can be used to
# cut off searches early.
newsum = sumsofar
for j in range(i, n): # pick smallest from column j
newsum += min(lists[lix][j] for lix in freelists)
if newsum >= smallest_sum:
return
# Simple greedy heuristic: try available lists in the order
# of smallest-to-largest at index i.
sortedlix = sorted(freelists, key=lambda lix: lists[lix][i])
# What's the next int in the previous slice? As soon as we
# hit an int at least that large, we can do at least as well
# by just returning, to let the caller extend the previous
# slice instead.
if i:
prev = lists[ixsofar[i-1]][i]
else:
prev = lists[sortedlix[-1]][i] + 1
for lix in sortedlix:
L = lists[lix]
if prev <= L[i]:
return
freelists.remove(lix)
newsum = sumsofar
# Try all non-empty slices in L starting at i.
for j in range(i, n):
newsum += L[j]
if newsum >= smallest_sum:
break
ixsofar[j] = lix
inner(j + 1, newsum, freelists)
freelists.add(lix)
inner(0, 0, set(range(len(lists))))
stop.set()
w.join()
Bounded by DP
I've had a lot of fun with this :-) Here's the approach they were probably looking for, using dynamic programming (DP). I have several programs that run faster in "smallish" cases, but none that can really compete on a non-contrived 20x50 case. The runtime is O(2**n * n**2 * p). Yes, that's more than exponential in n! But it's still a minuscule fraction of what brute force can require (see above), and is a hard upper bound.
Note: this is just a loop nest slinging machine-size integers, and using no "fancy" Python features. It would be easy to recode in C, where it would run much faster. As is, this code runs over 10x faster under PyPy (as opposed to the standard CPython interpreter).
Key insight: suppose we're going left to right, have reached column j, the last list we picked from was D, and before that we picked columns from lists A, B, and C. How can we proceed? Well, we can pick the next column from D too, and the "used" set {A, B, C} doesn't change. Or we can pick some other list E, the "used" set changes to {A, B, C, D}, and E becomes the last list we picked from.
Now in all these cases, the details of how we reached state "used set {A, B, C} with last list D at column j" make no difference to the collection of possible completions. It doesn't matter how many columns we picked from each, or the order in which A, B, C were used: all that matters to future choices is that A, B, and C can't be used again, and D can be but - if so - must be used immediately.
Since all ways of reaching this state have the same possible completions, the cheapest full solution must have the cheapest way of reaching this state.
So we just go left to right, one column at a time, and remember for each state in the column the smallest sum reaching that state.
This isn't cheap, but it's finite ;-) Since states are subsets of row indices, combined with (the index of) the last list used, there are 2**n * n possible states to keep track of. In fact, there are only half that, since the way sketched above never includes the index of the last-used list in the used set, but catering to that would probably cost more than it saves.
As is, states here are not represented explicitly. Instead there's just a large list of sums-so-far, of length 2**n * n. The state is implied by the list index: index i represents the state where:
i >> n is the index of the last-used list.
The last n bits of i are a bitset, where bit 2**j is set if and only if list index j is in the set of used list indices.
You could, e.g., represent these by dicts mapping (frozenset, index) pairs to sums instead, but then memory use explodes, runtime zooms, and PyPy becomes much less effective at speeding it.
Sad but true: like most DP algorithms, this finds "the best" answer but retains scant memory of how it was reached. Adding code to allow for that is harder than what's here, and typically explodes memory requirements. Probably easiest here: write new to disk at the end of each outer-loop iteration, one file per column. Then memory use isn't affected. When it's done, those files can be read back in again, in reverse order, and mildly tedious code can reconstruct the path it must have taken to reach the winning state, working backwards one column at a time from the end.
def dumbdp(lists):
import datetime
_min = min
now = datetime.datetime.now
start = now()
n = len(lists)
p = len(lists[0])
assert all(len(L) == p for L in lists)
rangen = range(n)
USEDMASK = (1 << n) - 1
HUGE = sum(sum(L) for L in lists) + 1
new = [HUGE] * (2**n * n)
for i in rangen:
new[i << n] = lists[i][0]
for j in range(1, p):
print("working on", j, now() - start)
old = new
new = [HUGE] * (2**n * n)
for key, g in enumerate(old):
if g == HUGE:
continue
i = key >> n
new[key] = _min(new[key], g + lists[i][j])
newused = (key & USEDMASK) | (1 << i)
for i in rangen:
mask = 1 << i
if newused & mask == 0:
newkey = newused | (i << n)
new[newkey] = _min(new[newkey],
g + lists[i][j])
result = min(new)
print("DONE", result, now() - start)
return result
We are given an unsorted array A of integers (duplicates allowed) with size N possibly large. We can count the number of pairs with indices i < j, for which A[i] < A[j], let's call this X.
We can change maximum one element from the array with a cost equal to the difference in absolute values (for instance, if we replace element on index k with the new number K, the cost Y is | A[k] - K |).
We can only replace this element with other elements found in the array.
We want to find the minimum possible value of X + Y.
Some examples:
[1,2,2] should return 1 (change the 1 to 2 such that the array becomes [2,2,2])
[2,2,3] should return 1 (change the 3 to 2)
[2,1,1] should return 0 (because no changes are necessary)
[1,2,3,4] should return 6 (this is already the minimum possible value)
[4,4,5,5] should return 3 (this can accomplished by changing the first 4 into a 5 or the last 5 in a 4)
The number of pairs can be found with a naive O(n²) solution, here in Python:
def calc_x(arr):
n = len(arr)
cnt = 0
for i in range(n):
for j in range(i+1, n):
if arr[j] > arr[i]:
cnt += 1
return cnt
A brute-force solution is easily written as for example:
def f(arr):
best_val = calc_x(arr)
used = set(arr)
for i, v in enumerate(arr):
for replacement in used:
if replacement == v:
continue
arr2 = arr[0:i] + replacement + arr[i:]
y = abs(replacement - v)
x = calc_x(arr2)
best_val = min(best_val, x + y)
return best_val
We can count for each element the number of items right of it larger than itself in O(n*log(n)) using for instance an AVL-tree or some variation on merge sort.
However, we still have to search which element to change and what improvement it can achieve.
This was given as an interview question and I would like some hints or insights as how to solve this problem efficiently (data structures or algorithm).
Definitely go for a O(n log n) complexity when counting inversions.
We can see that when you change a value at index k, you can either:
1) increase it, and then possibly reduce the number of inversions with elements bigger than k, but increase the number of inversions with elements smaller than k
2) decrease it (the opposite thing happens)
Let's try not to count x every time you change a value. What do you need to know?
In case 1):
You have to know how many elements on the left are smaller than your new value v and how many elements on the right are bigger than your value. You can pretty easily check that in O (n). So what is your x now? You can count it with the following formula:
prev_val - your previous value
prev_x - x that you've counted at the beginning of your program
prev_l - number of elements on the left smaller than prev_val
prev_r - number of elements on the right bigger than prev_val
v - new value
l - number of elements on the right smaller than v
r - number of elements on the right bigger than v
new_x = prev_x + r + l - prev_l - prev_r
In the second case you pretty much do the opposite thing.
Right now you get something like O( n^3 ) instead of O (n^3 log n), which is probably still bad. Unfortunately that's all what I came up for now. I'll definitely tell you if I come up with sth better.
EDIT: What about memory limit? Is there any? If not, you can just for each element in the array make two sets with elements before and after the current one. Then you can find the amount of smaller/bigger in O (log n), making your time complexity O (n^2 log n).
EDIT 2: We can also try to check, what element would be the best to change to a value v, for every possible value v. You can make then two sets and add/erase elements from them while checking for every element, making the time complexity O(n^2 log n) without using too much space. So the algorithm would be:
1) determine every value v that you can change any element, calculate x
2) for each possible value v:
make two sets, push all elements into the second one
for each element e in array:
add previous element (if there's any) to the first set and erase element e from the second set, then count number of bigger/smaller elements in set 1 and 2 and calculate new x
EDIT 3: Instead of making two sets, you could go with prefix sum for a value. That's O (n^2) already, but I think we can go even better than this.
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. :)
I was asked this question:
You are given two lists of integers, each of which is sorted in ascending order and each of which has length n. All integers in the two lists are different. You wish to find the n-th smallest element of the union of the two lists. (That is, if you concatenated the lists and sorted the resulting list in ascending order, the element which would be at the n-th position.)
My Solution:
Assume that lists are 0-indexed.
O(n) solution:
A straight-forward solution is to observe that the arrays are already sorted,so we can merge them, and stop after n steps. The first n-1 elements do not need to be copied
into a new array, so this solution takes O(n) time and O(1) memory.
O(log2 n) solution:
The O(log2 n) solution uses alternates binary search on each list. In short, it takes the middle element in the current search interval in the first list (l1[p1]) and searches for it in l2. Since the elements are unique, we will find at most 2 values closest to l1[p1]. Depending on their values relative to l1[p1-1] and l1[p1 + 1] and their indices p21 and p22, we either return the n-th element or recurse: If the sum of any out of the (at most) 3 indices in l1 can be combined with one of the (at most) 2 indices in l2 so that l1[p1'] and l2[p2'] would be right next to each other in the sorted union of the two lists and p1' + p2' = n or p1' + p2' = n + 1, we return one of the 5 elements. If p1 + p2 > n, we recurse to left half of the search interval in l1, otherwise we recurse to the right interval. This way, for out of the O(log n) possible midpoints in l1 we do an O(log n) binary search in l2. Therefore the running time is O(log2 n).
O(log n) solution:
Assuming the lists l1 and l2 have constant access time to any of their elements, we
can use a modified version of binary search to get an O(log n) solution. The easiest approach is to search for an index p1 in just one of the lists and calculate the corresponding index p2 in the other list so that p1 + p2 = n at all times. (This assumes the lists are indexed from 1.)
First we check for the special case when all elements of one list are smaller than any element in the other list:
If l1[n] < l2[0] return l1[n].
If l2[n] < l1[0] return l2[n].
If we do not find the n-th smallest element after this step, call findNth(1,n) with the approximate pseudocode:
findNth(start,end)
p1 = (start + end)/2
p2 = n-p1
if l1[p1] < l2[p2]:
if l1[p1 + 1] > l2[p2]:
return l2[p2]
else:
return findNth(p1+1, end)
else:
if l2[p2 + 1] > l1[p1]:
return l1[p1]
else:
return findNth(start,p1-1)
Element l2[p2] is returned when l2[p2] is greater than exactly p1 + p2-1 = n-1 elements
(and therefore is the n-th smallest). l1[p1] is returned under the same but symmetric conditions. If l1[p1] < l2[p2] and l1[p1+1] < l2[p2], the rank of l2[p2] is greater than n, so we need to take more elements from l1 and less from l2. Therefore we search for p1 in the upper half of the previous search interval. On the other hand, if l2[p2] < l1[p1] and l2[p2 + 1] < l1[p1], the rank of l1[p1] is greater than n. Therefore the real p1 will lie in the bottom half of our current search interval.Since we are halving the size of the problem at each call to findNth and we need to do only constant work to halve the problem size, the recurrence for this algorithm is T(n) = T(n/2) +O(1), which has an O(log n)-time solution.
Interviewer continually ask me different approaches for above problem.I have proposed above three approaches.Is they are correct?Is there any other best possible solution for this question? Actually this question asked lot of times so please provide some good stuff about it.
Not sure if you took a look at this: http://www.leetcode.com/2011/01/find-k-th-smallest-element-in-union-of.html
That solve a more generalized version of the problem you are asking about. Definitely log complexity is possible...
I think this will be the best solution . .
->1 2 3 4 5 6 7 8 9
->10 11 12 13 14 15 16 17 18
take two pointers i and j each pointing at start of arrays, increment i if a[i]< b[j]
increment j if a[i]>b[j]
do this n times.
linear O(n) O(1) space solution.