Now I have N different intergers, I need to find an interval that has the most numbers whose value is between the endpoints of the interval in O(NlogN) time. I call it a "divide and conquer" problem because it is in my final exam's "divide and conquer" category. I have been thinking about it for 2 weeks and have done a lot of experiments, none of them are right(compared to a brute force algorithm). Could someone help me?
examples:
8,1,3,4,7. The answer is 1-7.
2,6,5,4,9,8. The answer is 2-9 or 2-8.
I think the word "interval" doesn't express my meaning. I mean to find a subsequence of the array that has the most numbers whose value is between the endpoints of the subsequence. Eg.1: "1,3,4,7" has two numbers(3,4), and eg.2: both "2,6,5,4,9" and "2,6,5,4,9,8" have three numbers(6,5,4).
here is my code(O(n^2)). #Vaughn Cato I use this to compare to your code.
#! /usr/bin/env python
#coding=utf-8
import itertools
def n2(numbers):
a = [0]*len(numbers)
ans = -1
l = 0
r = 0
for j in range(1,len(numbers)):
t = 0
for i in range(j-1,-1,-1):
if numbers[i]<numbers[j]:
x = t - a[i]
if x>ans:
ans = x
l = i
r = j
t += 1
else:
a[i] += 1
return (numbers[l],numbers[r],ans)
def countBetween(numbers,left,right):
cnt = 0
for i in range(left+1,right):
if numbers[left]<numbers[i]<numbers[right]:
cnt += 1
return cnt
for numbers in itertools.permutations(range(5)):
ans1=n2(numbers)
ans2=longestInterval(numbers)
if(ans1[2]!=ans2[2]):
print ans1,ans2,numbers
NOTE: This doesn't actually work, but it might give you some ideas.
Think of it this way:
Let X be the array of numbers.
Let s be the index of the start of the subsequence.
Let e be the index of the end of the subsequence.
If you pick an arbitrary partition index p, then the longest subsequence either goes across this partition or it falls to the left or right of that partition. If the longest subsequence goes across this partition, then s < p <= e. To find s, find the index with the most numbers between s and p which are greater than X[s]. To find 'e', find the index with the most numbers between p and e which are less than X[e].
You can recursively check the left and right sides to see if you can find a longer subsequence.
Finding which index has the most greater numbers to the right or the most less than numbers to the left can be done in linear time if you have the indices of X sorted by value:
To find the start index, begin with the first index of your sorted list of indices and say it is the best so far. If the next index is greater than the best so far, then any future index will need to be even farther to the left than our current best to be the new best, so we subtract one from our best index (but remember what the best index really was). If the next index is to the left of our best index, then make it be the best index. Keep repeating this process, for each of the indices in order.
You can do a similar procedure to find the best index for the end on the right side.
The only remaining trick is to maintain the sorted list of indices for whatever range we are working on. This can be done by sorting the entire set of numbers initially and finding their indices, then at each level of the recursion we can split the sorted indices into two sublists in linear time.
Here is a python implementation of the idea:
# Find the index from the given indices that has the most numbers to the
# right of it which are greater in value. The indices are sorted by
# the value of the numbers at that index. We don't even need to know
# what the numbers are.
def longestLowerSequence(indices):
best_index=indices[0]
target_index=best_index
for i in range(0,len(indices)):
if indices[i]<target_index:
best_index=indices[i]
target_index=best_index
else:
target_index-=1
return best_index
# Find the index from the given indices that has the most numbers to the
# left of it which are less in value.
def longestUpperSequence(indices):
n=len(indices)
best_index=indices[n-1]
target_index=best_index
for i in range(0,n):
if indices[n-1-i]>target_index:
best_index=indices[n-1-i]
target_index=best_index
else:
target_index+=1
return best_index
# Return the pair of indices which has the most values between it.
def longestRangeFromSortedIndices(numbers,indices,begin,end):
assert end>begin
if end-begin<=2:
return (indices[begin],indices[end-1])
assert type(indices) is list
partition=(begin+end)/2
left_indices=filter(lambda index: index<partition,indices)
right_indices=filter(lambda index: index>=partition,indices)
assert len(left_indices)>0
assert len(right_indices)>0
left=longestLowerSequence(left_indices)
right=longestUpperSequence(right_indices)
left_range=longestRangeFromSortedIndices(numbers,indices,begin,partition)
right_range=longestRangeFromSortedIndices(numbers,indices,partition,end)
best_size=countBetween(numbers,left,right)
best_range=(left,right)
left_size=countBetween(numbers,left_range[0],left_range[1])
right_size=countBetween(numbers,right_range[0],right_range[1])
if left_size>best_size:
best_size=left_size
best_range=left_range
if right_size>best_size:
best_size=right_size
best_range=right_range
return best_range
def sortedIndices(numbers):
return sorted(range(len(numbers)),key=lambda i: numbers[i])
def longestInterval(numbers):
indices=sortedIndices(numbers)
longest_range=longestRangeFromSortedIndices(numbers,indices,0,len(numbers))
return (numbers[longest_range[0]],numbers[longest_range[1]])
I believe this is a variant of the maximum subarray problem.
It can be solved using divide and conquer as follows:
Divide the integer array into equal halves
Compute the results R1, R2 on both halves respectively(R1, R2 are lengths of the maximum intervals for each half, and the start and end points are stored as well)
Obtain the minimum integer MIN from the first half and the maximum integer MAX from the second half and compute result R3 as the distance from MIN to MAX in the original array (Min and Max are the start and end point respectively)
Return the largest of R1, R2 and R3 as the result of the entire problem
Why this works:
The largest interval comes from one of the three cases: 1) the first half 2） the second half 3) across the two halves. Thus, computing the largest of the three yields the optimal result.
Time complexity:
Solving the recurrence:
T(n) = 2T(n/2) + O(n)
gives T(n) = O(nlogn). Note: as the recurrence indicates, we solve two subproblems of half size（2T(n/2)）and find the minimum and maximum integers in two halves in linear time(O(n)).
Related
I'm looking for an algorithm to generate or iterate through all permutations of a list of objects such that:
They are generated by fewest to least positional changes from the original. So first all the permutations with a single pair of elements swapped, then all the permutations with only two pairs of elements swapped, etc.
The list generated is complete, so for n objects in a list there should be n! total, unique permutations.
Ideally (but not necessarily) there should be a way of specifying (and generating) a particular permutation without having to generate the full list first and then reference the index.
The speed of the algorithm is not particularly important.
I've looked through all the permutation algorithms that I can find, and none so far have met criteria 1 and 2, let alone 3.
I have an idea how I could write this algorithm myself using recursion, and filtering for duplicates to only get unique permutations. However, if there is any existing algorithm I'd much rather use something proven.
This code answers your requirement #3, which is to compute permutation at index N directly.
This code relies on the following principle:
The first permutation is the identity; then the next (n choose 2) permutations just swap two elements; then the next (n choose 3)(subfactorial(3)) permutations derange 3 elements; then the next (n choose 4)(subfactorial(4)) permutations derange 4 elements; etc. To find the Nth permutation, first figure out how many elements it deranges by finding the largest K such that sum[k = 0 ^ K] (n choose k) subfactorial(k) ⩽ N.
This number K is found by function number_of_derangements_for_permutation_at_index in the code.
Then, the relevant subset of indices which must be deranged is computed efficiently using more_itertools.nth_combination.
However, I didn't have a function nth_derangement to find the relevant derangement of the deranged subset of indices. Hence the last step of the algorithm, which computes this derangement, could be optimised if there exists an efficient function to find the nth derangement of a sequence efficiently.
As a result, this last step takes time proportional to idx_r, where idx_r is the index of the derangement, a number between 0 and factorial(k), where k is the number of elements which are deranged by the returned permutation.
from sympy import subfactorial
from math import comb
from itertools import count, accumulate, pairwise, permutations
from more_itertools import nth_combination, nth
def number_of_derangements_for_permutation_at_index(n, idx):
#n = len(seq)
for k, (low_acc, high_acc) in enumerate(pairwise(accumulate((comb(n,k) * subfactorial(k) for k in count(2)), initial=1)), start=2):
if low_acc <= idx < high_acc:
return k, low_acc
def is_derangement(seq, perm):
return all(i != j for i,j in zip(seq, perm))
def lift_permutation(seq, deranged, permutation):
result = list(seq)
for i,j in zip(deranged, permutation):
result[i] = seq[j]
return result
# THIS FUNCTION NOT EFFICIENT
def nth_derangement(seq, idx):
return nth((p for p in permutations(seq) if is_derangement(seq, p)),
idx)
def nth_permutation(seq, idx):
if idx == 0:
return list(seq)
n = len(seq)
k, acc = number_of_derangements_for_permutation_at_index(n, idx)
idx_q, idx_r = divmod(idx - acc, subfactorial(k))
deranged = nth_combination(range(n), k, idx_q)
derangement = nth_derangement(deranged, idx_r) # TODO: FIND EFFICIENT VERSION
return lift_permutation(seq, deranged, derangement)
Testing for correctness on small data:
print( [''.join(nth_permutation('abcd', i)) for i in range(24)] )
# ['abcd',
# 'bacd', 'cbad', 'dbca', 'acbd', 'adcb', 'abdc',
# 'bcad', 'cabd', 'bdca', 'dacb', 'cbda', 'dbac', 'acdb', 'adbc',
# 'badc', 'bcda', 'bdac', 'cadb', 'cdab', 'cdba', 'dabc', 'dcab', 'dcba']
Testing for speed on medium data:
from math import factorial
seq = 'abcdefghij'
n = len(seq) # 10
N = factorial(n) // 2 # 1814400
perm = ''.join(nth_permutation(seq, N))
print(perm)
# fcjdibaehg
Imagine a graph with n! nodes labeled with every permutation of n elements. If we add edges to this graph such that nodes which can be obtained by swapping one pair of elements are connected, an answer to your problem is obtained by doing a breadth-first search from whatever node you like.
You can actually generate the graph or just let it be implied and just deduce at each stage what nodes should be adjacent (and of course, keep track of ones you've already visited, to avoid revisiting them).
I concede this probably doesn't help with point 3, but maybe is a viable strategy for getting points 1 and 2 answered.
To solve 1 & 2, you could first generate all possible permutations, keeping track of how many swaps occurred during generation for each list. Then sort them by number of swaps. Which I think is O(n! + nlgn) = O(n!)
Question:
Given an array A of integers and a score S = 0. For each place in the array, you can do one of the following:
Place a "(". The score would be S += Ai
Place a ")". The score would be S -= Ai
Skip it
What is the highest score you can get so that the brackets are balanced?
Limits:
|Ai| <= 10^9
Size of array A: <= 10^5
P/S:
I have tried many ways but my best take is a brute force that takes O(3^n). Is there a way to do this problem in O(n.logn) or less?
You can do this in O(n log n) time with a max-heap.
First, remove the asymmetry in the operations. Rather than having open and closed brackets, assume we start off with a running sum of -sum(A), i.e. all closed brackets. Now, for every element in A, we can add it to our running sum either zero, one or two times, corresponding to leaving a closed bracket, removing the closed bracket, or adding an open bracket, respectively. The balance constraint now says that after processing the first k elements, we have:
Made at least k additions, for all integers k,
We make length(A) total additions.
We have added the final element to our sum either zero or one times.
Suppose that after processing the first k elements, we have made k additions, and that we have the maximum score possible of all such configurations. We can extend this to a maximum score configuration of the first k+1 elements with k+1 additions, greedily. We have a new choice going forward of adding the k+1-th element to our sum up to two times, but can only add it at most once now. Simply choose the largest seen element that has not yet been added to our sum two times, and add it to our sum: this must also be a maximum-score configuration, or we can show the old configuration wasn't maximum either.
Python Code: (All values are negated because Python only has a min-heap)
def solve(nums: List[int]) -> int:
"""Given an array of integers, return the maximum sum achievable.
We must add k elements from nums and subtract k elements from nums,
left to right and all distinct, so that at no point have we subtracted
more elements than we have added.
"""
max_heap = []
running_sum = 0
# Balance will be 0 after all loop iterations.
for value in nums:
running_sum -= value # Assume value is subtracted
heapq.heappush(max_heap, -value) # Option to not subtract value
heapq.heappush(max_heap, -value) # Option to add value
# Either un-subtract or add the largest previous free element
running_sum -= heapq.heappop(max_heap)
return running_sum
You can do this in O(n2) time by using a two-dimensional array highest_score, where highest_score[i][b] is the highest score achievable after position i with b open brackets yet to be closed. Each element highest_score[i][b] depends only on highest_score[i−1][b−1], highest_score[i−1][b], and highest_score[i−1][b+1] (and of course A[i]), so each row highest_score[i] can be computed in O(n) time from the previous row highest_score[i−1], and the final answer is highest_score[n][0].
(Note: that uses O(n2) space, but since each row of highest_score depends only on the previous row, you can actually do it in O(n) by reusing rows. But the asymptotic runtime complexity will be the same either way.)
Given two sorted arrays of numbers, we want to find the pair with the kth largest possible sum. (A pair is one element from the first array and one element from the second array). For example, with arrays
[2, 3, 5, 8, 13]
[4, 8, 12, 16]
The pairs with largest sums are
13 + 16 = 29
13 + 12 = 25
8 + 16 = 24
13 + 8 = 21
8 + 12 = 20
So the pair with the 4th largest sum is (13, 8). How to find the pair with the kth largest possible sum?
Also, what is the fastest algorithm? The arrays are already sorted and sizes M and N.
I am already aware of the O(Klogk) solution , using Max-Heap given here .
It also is one of the favorite Google interview question , and they demand a O(k) solution .
I've also read somewhere that there exists a O(k) solution, which i am unable to figure out .
Can someone explain the correct solution with a pseudocode .
P.S.
Please DON'T post this link as answer/comment.It DOESN'T contain the answer.
I start with a simple but not quite linear-time algorithm. We choose some value between array1[0]+array2[0] and array1[N-1]+array2[N-1]. Then we determine how many pair sums are greater than this value and how many of them are less. This may be done by iterating the arrays with two pointers: pointer to the first array incremented when sum is too large and pointer to the second array decremented when sum is too small. Repeating this procedure for different values and using binary search (or one-sided binary search) we could find Kth largest sum in O(N log R) time, where N is size of the largest array and R is number of possible values between array1[N-1]+array2[N-1] and array1[0]+array2[0]. This algorithm has linear time complexity only when the array elements are integers bounded by small constant.
Previous algorithm may be improved if we stop binary search as soon as number of pair sums in binary search range decreases from O(N2) to O(N). Then we fill auxiliary array with these pair sums (this may be done with slightly modified two-pointers algorithm). And then we use quickselect algorithm to find Kth largest sum in this auxiliary array. All this does not improve worst-case complexity because we still need O(log R) binary search steps. What if we keep the quickselect part of this algorithm but (to get proper value range) we use something better than binary search?
We could estimate value range with the following trick: get every second element from each array and try to find the pair sum with rank k/4 for these half-arrays (using the same algorithm recursively). Obviously this should give some approximation for needed value range. And in fact slightly improved variant of this trick gives range containing only O(N) elements. This is proven in following paper: "Selection in X + Y and matrices with sorted rows and columns" by A. Mirzaian and E. Arjomandi. This paper contains detailed explanation of the algorithm, proof, complexity analysis, and pseudo-code for all parts of the algorithm except Quickselect. If linear worst-case complexity is required, Quickselect may be augmented with Median of medians algorithm.
This algorithm has complexity O(N). If one of the arrays is shorter than other array (M < N) we could assume that this shorter array is extended to size N with some very small elements so that all calculations in the algorithm use size of the largest array. We don't actually need to extract pairs with these "added" elements and feed them to quickselect, which makes algorithm a little bit faster but does not improve asymptotic complexity.
If k < N we could ignore all the array elements with index greater than k. In this case complexity is equal to O(k). If N < k < N(N-1) we just have better complexity than requested in OP. If k > N(N-1), we'd better solve the opposite problem: k'th smallest sum.
I uploaded simple C++11 implementation to ideone. Code is not optimized and not thoroughly tested. I tried to make it as close as possible to pseudo-code in linked paper. This implementation uses std::nth_element, which allows linear complexity only on average (not worst-case).
A completely different approach to find K'th sum in linear time is based on priority queue (PQ). One variation is to insert largest pair to PQ, then repeatedly remove top of PQ and instead insert up to two pairs (one with decremented index in one array, other with decremented index in other array). And take some measures to prevent inserting duplicate pairs. Other variation is to insert all possible pairs containing largest element of first array, then repeatedly remove top of PQ and instead insert pair with decremented index in first array and same index in second array. In this case there is no need to bother about duplicates.
OP mentions O(K log K) solution where PQ is implemented as max-heap. But in some cases (when array elements are evenly distributed integers with limited range and linear complexity is needed only on average, not worst-case) we could use O(1) time priority queue, for example, as described in this paper: "A Complexity O(1) Priority Queue for Event Driven Molecular Dynamics Simulations" by Gerald Paul. This allows O(K) expected time complexity.
Advantage of this approach is a possibility to provide first K elements in sorted order. Disadvantages are limited choice of array element type, more complex and slower algorithm, worse asymptotic complexity: O(K) > O(N).
EDIT: This does not work. I leave the answer, since apparently I am not the only one who could have this kind of idea; see the discussion below.
A counter-example is x = (2, 3, 6), y = (1, 4, 5) and k=3, where the algorithm gives 7 (3+4) instead of 8 (3+5).
Let x and y be the two arrays, sorted in decreasing order; we want to construct the K-th largest sum.
The variables are: i the index in the first array (element x[i]), j the index in the second array (element y[j]), and k the "order" of the sum (k in 1..K), in the sense that S(k)=x[i]+y[j] will be the k-th greater sum satisfying your conditions (this is the loop invariant).
Start from (i, j) equal to (0, 0): clearly, S(1) = x[0]+y[0].
for k from 1 to K-1, do:
if x[i+1]+ y[j] > x[i] + y[j+1], then i := i+1 (and j does not change) ; else j:=j+1
To see that it works, consider you have S(k) = x[i] + y[j]. Then, S(k+1) is the greatest sum which is lower (or equal) to S(k), and such as at least one element (i or j) changes. It is not difficult to see that exactly one of i or j should change.
If i changes, the greater sum you can construct which is lower than S(k) is by setting i=i+1, because x is decreasing and all the x[i'] + y[j] with i' < i are greater than S(k). The same holds for j, showing that S(k+1) is either x[i+1] + y[j] or x[i] + y[j+1].
Therefore, at the end of the loop you found the K-th greater sum.
tl;dr: If you look ahead and look behind at each iteration, you can start with the end (which is highest) and work back in O(K) time.
Although the insight underlying this approach is, I believe, sound, the code below is not quite correct at present (see comments).
Let's see: first of all, the arrays are sorted. So, if the arrays are a and b with lengths M and N, and as you have arranged them, the largest items are in slots M and N respectively, the largest pair will always be a[M]+b[N].
Now, what's the second largest pair? It's going to have perhaps one of {a[M],b[N]} (it can't have both, because that's just the largest pair again), and at least one of {a[M-1],b[N-1]}. BUT, we also know that if we choose a[M-1]+b[N-1], we can make one of the operands larger by choosing the higher number from the same list, so it will have exactly one number from the last column, and one from the penultimate column.
Consider the following two arrays: a = [1, 2, 53]; b = [66, 67, 68]. Our highest pair is 53+68. If we lose the smaller of those two, our pair is 68+2; if we lose the larger, it's 53+67. So, we have to look ahead to decide what our next pair will be. The simplest lookahead strategy is simply to calculate the sum of both possible pairs. That will always cost two additions, and two comparisons for each transition (three because we need to deal with the case where the sums are equal);let's call that cost Q).
At first, I was tempted to repeat that K-1 times. BUT there's a hitch: the next largest pair might actually be the other pair we can validly make from {{a[M],b[N]}, {a[M-1],b[N-1]}. So, we also need to look behind.
So, let's code (python, should be 2/3 compatible):
def kth(a,b,k):
M = len(a)
N = len(b)
if k > M*N:
raise ValueError("There are only %s possible pairs; you asked for the %sth largest, which is impossible" % M*N,k)
(ia,ib) = M-1,N-1 #0 based arrays
# we need this for lookback
nottakenindices = (0,0) # could be any value
nottakensum = float('-inf')
for i in range(k-1):
optionone = a[ia]+b[ib-1]
optiontwo = a[ia-1]+b[ib]
biggest = max((optionone,optiontwo))
#first deal with look behind
if nottakensum > biggest:
if optionone == biggest:
newnottakenindices = (ia,ib-1)
else: newnottakenindices = (ia-1,ib)
ia,ib = nottakenindices
nottakensum = biggest
nottakenindices = newnottakenindices
#deal with case where indices hit 0
elif ia <= 0 and ib <= 0:
ia = ib = 0
elif ia <= 0:
ib-=1
ia = 0
nottakensum = float('-inf')
elif ib <= 0:
ia-=1
ib = 0
nottakensum = float('-inf')
#lookahead cases
elif optionone > optiontwo:
#then choose the first option as our next pair
nottakensum,nottakenindices = optiontwo,(ia-1,ib)
ib-=1
elif optionone < optiontwo: # choose the second
nottakensum,nottakenindices = optionone,(ia,ib-1)
ia-=1
#next two cases apply if options are equal
elif a[ia] > b[ib]:# drop the smallest
nottakensum,nottakenindices = optiontwo,(ia-1,ib)
ib-=1
else: # might be equal or not - we can choose arbitrarily if equal
nottakensum,nottakenindices = optionone,(ia,ib-1)
ia-=1
#+2 - one for zero-based, one for skipping the 1st largest
data = (i+2,a[ia],b[ib],a[ia]+b[ib],ia,ib)
narrative = "%sth largest pair is %s+%s=%s, with indices (%s,%s)" % data
print (narrative) #this will work in both versions of python
if ia <= 0 and ib <= 0:
raise ValueError("Both arrays exhausted before Kth (%sth) pair reached"%data[0])
return data, narrative
For those without python, here's an ideone: http://ideone.com/tfm2MA
At worst, we have 5 comparisons in each iteration, and K-1 iterations, which means that this is an O(K) algorithm.
Now, it might be possible to exploit information about differences between values to optimise this a little bit, but this accomplishes the goal.
Here's a reference implementation (not O(K), but will always work, unless there's a corner case with cases where pairs have equal sums):
import itertools
def refkth(a,b,k):
(rightia,righta),(rightib,rightb) = sorted(itertools.product(enumerate(a),enumerate(b)), key=lamba((ia,ea),(ib,eb):ea+eb)[k-1]
data = k,righta,rightb,righta+rightb,rightia,rightib
narrative = "%sth largest pair is %s+%s=%s, with indices (%s,%s)" % data
print (narrative) #this will work in both versions of python
return data, narrative
This calculates the cartesian product of the two arrays (i.e. all possible pairs), sorts them by sum, and takes the kth element. The enumerate function decorates each item with its index.
The max-heap algorithm in the other question is simple, fast and correct. Don't knock it. It's really well explained too. https://stackoverflow.com/a/5212618/284795
Might be there isn't any O(k) algorithm. That's okay, O(k log k) is almost as fast.
If the last two solutions were at (a1, b1), (a2, b2), then it seems to me there are only four candidate solutions (a1-1, b1) (a1, b1-1) (a2-1, b2) (a2, b2-1). This intuition could be wrong. Surely there are at most four candidates for each coordinate, and the next highest is among the 16 pairs (a in {a1,a2,a1-1,a2-1}, b in {b1,b2,b1-1,b2-1}). That's O(k).
(No it's not, still not sure whether that's possible.)
[2, 3, 5, 8, 13]
[4, 8, 12, 16]
Merge the 2 arrays and note down the indexes in the sorted array. Here is the index array looks like (starting from 1 not 0)
[1, 2, 4, 6, 8]
[3, 5, 7, 9]
Now start from end and make tuples. sum the elements in the tuple and pick the kth largest sum.
public static List<List<Integer>> optimization(int[] nums1, int[] nums2, int k) {
// 2 * O(n log(n))
Arrays.sort(nums1);
Arrays.sort(nums2);
List<List<Integer>> results = new ArrayList<>(k);
int endIndex = 0;
// Find the number whose square is the first one bigger than k
for (int i = 1; i <= k; i++) {
if (i * i >= k) {
endIndex = i;
break;
}
}
// The following Iteration provides at most endIndex^2 elements, and both arrays are in ascending order,
// so k smallest pairs must can be found in this iteration. To flatten the nested loop, refer
// 'https://stackoverflow.com/questions/7457879/algorithm-to-optimize-nested-loops'
for (int i = 0; i < endIndex * endIndex; i++) {
int m = i / endIndex;
int n = i % endIndex;
List<Integer> item = new ArrayList<>(2);
item.add(nums1[m]);
item.add(nums2[n]);
results.add(item);
}
results.sort(Comparator.comparing(pair->pair.get(0) + pair.get(1)));
return results.stream().limit(k).collect(Collectors.toList());
}
Key to eliminate O(n^2):
Avoid cartesian product(or 'cross join' like operation) of both arrays, which means flattening the nested loop.
Downsize iteration over the 2 arrays.
So:
Sort both arrays (Arrays.sort offers O(n log(n)) performance according to Java doc)
Limit the iteration range to the size which is just big enough to support k smallest pairs searching.
Yet another interview question asked me to find the maximum possible subarray of repeated values given a sorted array in shortest computational time possible.
Let input array be A[1 ... n]
Find an array B of consecutive integers in A such that:
for x in range(len(B)-1):
B[x] == B[x+1]
I believe that the best algorithm is dividing the array in half and going from the middle outwards and comparing from the middle the integers with one another and finding the longest strain of the same integers from the middle. Then I would call the method recursively by dividing the array in half and calling the method on the two halves.
My interviewer said my algorithm is good but my analysis that the algorithm is O(logn) is incorrect but never got around to telling me what the correct answer is. My first question is what is the Big-O analysis of this algorithm? (Show as much work as possible please! Big-O is not my forte.) And my second question is purely for my curiosity whether there is an even more time efficient algorithm?
The best you can do for this problem is an O(n) solution, so your algorithm cannot possibly be both correct and O(lg n).
Consider for example, the case where the array contains no repeated elements. To determine this, one needs to examine every element, and examining every element is O(n).
This is a simple algorithm that will find the longest subsequence of a repeated element:
start = end = 0
maxLength = 0
i = 0
while i + maxLength < a.length:
if a[i] == a[i + maxLength]:
while i + maxLength < a.length and a[i] == a[i + maxLength]:
maxLength += 1
start = i
end = i + maxLength
i += maxLength
return a[start:end]
If you have reason to believe the subsequence will be long, you can set the initial value of maxLength to some heuristically selected value to speed things along, and then only look for shorter sequences if you don't find one (i.e. you end up with end == 0 after the first pass.)
I think we all agree that in the worst case scenario, where all of A is unique or where all of A is the same, you have to examine every element in the array to either determine there are no duplicates or determine all the array contains one number. Like the other posters have said, that's going to be O(N). I'm not sure divide & conquer helps you much with algorithmic complexity on this one, though you may be able to simplify the code a bit by using recursion. Divide & conquer really helps cut down on Big O when you can throw away large portions of the input (e.g. Binary Search), but in the case where you potentially have to examine all the input, it's not going to be much different.
I'm assuming the result here is you're just returning the size of the largest B you've found, though you could easily modify this to return B instead.
So on the algorithm front, given that A is sorted, I'm not sure there's going to be any answer faster/simpler answer than just walking through the array in order. It seems like the simplest answer is to have 2 pointers, one starting at index 0 and one starting at index 1. Compare them and then increment them both; each time they're the same you tick a counter upward to give you the current size of B and when they differ you reset that counter to zero. You also keep around a variable for the max size of a B you've found so far and update it every time you find a bigger B.
In this algorithm, n elements are visited with a constant number of calculations per each visited element, so the running time is O(n).
Given sorted array A[1..n]:
max_start = max_end = 1
max_length = 1
start = end = 1
while start < n
while A[start] == A[end] && end < n
end++
if end - start > max_length
max_start = start
max_end = end - 1
max_length = end - start
start = end
Assuming that the longest consecutive integers is only of length 1, you'll be scanning through the entire array A of n items. Thus, the complexity is not in terms of n, but in terms of len(B).
Not sure if the complexity is O(n/len(B)).
Checking the 2 edge case
- When n == len(B), you get instant result (only checking A[0] and A[n-1]
- When n == 1, you get O(n), checking all elements
- When normal case, I'm too lazy to write the algo to analyze...
Edit
Given that len(B) is not known in advance, we must take the worst case, i.e. O(n)
I have an interview question that I can't seem to figure out. Given an array of size N, find the subset of size k such that the elements in the subset are the furthest apart from each other. In other words, maximize the minimum pairwise distance between the elements.
Example:
Array = [1,2,6,10]
k = 3
answer = [1,6,10]
The bruteforce way requires finding all subsets of size k which is exponential in runtime.
One idea I had was to take values evenly spaced from the array. What I mean by this is
Take the 1st and last element
find the difference between them (in this case 10-1) and divide that by k ((10-1)/3=3)
move 2 pointers inward from both ends, picking out elements that are +/- 3 from your previous pick. So in this case, you start from 1 and 10 and find the closest elements to 4 and 7. That would be 6.
This is based on the intuition that the elements should be as evenly spread as possible. I have no idea how to prove it works/doesn't work. If anyone knows how or has a better algorithm please do share. Thanks!
This can be solved in polynomial time using DP.
The first step is, as you mentioned, sort the list A. Let X[i,j] be the solution for selecting j elements from first i elements A.
Now, X[i+1, j+1] = max( min( X[k,j], A[i+1]-A[k] ) ) over k<=i.
I will leave initialization step and memorization of subset step for you to work on.
In your example (1,2,6,10) it works the following way:
1 2 6 10
1 - - - -
2 - 1 5 9
3 - - 1 4
4 - - - 1
The basic idea is right, I think. You should start by sorting the array, then take the first and the last elements, then determine the rest.
I cannot think of a polynomial algorithm to solve this, so I would suggest one of the two options.
One is to use a search algorithm, branch-and-bound style, since you have a nice heuristic at hand: the upper bound for any solution is the minimum size of the gap between the elements picked so far, so the first guess (evenly spaced cells, as you suggested) can give you a good baseline, which will help prune most of the branches right away. This will work fine for smaller values of k, although the worst case performance is O(N^k).
The other option is to start with the same baseline, calculate the minimum pairwise distance for it and then try to improve it. Say you have a subset with minimum distance of 10, now try to get one with 11. This can be easily done by a greedy algorithm -- pick the first item in the sorted sequence such that the distance between it and the previous item is bigger-or-equal to the distance you want. If you succeed, try increasing further, if you fail -- there is no such subset.
The latter solution can be faster when the array is large and k is relatively large as well, but the elements in the array are relatively small. If they are bound by some value M, this algorithm will take O(N*M) time, or, with a small improvement, O(N*log(M)), where N is the size of the array.
As Evgeny Kluev suggests in his answer, there is also a good upper bound on the maximum pairwise distance, which can be used in either one of these algorithms. So the complexity of the latter is actually O(N*log(M/k)).
You can do this in O(n*(log n) + n*log(M)), where M is max(A) - min(A).
The idea is to use binary search to find the maximum separation possible.
First, sort the array. Then, we just need a helper function that takes in a distance d, and greedily builds the longest subarray possible with consecutive elements separated by at least d. We can do this in O(n) time.
If the generated array has length at least k, then the maximum separation possible is >=d. Otherwise, it's strictly less than d. This means we can use binary search to find the maximum value. With some cleverness, you can shrink the 'low' and 'high' bounds of the binary search, but it's already so fast that sorting would become the bottleneck.
Python code:
def maximize_distance(nums: List[int], k: int) -> List[int]:
"""Given an array of numbers and size k, uses binary search
to find a subset of size k with maximum min-pairwise-distance"""
assert len(nums) >= k
if k == 1:
return [nums[0]]
nums.sort()
def longest_separated_array(desired_distance: int) -> List[int]:
"""Given a distance, returns a subarray of nums
of length k with pairwise differences at least that distance (if
one exists)."""
answer = [nums[0]]
for x in nums[1:]:
if x - answer[-1] >= desired_distance:
answer.append(x)
if len(answer) == k:
break
return answer
low, high = 0, (nums[-1] - nums[0])
while low < high:
mid = (low + high + 1) // 2
if len(longest_separated_array(mid)) == k:
low = mid
else:
high = mid - 1
return longest_separated_array(low)
I suppose your set is ordered. If not, my answer will be changed slightly.
Let's suppose you have an array X = (X1, X2, ..., Xn)
Energy(Xi) = min(|X(i-1) - Xi|, |X(i+1) - Xi|), 1 < i <n
j <- 1
while j < n - k do
X.Exclude(min(Energy(Xi)), 1 < i < n)
j <- j + 1
n <- n - 1
end while
$length = length($array);
sort($array); //sorts the list in ascending order
$differences = ($array << 1) - $array; //gets the difference between each value and the next largest value
sort($differences); //sorts the list in ascending order
$max = ($array[$length-1]-$array[0])/$M; //this is the theoretical max of how large the result can be
$result = array();
for ($i = 0; i < $length-1; $i++){
$count += $differences[i];
if ($length-$i == $M - 1 || $count >= $max){ //if there are either no more coins that can be taken or we have gone above or equal to the theoretical max, add a point
$result.push_back($count);
$count = 0;
$M--;
}
}
return min($result)
For the non-code people: sort the list, find the differences between each 2 sequential elements, sort that list (in ascending order), then loop through it summing up sequential values until you either pass the theoretical max or there arent enough elements remaining; then add that value to a new array and continue until you hit the end of the array. then return the minimum of the newly created array.
This is just a quick draft though. At a quick glance any operation here can be done in linear time (radix sort for the sorts).
For example, with 1, 4, 7, 100, and 200 and M=3, we get:
$differences = 3, 3, 93, 100
$max = (200-1)/3 ~ 67
then we loop:
$count = 3, 3+3=6, 6+93=99 > 67 so we push 99
$count = 100 > 67 so we push 100
min(99,100) = 99
It is a simple exercise to convert this to the set solution that I leave to the reader (P.S. after all the times reading that in a book, I've always wanted to say it :P)