python - Longest equally-spaced subsequence

Question

I have a million integers in sorted order and I would like to find the longest subsequence where the difference between consecutive pairs is equal. For example

1, 4, 5, 7, 8, 12

has a subsequence

   4,       8, 12

My naive method is greedy and just checks how far you can extend a subsequence from each point. This takes O(n²) time per point it seems.

Is there a faster way to solve this problem?

Update. I will test the code given in the answers as soon as possible (thank you). However it is clear already that using n^2 memory will not work. So far there is no code that terminates with the input as [random.randint(0,100000) for r in xrange(200000)] .

Timings. I tested with the following input data on my 32 bit system.

a= [random.randint(0,10000) for r in xrange(20000)] 
a.sort()

• The dynamic programming method of ZelluX uses 1.6G of RAM and takes 2 minutes and 14 seconds. With pypy it takes only 9 seconds! However it crashes with a memory error on large inputs.
• The O(nd) time method of Armin took 9 seconds with pypy but only 20MB of RAM. Of course this would be much worse if the range were much larger. The low memory usage meant I could also test it with a= [random.randint(0,100000) for r in xrange(200000)] but it didn't finish in the few minutes I gave it with pypy.

In order to be able to test the method of Kluev's I reran with

a= [random.randint(0,40000) for r in xrange(28000)] 
a = list(set(a))
a.sort()

to make a list of length roughly 20000. All timings with pypy

• ZelluX, 9 seconds
• Kluev, 20 seconds
• Armin, 52 seconds

It seems that if the ZelluX method could be made linear space it would be the clear winner.

Answer 1

Update: First algorithm described here is obsoleted by Armin Rigo's second answer, which is much simpler and more efficient. But both these methods have one disadvantage. They need many hours to find the result for one million integers. So I tried two more variants (see second half of this answer) where the range of input integers is assumed to be limited. Such limitation allows much faster algorithms. Also I tried to optimize Armin Rigo's code. See my benchmarking results at the end.

---

Here is an idea of algorithm using O(N) memory. Time complexity is O(N² log N), but may be decreased to O(N²).

Algorithm uses the following data structures:

1. prev: array of indexes pointing to previous element of (possibly incomplete) subsequence.
2. hash: hashmap with key = difference between consecutive pairs in subsequence and value = two other hashmaps. For these other hashmaps: key = starting/ending index of the subsequence, value = pair of (subsequence length, ending/starting index of the subsequence).
3. pq: priority queue for all possible "difference" values for subsequences stored in prev and hash.

Algorithm:

1. Initialize prev with indexes i-1. Update hash and pq to register all (incomplete) subsequences found on this step and their "differences".
2. Get (and remove) smallest "difference" from pq. Get corresponding record from hash and scan one of second-level hash maps. At this time all subsequences with given "difference" are complete. If second-level hash map contains subsequence length better than found so far, update the best result.
3. In the array prev: for each element of any sequence found on step #2, decrement index and update hash and possibly pq. While updating hash, we could perform one of the following operations: add a new subsequence of length 1, or grow some existing subsequence by 1, or merge two existing subsequences.
4. Remove hash map record found on step #2.
5. Continue from step #2 while pq is not empty.

This algorithm updates O(N) elements of prev O(N) times each. And each of these updates may require to add a new "difference" to pq. All this means time complexity of O(N² log N) if we use simple heap implementation for pq. To decrease it to O(N²) we might use more advanced priority queue implementations. Some of the possibilities are listed on this page: Priority Queues.

See corresponding Python code on Ideone. This code does not allow duplicate elements in the list. It is possible to fix this, but it would be a good optimization anyway to remove duplicates (and to find the longest subsequence beyond duplicates separately).

And the same code after a little optimization. Here search is terminated as soon as subsequence length multiplied by possible subsequence "difference" exceeds source list range.

---

Armin Rigo's code is simple and pretty efficient. But in some cases it does some extra computations that may be avoided. Search may be terminated as soon as subsequence length multiplied by possible subsequence "difference" exceeds source list range:

def findLESS(A):
  Aset = set(A)
  lmax = 2
  d = 1
  minStep = 0

  while (lmax - 1) * minStep <= A[-1] - A[0]:
    minStep = A[-1] - A[0] + 1
    for j, b in enumerate(A):
      if j+d < len(A):
        a = A[j+d]
        step = a - b
        minStep = min(minStep, step)
        if a + step in Aset and b - step not in Aset:
          c = a + step
          count = 3
          while c + step in Aset:
            c += step
            count += 1
          if count > lmax:
            lmax = count
    d += 1

  return lmax

print(findLESS([1, 4, 5, 7, 8, 12]))

---

If range of integers in source data (M) is small, a simple algorithm is possible with O(M²) time and O(M) space:

def findLESS(src):
  r = [False for i in range(src[-1]+1)]
  for x in src:
    r[x] = True

  d = 1
  best = 1

  while best * d < len(r):
    for s in range(d):
      l = 0

      for i in range(s, len(r), d):
        if r[i]:
          l += 1
          best = max(best, l)
        else:
          l = 0

    d += 1

  return best


print(findLESS([1, 4, 5, 7, 8, 12]))

It is similar to the first method by Armin Rigo, but it doesn't use any dynamic data structures. I suppose source data has no duplicates. And (to keep the code simple) I also suppose that minimum input value is non-negative and close to zero.

---

Previous algorithm may be improved if instead of the array of booleans we use a bitset data structure and bitwise operations to process data in parallel. The code shown below implements bitset as a built-in Python integer. It has the same assumptions: no duplicates, minimum input value is non-negative and close to zero. Time complexity is O(M² * log L) where L is the length of optimal subsequence, space complexity is O(M):

def findLESS(src):
  r = 0
  for x in src:
    r |= 1 << x

  d = 1
  best = 1

  while best * d < src[-1] + 1:
    c = best
    rr = r

    while c & (c-1):
      cc = c & -c
      rr &= rr >> (cc * d)
      c &= c-1

    while c != 1:
      c = c >> 1
      rr &= rr >> (c * d)

    rr &= rr >> d

    while rr:
      rr &= rr >> d
      best += 1

    d += 1

  return best

---

Benchmarks:

Input data (about 100000 integers) is generated this way:

random.seed(42)
s = sorted(list(set([random.randint(0,200000) for r in xrange(140000)])))

And for fastest algorithms I also used the following data (about 1000000 integers):

s = sorted(list(set([random.randint(0,2000000) for r in xrange(1400000)])))

All results show time in seconds:

Size:                         100000   1000000
Second answer by Armin Rigo:     634         ?
By Armin Rigo, optimized:         64     >5000
O(M^2) algorithm:                 53      2940
O(M^2*L) algorithm:                7       711

Answer 2

We can have a solution O(n*m) in time with very little memory needs, by adapting yours. Here n is the number of items in the given input sequence of numbers, and m is the range, i.e. the highest number minus the lowest one.

Call A the sequence of all input numbers (and use a precomputed set() to answer in constant time the question "is this number in A?"). Call d the step of the subsequence we're looking for (the difference between two numbers of this subsequence). For every possible value of d, do the following linear scan over all input numbers: for every number n from A in increasing order, if the number was not already seen, look forward in A for the length of the sequence starting at n with a step d. Then mark all items in that sequence as already seen, so that we avoid searching again from them, for the same d. Because of this, the complexity is just O(n) for every value of d.

A = [1, 4, 5, 7, 8, 12]    # in sorted order
Aset = set(A)

for d in range(1, 12):
    already_seen = set()
    for a in A:
        if a not in already_seen:
            b = a
            count = 1
            while b + d in Aset:
                b += d
                count += 1
                already_seen.add(b)
            print "found %d items in %d .. %d" % (count, a, b)
            # collect here the largest 'count'

Updates:

• This solution might be good enough if you're only interested in values of d that are relatively small; for example, if getting the best result for d <= 1000 would be good enough. Then the complexity goes down to O(n*1000). This makes the algorithm approximative, but actually runnable for n=1000000. (Measured at 400-500 seconds with CPython, 80-90 seconds with PyPy, with a random subset of numbers between 0 and 10'000'000.)

• If you still want to search for the whole range, and if the common case is that long sequences exist, a notable improvement is to stop as soon as d is too large for an even longer sequence to be found.

Answer 3

UPDATE: I've found a paper on this problem, you can download it here.

Here is a solution based on dynamic programming. It requires O(n^2) time complexity and O(n^2) space complexity, and does not use hashing.

We assume all numbers are saved in array a in ascending order, and n saves its length. 2D array l[i][j] defines length of longest equally-spaced subsequence ending with a[i] and a[j], and l[j][k] = l[i][j] + 1 if a[j] - a[i] = a[k] - a[j] (i < j < k).

lmax = 2
l = [[2 for i in xrange(n)] for j in xrange(n)]
for mid in xrange(n - 1):
    prev = mid - 1
    succ = mid + 1
    while (prev >= 0 and succ < n):
        if a[prev] + a[succ] < a[mid] * 2:
            succ += 1
        elif a[prev] + a[succ] > a[mid] * 2:
            prev -= 1
        else:
            l[mid][succ] = l[prev][mid] + 1
            lmax = max(lmax, l[mid][succ])
            prev -= 1
            succ += 1

print lmax

Answer 4

Algorithm

• Main loop traversing the list
• If number found in precalculate list, then it's belong to all sequences which are in that list, recalculate all the sequences with count + 1
• Remove all precalculated for current element
• Recalculate new sequences where first element is from range from 0 to current, and second is current element of traversal (actually, not from 0 to current, we can use the fact that new element shouldn't be more that max(a) and new list should have possibility to become longer that already found one)

So for list [1, 2, 4, 5, 7] output would be (it's a little messy, try code yourself and see)

• index 0, element 1:
  • if 1 in precalc? No - do nothing
  • Do nothing
• index 1, element 2:
  • if 2 in precalc? No - do nothing
  • check if 3 = 1 + (2 - 1) * 2 in our set? No - do nothing
• index 2, element 4:
  • if 4 in precalc? No - do nothing
    • check if 6 = 2 + (4 - 2) * 2 in our set? No
    • check if 7 = 1 + (4 - 1) * 2 in our set? Yes - add new element {7: {3: {'count': 2, 'start': 1}}} 7 - element of the list, 3 is step.
• index 3, element 5:
  • if 5 in precalc? No - do nothing
    • do not check 4 because 6 = 4 + (5 - 4) * 2 is less that calculated element 7
    • check if 8 = 2 + (5 - 2) * 2 in our set? No
    • check 10 = 2 + (5 - 1) * 2 - more than max(a) == 7
• index 4, element 7:
  • if 7 in precalc? Yes - put it into result
    • do not check 5 because 9 = 5 + (7 - 5) * 2 is more than max(a) == 7

result = (3, {'count': 3, 'start': 1}) # step 3, count 3, start 1, turn it into sequence

Complexity

It shouldn't be more than O(N^2), and I think it's less because of earlier termination of searching new sequencies, I'll try to provide detailed analysis later

Code

def add_precalc(precalc, start, step, count, res, N):
    if step == 0: return True
    if start + step * res[1]["count"] > N: return False

    x = start + step * count
    if x > N or x < 0: return False

    if precalc[x] is None: return True

    if step not in precalc[x]:
        precalc[x][step] = {"start":start, "count":count}

    return True

def work(a):
    precalc = [None] * (max(a) + 1)
    for x in a: precalc[x] = {}
    N, m = max(a), 0
    ind = {x:i for i, x in enumerate(a)}

    res = (0, {"start":0, "count":0})
    for i, x in enumerate(a):
        for el in precalc[x].iteritems():
            el[1]["count"] += 1
            if el[1]["count"] > res[1]["count"]: res = el
            add_precalc(precalc, el[1]["start"], el[0], el[1]["count"], res, N)
            t = el[1]["start"] + el[0] * el[1]["count"]
            if t in ind and ind[t] > m:
                m = ind[t]
        precalc[x] = None

        for y in a[i - m - 1::-1]:
            if not add_precalc(precalc, y, x - y, 2, res, N): break

    return [x * res[0] + res[1]["start"] for x in range(res[1]["count"])]

Answer 5

Here is another answer, working in time O(n^2) and without any notable memory requirements beyond that of turning the list into a set.

The idea is quite naive: like the original poster, it is greedy and just checks how far you can extend a subsequence from each pair of points --- however, checking first that we're at the start of a subsequence. In other words, from points a and b you check how far you can extend to b + (b-a), b + 2*(b-a), ... but only if a - (b-a) is not already in the set of all points. If it is, then you already saw the same subsequence.

The trick is to convince ourselves that this simple optimization is enough to lower the complexity to O(n^2) from the original O(n^3). That's left as an exercice to the reader :-) The time is competitive with other O(n^2) solutions here.

A = [1, 4, 5, 7, 8, 12]    # in sorted order
Aset = set(A)

lmax = 2
for j, b in enumerate(A):
    for i in range(j):
        a = A[i]
        step = b - a
        if b + step in Aset and a - step not in Aset:
            c = b + step
            count = 3
            while c + step in Aset:
                c += step
                count += 1
            #print "found %d items in %d .. %d" % (count, a, c)
            if count > lmax:
                lmax = count

print lmax

Answer 6

Your solution is O(N^3) now (you said O(N^2) per index). Here it is O(N^2) of time and O(N^2) of memory solution.

Idea

If we know subsequence that goes through indices i[0],i[1],i[2],i[3] we shouldn't try subsequence that starts with i[1] and i[2] or i[2] and i[3]

Note I edited that code to make it a bit easier using that a sorted but it will not work for equal elements. You may check number max number of equal elements in O(N) easily

Pseudocode

I'm seeking only for max length but that doesn't change anything

whereInA = {}
for i in range(n):
   whereInA[a[i]] = i; // It doesn't matter which of same elements it points to

boolean usedPairs[n][n];

for i in range(n):
    for j in range(i + 1, n):
       if usedPair[i][j]:
          continue; // do not do anything. It was in one of prev sequences.

    usedPair[i][j] = true;

    //here quite stupid solution:
    diff = a[j] - a[i];
    if diff == 0:
       continue; // we can't work with that
    lastIndex = j
    currentLen = 2
    while whereInA contains index a[lastIndex] + diff :
        nextIndex = whereInA[a[lastIndex] + diff]
        usedPair[lastIndex][nextIndex] = true
        ++currentLen
        lastIndex = nextIndex

    // you may store all indicies here
    maxLen = max(maxLen, currentLen)

Thoughts about memory usage

O(n^2) time is very slow for 1000000 elements. But if you are going to run this code on such number of elements the biggest problem will be memory usage.
What can be done to reduce it?

• Change boolean arrays to bitfields to store more booleans per bit.
• Make each next boolean array shorter because we only use usedPairs[i][j] if i < j

Few heuristics:

• Store only pairs of used indicies. (Conflicts with the first idea)
• Remove usedPairs that will never used more (that are for such i,j that was already chosen in the loop)

Answer 7

This is my 2 cents.

If you have a list called input:

input = [1, 4, 5, 7, 8, 12]

You can build a data structure that for each one of this points (excluding the first one), will tell you how far is that point from anyone of its predecessors:

[1, 4, 5, 7, 8, 12]
 x  3  4  6  7  11   # distance from point i to point 0
 x  x  1  3  4   8   # distance from point i to point 1
 x  x  x  2  3   7   # distance from point i to point 2
 x  x  x  x  1   5   # distance from point i to point 3
 x  x  x  x  x   4   # distance from point i to point 4

Now that you have the columns, you can consider the i-th item of input (which is input[i]) and each number n in its column.

The numbers that belong to a series of equidistant numbers that include input[i], are those which have n * j in the i-th position of their column, where j is the number of matches already found when moving columns from left to right, plus the k-th predecessor of input[i], where k is the index of n in the column of input[i].

Example: if we consider i = 1, input[i] = 4, n = 3, then, we can identify a sequence comprehending 4 (input[i]), 7 (because it has a 3 in position 1 of its column) and 1, because k is 0, so we take the first predecessor of i.

Possible implementation (sorry if the code is not using the same notation as the explanation):

def build_columns(l):
    columns = {}
    for x in l[1:]:
        col = []
        for y in l[:l.index(x)]:
            col.append(x - y)
        columns[x] = col
    return columns

def algo(input, columns):
    seqs = []
    for index1, number in enumerate(input[1:]):
        index1 += 1 #first item was sliced
        for index2, distance in enumerate(columns[number]):
            seq = []
            seq.append(input[index2]) # k-th pred
            seq.append(number)
            matches = 1
            for successor in input[index1 + 1 :]:
                column = columns[successor]
                if column[index1] == distance * matches:
                    matches += 1
                    seq.append(successor)
            if (len(seq) > 2):
                seqs.append(seq)
    return seqs

The longest one:

print max(sequences, key=len)

Answer 8

Traverse the array, keeping a record of the optimal result/s and a table with

(1) index - the element difference in the sequence,
(2) count - number of elements in the sequence so far, and
(3) the last recorded element.

For each array element look at the difference from each previous array element; if that element is last in a sequence indexed in the table, adjust that sequence in the table, and update the best sequence if applicable, otherwise start a new sequence, unless the current max is greater than the length of the possible sequence.

Scanning backwards we can stop our scan when d is greater than the middle of the array's range; or when the current max is greater than the length of the possible sequence, for d greater than the largest indexed difference. Sequences where s[j] is greater than the last element in the sequence are deleted.

I converted my code from JavaScript to Python (my first python code):

import random
import timeit
import sys

#s = [1,4,5,7,8,12]
#s = [2, 6, 7, 10, 13, 14, 17, 18, 21, 22, 23, 25, 28, 32, 39, 40, 41, 44, 45, 46, 49, 50, 51, 52, 53, 63, 66, 67, 68, 69, 71, 72, 74, 75, 76, 79, 80, 82, 86, 95, 97, 101, 110, 111, 112, 114, 115, 120, 124, 125, 129, 131, 132, 136, 137, 138, 139, 140, 144, 145, 147, 151, 153, 157, 159, 161, 163, 165, 169, 172, 173, 175, 178, 179, 182, 185, 186, 188, 195]
#s = [0, 6, 7, 10, 11, 12, 16, 18, 19]

m = [random.randint(1,40000) for r in xrange(20000)]
s = list(set(m))
s.sort()

lenS = len(s)
halfRange = (s[lenS-1] - s[0]) // 2

while s[lenS-1] - s[lenS-2] > halfRange:
    s.pop()
    lenS -= 1
    halfRange = (s[lenS-1] - s[0]) // 2

while s[1] - s[0] > halfRange:
    s.pop(0)
    lenS -=1
    halfRange = (s[lenS-1] - s[0]) // 2

n = lenS

largest = (s[n-1] - s[0]) // 2
#largest = 1000 #set the maximum size of d searched

maxS = s[n-1]
maxD = 0
maxSeq = 0
hCount = [None]*(largest + 1)
hLast = [None]*(largest + 1)
best = {}

start = timeit.default_timer()

for i in range(1,n):

    sys.stdout.write(repr(i)+"\r")

    for j in range(i-1,-1,-1):
        d = s[i] - s[j]
        numLeft = n - i
        if d != 0:
            maxPossible = (maxS - s[i]) // d + 2
        else:
            maxPossible = numLeft + 2
        ok = numLeft + 2 > maxSeq and maxPossible > maxSeq

        if d > largest or (d > maxD and not ok):
            break

        if hLast[d] != None:
            found = False
            for k in range (len(hLast[d])-1,-1,-1):
                tmpLast = hLast[d][k]
                if tmpLast == j:
                    found = True
                    hLast[d][k] = i
                    hCount[d][k] += 1
                    tmpCount = hCount[d][k]
                    if tmpCount > maxSeq:
                        maxSeq = tmpCount
                        best = {'len': tmpCount, 'd': d, 'last': i}
                elif s[tmpLast] < s[j]:
                    del hLast[d][k]
                    del hCount[d][k]
            if not found and ok:
                hLast[d].append(i)
                hCount[d].append(2)
        elif ok:
            if d > maxD: 
                maxD = d
            hLast[d] = [i]
            hCount[d] = [2]


end = timeit.default_timer()
seconds = (end - start)

#print (hCount)
#print (hLast)
print(best)
print(seconds)

Answer 9

This is a particular case for the more generic problem described here: Discover long patterns where K=1 and is fixed. It is demostrated there that it can be solved in O(N^2). Runnig my implementation of the C algorithm proposed there it takes 3 seconds to find the solution for N=20000 and M=28000 in my 32bit machine.

Answer 10

Greedy method
1 .Only one sequence of decision is generated.
2. Many number of decisions are generated. Dynamic programming 1. It does not guarantee to give an optimal solution always.
2. It definitely gives an optimal solution.