Algorithm to get unique combination of integers using multiplication - algorithm

I got bunch of 2s and 3s that i have to multiply together. Now i want to generate every unique combination of these numbers such that when these combination are multiplied does not exceed 10.
For example, I have something like this.
2*2*2*2*3*3*3
I can have following valid combination from above.
4*4*9*3
8*6*9
4*2*6*9
But the following combination are wrong. 16*3*9 and 4*4*27.
Can somebody suggest an algorithm to do this?


The solution can be built up recursively. Consider the input as a list of numbers such as [2,2,2,2,3,3,3]. Divide the list into a prefix (such as [2,2]) and the corresponding suffix ([2,2,3,3,3] in this case). Now multiple the entries in the prefix (and we get 4 in this example), and recursively solve the same problem for the suffix. Inserting the value from the multiplicity to the beginning of each of the solution for the suffix, we get the answer for the original problem.
In the following Python code, the recursive logic is defined in the function collapse, which finds all valid prefix (whose multiplicity is less than 10) and insert the multiplicity to all the results returned in collapsing the remaining data after cutting out the prefix (collapse(d[prefix_len:])).
a = [2,2,2,2,3,3,3]
def collapse(d):
if len(d) > 0:
for prefix_len in range(1, len(d) + 1):
prefix = reduce(lambda x,y: x*y, d[:prefix_len], 1)
if prefix > 10:
break
for suffix in collapse(d[prefix_len:]):
yield [prefix] + suffix
else:
yield d
for i in collapse(a):
print i
Output is
[2, 2, 2, 2, 3, 3, 3]
[2, 2, 2, 2, 3, 9]
[2, 2, 2, 2, 9, 3]
[2, 2, 2, 6, 3, 3]
[2, 2, 2, 6, 9]
[2, 2, 4, 3, 3, 3]
[2, 2, 4, 3, 9]
[2, 2, 4, 9, 3]
[2, 4, 2, 3, 3, 3]
[2, 4, 2, 3, 9]
[2, 4, 2, 9, 3]
[2, 4, 6, 3, 3]
[2, 4, 6, 9]
[2, 8, 3, 3, 3]
[2, 8, 3, 9]
[2, 8, 9, 3]
[4, 2, 2, 3, 3, 3]
[4, 2, 2, 3, 9]
[4, 2, 2, 9, 3]
[4, 2, 6, 3, 3]
[4, 2, 6, 9]
[4, 4, 3, 3, 3]
[4, 4, 3, 9]
[4, 4, 9, 3]
[8, 2, 3, 3, 3]
[8, 2, 3, 9]
[8, 2, 9, 3]
[8, 6, 3, 3]
[8, 6, 9]


If order matters (i.e. 2*4*2 is not the same as 2*2*4) and you have to list them (i.e. "generate") then you should just do it all recursively. In Scala:
def combine(who: List[Int], limit: Int=10): List[List[Int]] = who match {
case x :: y :: more =>
combine(y :: more, limit).map(x :: _) :::
(if (x*y<limit) combine((x*y) :: more, limit) else Nil)
case x :: Nil => List(who)
case Nil => List()
}
You may not know Scala, so here's how the three cases work. First case: list has at least two elements remaining. Pick off the first element and add it to all possible later combinations. Then, if you can merge the first and second elements, do so, and find all combinations of the list that starts with that. Second case: trivial list with only one element; return that as the only thing in the list. Third case: degenerate input (no values given); return an empty list.
(In Scala, ::: concatenates two lists together, while x :: list sticks x on the front of list. When you're matching, it goes the other way around: case x :: stuff is used if the list can be broken into an element x and the rest, stuff. Nil is the empty list.)
Here it is in action:
scala> combine( List(2,2,2,2,3,3,3) )
res18: List[List[Int]] = List(List(2, 2, 2, 2, 3, 3, 3), List(2, 2, 2, 2, 3, 9),
List(2, 2, 2, 2, 9, 3), List(2, 2, 2, 6, 3, 3), List(2, 2, 2, 6, 9),
List(2, 2, 4, 3, 3, 3), List(2, 2, 4, 3, 9), List(2, 2, 4, 9, 3),
List(2, 4, 2, 3, 3, 3), List(2, 4, 2, 3, 9), List(2, 4, 2, 9, 3), List(2, 4, 6, 3, 3),
List(2, 4, 6, 9), List(2, 8, 3, 3, 3), List(2, 8, 3, 9), List(2, 8, 9, 3),
List(4, 2, 2, 3, 3, 3), List(4, 2, 2, 3, 9), List(4, 2, 2, 9, 3), List(4, 2, 6, 3, 3),
List(4, 2, 6, 9), List(4, 4, 3, 3, 3), List(4, 4, 3, 9), List(4, 4, 9, 3),
List(8, 2, 3, 3, 3), List(8, 2, 3, 9), List(8, 2, 9, 3), List(8, 6, 3, 3), List(8, 6, 9))
Edit: if you just wanted to count them, you'd use a different type of recurrence. Let S(n) be the number of combinations taken from the nth onwards, and let L(n) be the value of the nth item in your list. Then
S(i) = S(i+1) +
if (L(i)+L(i+1)<10) S(i+2) +
if (L(i)+...+L(i+2)<10) S(i+3) +
....
So you start with the last item--only one possibility there--and work your way backwards in order using this formula. (If this is what you're after, I'll write code that does it, but hopefully the algorithm is clear enough as is.)

Related

What type of sorting algorithm it will be called? [duplicate]

what is the name of this sort? its just like bubble sort but its easy to write but harder in term of complexity.codes are in python language.
def sort(arr):
n = len(arr)
for i in range(n):
for j in range(n):
if arr[i] < arr[j] :
arr[j], arr[i] = arr[i], arr[j]
but bubble sort is like this :
def bubbleSort(arr):
n = len(arr)
# Traverse through all array elements
for i in range(n):
# Last i elements are already in place
for j in range(0, n-i-1):
# traverse the array from 0 to n-i-1
# Swap if the element found is greater
# than the next element
if arr[j] > arr[j+1] :
arr[j], arr[j+1] = arr[j+1], arr[j]

It is neither selection sort nor bubble sort, but an unnecessary bad sort. All of them, your sort, selection sort and bubble sort are of the same complexity, O(n^2).
Your sort sweeps over the array in two loops without consideration for if an element has reached its proper place (as in selection sort) or if an element has gained a place relative to the next (as in bubble sort). The extra code in bubble and selection sort make them far better than yours.
Compare, yourself:
def sort(arr):
print(arr)
n = len(arr)
for i in range(n):
for j in range(n):
if arr[i] < arr[j] :
arr[j], arr[i] = arr[i], arr[j]
print(arr)
def selectionsort(arr):
print(arr)
n = len(arr)
for i in range(n-1):
min_index = i
for j in range(i+1, n):
if arr[j] < arr[min_index]:
min_index = j
arr[i], arr[min_index] = arr[min_index], arr[i]
print(arr)
def bubblesort(arr):
print(arr)
n = len(arr)
for i in range(n):
for j in range(0, n-i-1):
if arr[j] > arr[j+1] :
arr[j], arr[j+1] = arr[j+1], arr[j]
print(arr)
print("sort:")
sort([4,6,3,2,7,1,8,5])
print("selectionsort:")
selectionsort([4,6,3,2,7,1,8,5])
print("bubblesort:")
bubblesort([4,6,3,2,7,1,8,5])
with the results:
sort:
[4, 6, 3, 2, 7, 1, 8, 5]
[6, 4, 3, 2, 7, 1, 8, 5]
[7, 4, 3, 2, 6, 1, 8, 5]
[8, 4, 3, 2, 6, 1, 7, 5]
[4, 8, 3, 2, 6, 1, 7, 5]
[3, 8, 4, 2, 6, 1, 7, 5]
[3, 4, 8, 2, 6, 1, 7, 5]
[2, 4, 8, 3, 6, 1, 7, 5]
[2, 3, 8, 4, 6, 1, 7, 5]
[2, 3, 4, 8, 6, 1, 7, 5]
[2, 3, 4, 6, 8, 1, 7, 5]
[1, 3, 4, 6, 8, 2, 7, 5]
[1, 2, 4, 6, 8, 3, 7, 5]
[1, 2, 3, 6, 8, 4, 7, 5]
[1, 2, 3, 4, 8, 6, 7, 5]
[1, 2, 3, 4, 6, 8, 7, 5]
[1, 2, 3, 4, 6, 7, 8, 5]
[1, 2, 3, 4, 5, 7, 8, 6]
[1, 2, 3, 4, 5, 6, 8, 7]
[1, 2, 3, 4, 5, 6, 7, 8]
selectionsort:
[4, 6, 3, 2, 7, 1, 8, 5]
[1, 6, 3, 2, 7, 4, 8, 5]
[1, 2, 3, 6, 7, 4, 8, 5]
[1, 2, 3, 6, 7, 4, 8, 5]
[1, 2, 3, 4, 7, 6, 8, 5]
[1, 2, 3, 4, 5, 6, 8, 7]
[1, 2, 3, 4, 5, 6, 8, 7]
[1, 2, 3, 4, 5, 6, 7, 8]
bubblesort:
[4, 6, 3, 2, 7, 1, 8, 5]
[4, 3, 6, 2, 7, 1, 8, 5]
[4, 3, 2, 6, 7, 1, 8, 5]
[4, 3, 2, 6, 1, 7, 8, 5]
[4, 3, 2, 6, 1, 7, 5, 8]
[3, 4, 2, 6, 1, 7, 5, 8]
[3, 2, 4, 6, 1, 7, 5, 8]
[3, 2, 4, 1, 6, 7, 5, 8]
[3, 2, 4, 1, 6, 5, 7, 8]
[2, 3, 4, 1, 6, 5, 7, 8]
[2, 3, 1, 4, 6, 5, 7, 8]
[2, 3, 1, 4, 5, 6, 7, 8]
[2, 1, 3, 4, 5, 6, 7, 8]
[1, 2, 3, 4, 5, 6, 7, 8]
It doesn't make a big difference on small arrays but on large arrays it make a lot of difference (but then there are other methods).

Finding duplicate columns in a nested array

0 1 2 3 4 5 6
0{1,2,1,2,1,5,5}
1{5,4,5,4,5,1,1}
2{2,4,2,4,2,1,1}
3{1,2,1,2,1,1,1}
4{4,4,4,4,4,1,1}
5{2,4,2,4,2,2,2}
output: {{0,2,4}, {1,3}, {5,6}} (can use any data structure)
Let's say there is a nested array like above. If we wanted to find column indices that contain the same exact numbers in the same order (for example, column 0, 2, 4 with (1,5,2,1,4,2) and column 1, 3 with (2,4,4,2,4,4), and column 5, 6 with (5,1,1,1,1,2), how can we go about with this efficiently? Will it require dynamic programming?
Thanks in advance.

You can just iterate through the columns, keeping a hashmap of the columns that you've seen so far. Here's an implementation in python:
x = [[1, 2, 1, 2, 1, 5, 5],
[5, 4, 5, 4, 5, 1, 1],
[2, 4, 2, 4, 2, 1, 1],
[1, 2, 1, 2, 1, 1, 1],
[4, 4, 4, 4, 4, 1, 1],
[2, 4, 2, 4, 2, 2, 2]]
seen_before = {}
for v, col in enumerate(zip(*x)):
if tuple(col) not in seen_before:
seen_before[tuple(col)] = [v]
else:
seen_before[tuple(col)].append(v)
This solves the problem in linear time. I hope that's good enough for you.

Transform one list into another [duplicate]

This question already has answers here:
Algorithm: optimal way to rearrange a list from one order to another?
(4 answers)
Closed 4 years ago.
Given two lists, for example:
a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
b = [2, 4, 6, 7, 0, 1, 3, 5, 8, 9]
I wish to find a series of moves which will transform list a into list b, where each move is an operation:
move(from_index, to_index)
which moves the element at location from_index and places it at location to_index. So if:
a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
then the operation move(3,1) on the list a will transform a into:
a = [0, 3, 1, 2, 4, 5, 6, 7, 8, 9]

a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
b = [2, 4, 6, 7, 0, 1, 3, 5, 8, 9]
move(0, 8)
a = [1, 2, 3, 4, 5, 6, 7, 0, 8, 9]
move(0, 8)
a = [2, 3, 4, 5, 6, 7, 0, 1, 8, 9]
move(1, 8)
a = [2, 4, 5, 6, 7, 0, 1, 3, 8, 9]
move(2, 8)
a = [2, 4, 6, 7, 0, 1, 3, 5, 8, 9]
a==b
Hopefully that's what you're looking for.
Basically, start with the left- most element and move it to where it should be. For example, I took 0 and placed it right after the value that it is supposed to eventually end up behind, which is 7. I continued moving from left to right until all of the elements were in the desired order.

I'd iterate over the second sequence (the sorted list) and swap items in the first. I wrote this pseudo-code in python:
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> b = [2, 4, 6, 7, 0, 1, 3, 5, 8, 9]
>>> def swap(seq, i, j):
... a = seq[i]
... seq[i] = seq[j]
... seq[j] = a
...
>>> for index_in_b, value in enumerate(b):
... index_in_a = a.index(value)
... if index_in_b != index_in_a:
... swap(a, index_in_a, index_in_b)
... print('move {} to {}'.format(index_in_a, index_in_b))
move 0 to 2
move 1 to 4
move 2 to 6
move 3 to 7
move 4 to 6
move 5 to 6
move 6 to 7
In this case I'm moving the items in the first sequence by swapping them.
Update
We can slightly improve the performance in python by removing the move inside swap function and also removing the function call. Here is a performance comparison:
import timeit
s1 = """
a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
b = [2, 4, 6, 7, 0, 1, 3, 5, 8, 9]
def swap(seq, i, j):
a = seq[i]
seq[i] = seq[j]
seq[j] = a
for index_in_b, value in enumerate(b):
index_in_a = a.index(value)
if index_in_b != index_in_a:
swap(a, index_in_a, index_in_b)"""
s2 = """
a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
b = [2, 4, 6, 7, 0, 1, 3, 5, 8, 9]
for index_in_b, value in enumerate(b):
index_in_a = a.index(value)
if index_in_b != index_in_a:
a[index_in_a], a[index_in_b] = a[index_in_b], a[index_in_a]"""
# on an i7 macbook pro
timeit.timeit(s1)
4.087386846542358
timeit.timeit(s2)
3.5381240844726562
Slightly better, but for sure there are better ways to achieve this.

Algorithm generating k-tuples of n items such that each is used at least once (k>n)

Given a set of n items, I'd like to generate all the ways we can select from this set (with replacement) k times such that order matters and each element is used as least once. So k>=n to have any valid arrangements. If k=n, this is just a permutation. So k>n is kind of like an extension of a permutation, but I do not know what it is called.
It's of course easy to get an algorithm, albeit a horribly slow one: just iterate through all possible selections and toss the ones that don't have each element at least once. So to make something efficient will require tricks similar to iterating through permutations, or break it into sub-problems where we can use existing permutation algorithms directly.
I tried to break this into a permutation and combination problem by doing something like the following using python to play with some ideas.
import itertools
def func(inputSet,k):
n = len(inputSet)
assert(k>n)
# first, guarantee we have each element once
for p in itertools.permutations(inputSet):
# now select (k-n) locations to insert other elements
for c in itertools.combinations_with_replacement(range(n+1),k-n):
insertions = [c[i]+i for i in range(len(c))]
out = list(p)
for index in insertions:
out.insert(index,0)
# now select values to put in those locations
for vals in itertools.product(inputSet,repeat=len(insertions)):
for i in xrange(len(insertions)):
out[c[i]] = vals[i]
yield tuple(out)
But different insertions can yield the same result, so this first stab is likely not starting down the correct path. I could add conditionals to check for these cases and filter out some results, but an algorithm for a combinatoric iteration problem that resorts to filtering is likely not the most efficient algorithm.
Does this "permutation extension" have a name?
What is an efficient algorithm for iterating through the arrangements?

For what it's called, this is the same as a permutation, just framed a little differently. Consider the set of elements P, you're essentially asking to generate all of the permutations in the set P unioned with (k-n) elements of P, which can be found with itertools.combinations_with_replacement.
To generate the actual permutations, you can then either use list(set(itertools.permutations)) or more_itertools.distinct_permutations: https://more-itertools.readthedocs.io/en/latest/api.html#more_itertools.distinct_permutations
Putting this into actual code
>>> x = [1,2,3]
>>> k = 5
>>> results = set()
>>> for y in itertools.combinations_with_replacement(x, k - len(x)):
... for z in itertools.permutations(x + list(y)):
... results.add(z)
...
>>> results
set([(1, 1, 1, 2, 3), (1, 3, 3, 1, 2), (1, 2, 3, 3, 2), (3, 3, 2, 1, 3), (1, 3, 3, 2, 2), (1, 1, 2, 2, 3), (3, 1, 2, 3, 2), (1, 1, 3, 2, 3), (1, 3, 1, 2, 2), (1, 2, 2, 1, 3), (3, 2, 1, 2, 2), (3, 1, 2, 1, 2), (3, 2, 1, 3, 1), (3, 1, 1, 3, 2), (2, 3, 3, 2, 1), (1, 2, 1, 3, 3), (3, 1, 2, 3, 3), (2, 3, 2, 1, 1), (2, 3, 2, 2, 1), (1, 2, 2, 3, 3), (2, 1, 3, 2, 3), (2, 2, 2, 3, 1), (1, 3, 2, 2, 3), (2, 3, 3, 1, 1), (1, 2, 1, 1, 3), (3, 2, 2, 1, 1), (2, 1, 2, 2, 3), (2, 2, 1, 3, 1), (1, 3, 3, 2, 3), (2, 3, 3, 1, 3), (3, 2, 3, 2, 1), (3, 2, 3, 1, 1), (2, 1, 1, 2, 3), (3, 3, 1, 1, 2), (3, 2, 2, 2, 1), (2, 2, 3, 2, 1), (1, 3, 1, 2, 3), (2, 2, 3, 1, 2), (3, 1, 2, 1, 3), (2, 1, 1, 3, 3), (3, 3, 2, 2, 1), (1, 1, 2, 3, 1), (1, 2, 2, 3, 2), (3, 2, 1, 1, 2), (2, 1, 3, 2, 2), (1, 3, 2, 2, 2), (3, 1, 1, 1, 2), (3, 3, 2, 1, 1), (2, 3, 3, 1, 2), (3, 2, 1, 3, 2), (1, 2, 1, 3, 1), (2, 3, 1, 2, 3), (1, 2, 3, 2, 1), (3, 1, 3, 1, 2), (3, 3, 1, 2, 2), (1, 2, 3, 1, 1), (2, 2, 1, 1, 3), (2, 1, 1, 3, 2), (1, 1, 2, 3, 2), (3, 2, 1, 1, 3), (2, 1, 3, 2, 1), (2, 1, 1, 3, 1), (1, 3, 2, 2, 1), (1, 3, 2, 1, 1), (3, 2, 2, 1, 3), (2, 2, 3, 3, 1), (3, 1, 1, 2, 1), (2, 2, 1, 3, 3), (1, 3, 3, 2, 1), (3, 2, 3, 1, 2), (3, 1, 2, 3, 1), (2, 2, 2, 1, 3), (1, 1, 3, 1, 2), (1, 1, 2, 1, 3), (2, 1, 3, 3, 2), (3, 3, 1, 2, 3), (1, 3, 2, 3, 2), (3, 3, 2, 3, 1), (3, 1, 3, 2, 2), (2, 1, 3, 1, 1), (1, 1, 2, 3, 3), (2, 1, 3, 1, 2), (1, 3, 2, 1, 2), (1, 2, 1, 2, 3), (3, 2, 2, 1, 2), (3, 1, 1, 2, 2), (2, 2, 1, 3, 2), (2, 3, 2, 3, 1), (1, 1, 1, 3, 2), (2, 3, 1, 2, 1), (1, 2, 3, 1, 3), (2, 3, 1, 1, 1), (1, 2, 3, 2, 3), (2, 1, 2, 3, 3), (3, 2, 1, 2, 3), (1, 2, 2, 2, 3), (3, 2, 1, 2, 1), (2, 1, 1, 1, 3), (1, 3, 2, 3, 3), (1, 1, 3, 3, 2), (3, 2, 2, 3, 1), (3, 1, 3, 2, 3), (2, 1, 2, 1, 3), (1, 3, 3, 3, 2), (3, 2, 3, 3, 1), (2, 2, 3, 1, 3), (3, 2, 1, 1, 1), (2, 1, 3, 1, 3), (1, 2, 1, 3, 2), (3, 3, 1, 3, 2), (1, 3, 2, 1, 3), (2, 3, 1, 3, 2), (3, 1, 1, 2, 3), (2, 3, 3, 3, 1), (1, 1, 3, 2, 1), (2, 3, 1, 1, 2), (1, 2, 3, 2, 2), (2, 1, 2, 3, 2), (1, 3, 1, 1, 2), (3, 1, 3, 3, 2), (3, 1, 2, 2, 2), (3, 3, 1, 2, 1), (3, 3, 3, 1, 2), (3, 2, 3, 1, 3), (1, 3, 1, 3, 2), (2, 3, 2, 1, 3), (2, 1, 3, 3, 3), (1, 2, 2, 3, 1), (2, 3, 1, 3, 3), (1, 2, 3, 3, 1), (2, 3, 1, 2, 2), (3, 3, 2, 1, 2), (2, 3, 1, 3, 1), (3, 2, 1, 3, 3), (1, 1, 3, 2, 2), (2, 3, 1, 1, 3), (2, 1, 2, 3, 1), (1, 2, 3, 3, 3), (1, 3, 1, 2, 1), (3, 1, 2, 2, 3), (3, 1, 2, 2, 1), (2, 1, 3, 3, 1), (3, 1, 2, 1, 1), (1, 3, 2, 3, 1), (1, 2, 3, 1, 2), (3, 1, 3, 2, 1), (2, 2, 1, 2, 3), (2, 3, 2, 1, 2), (3, 3, 3, 2, 1), (2, 2, 3, 1, 1)])
Be aware that this explodes pretty quickly combinatorially, however because both itertools.combinations_with_replacement and itertools.permutations return generators, you can also yield the results. You could also write it yourself recursively, however I personally think that's a lot less satisfying.
I believe it's also sufficient to use distinct_permutations here and you'll end up with a list of wholly distinct results, as each iteration of the outer loop results in a different frequency signature for the elements.

Here's an additional solution, which also iterates in lexigraphical order.
import itertools
def _recurse_extended_perm(remaining,inputList,requireList):
if remaining==len(requireList):
for p in itertools.permutations(requireList):
yield p
elif remaining==1:
for x in inputList:
yield [x]
else:
for x in inputList:
req = list(requireList)
if x in req:
req.remove(x)
for seq in _recurse_extended_perm(remaining-1,inputList,req):
out = [x]
out.extend(seq)
yield out
def extended_permutation(seq,k):
inputList = list(seq)
inputList.sort()
assert(k>len(inputList))
for seq in _recurse_extended_perm(k,inputList,inputList):
yield tuple(seq)

We can generate the list recursively. Here's a module-less implementation that also gives us a little flexibility:
def multisets_with_at_least_m(inputList, k, m):
def f(idx, n, multiset):
if idx == len(inputList):
return [multiset]
if (len(inputList) - idx) * m > n:
return []
minNum = n if idx == len(inputList) - 1 else m
maxNum = n - (len(inputList) - idx) * m + m
results = []
for i in xrange(minNum, maxNum + 1):
multisetCopy = multiset.copy()
multisetCopy[inputList[idx]] = i
results = results + f(idx + 1, n - i, multisetCopy)
return results
return f(0, k, {})
def distinct_permutations_from_multiset(multiset):
def f(multiset, permutation):
if multiset == {}:
return [permutation]
results = []
for k in multiset:
multisetCopy = multiset.copy()
if multiset[k] == 1:
del multisetCopy[k]
else:
multisetCopy[k] = multisetCopy[k] - 1
results = results + f(multisetCopy, permutation + [k])
return results
return f(multiset, [])
Output:
print [distinct_permutations_from_multiset(x) for x in multisets_with_at_least_m([1,2,3], 5, 1)]
"""
[[[1, 2, 3, 3, 3], [1, 3, 2, 3, 3], [1, 3, 3, 2, 3], [1, 3, 3, 3, 2]
, [2, 1, 3, 3, 3], [2, 3, 1, 3, 3], [2, 3, 3, 1, 3], [2, 3, 3, 3, 1]
, [3, 1, 2, 3, 3], [3, 1, 3, 2, 3], [3, 1, 3, 3, 2], [3, 2, 1, 3, 3]
, [3, 2, 3, 1, 3], [3, 2, 3, 3, 1], [3, 3, 1, 2, 3], [3, 3, 1, 3, 2]
, [3, 3, 2, 1, 3], [3, 3, 2, 3, 1], [3, 3, 3, 1, 2], [3, 3, 3, 2, 1]]
, [[1, 2, 2, 3, 3], [1, 2, 3, 2, 3], [1, 2, 3, 3, 2], [1, 3, 2, 2, 3]
, [1, 3, 2, 3, 2], [1, 3, 3, 2, 2], [2, 1, 2, 3, 3], [2, 1, 3, 2, 3]
, [2, 1, 3, 3, 2], [2, 2, 1, 3, 3], [2, 2, 3, 1, 3], [2, 2, 3, 3, 1]
, [2, 3, 1, 2, 3], [2, 3, 1, 3, 2], [2, 3, 2, 1, 3], [2, 3, 2, 3, 1]
, [2, 3, 3, 1, 2], [2, 3, 3, 2, 1], [3, 1, 2, 2, 3], [3, 1, 2, 3, 2]
, [3, 1, 3, 2, 2], [3, 2, 1, 2, 3], [3, 2, 1, 3, 2], [3, 2, 2, 1, 3]
, [3, 2, 2, 3, 1], [3, 2, 3, 1, 2], [3, 2, 3, 2, 1], [3, 3, 1, 2, 2]
, [3, 3, 2, 1, 2], [3, 3, 2, 2, 1]]
, [[1, 2, 2, 2, 3], [1, 2, 2, 3, 2], [1, 2, 3, 2, 2], [1, 3, 2, 2, 2]
, [2, 1, 2, 2, 3], [2, 1, 2, 3, 2], [2, 1, 3, 2, 2], [2, 2, 1, 2, 3]
, [2, 2, 1, 3, 2], [2, 2, 2, 1, 3], [2, 2, 2, 3, 1], [2, 2, 3, 1, 2]
, [2, 2, 3, 2, 1], [2, 3, 1, 2, 2], [2, 3, 2, 1, 2], [2, 3, 2, 2, 1]
, [3, 1, 2, 2, 2], [3, 2, 1, 2, 2], [3, 2, 2, 1, 2], [3, 2, 2, 2, 1]]
, [[1, 1, 2, 3, 3], [1, 1, 3, 2, 3], [1, 1, 3, 3, 2], [1, 2, 1, 3, 3]
, [1, 2, 3, 1, 3], [1, 2, 3, 3, 1], [1, 3, 1, 2, 3], [1, 3, 1, 3, 2]
, [1, 3, 2, 1, 3], [1, 3, 2, 3, 1], [1, 3, 3, 1, 2], [1, 3, 3, 2, 1]
, [2, 1, 1, 3, 3], [2, 1, 3, 1, 3], [2, 1, 3, 3, 1], [2, 3, 1, 1, 3]
, [2, 3, 1, 3, 1], [2, 3, 3, 1, 1], [3, 1, 1, 2, 3], [3, 1, 1, 3, 2]
, [3, 1, 2, 1, 3], [3, 1, 2, 3, 1], [3, 1, 3, 1, 2], [3, 1, 3, 2, 1]
, [3, 2, 1, 1, 3], [3, 2, 1, 3, 1], [3, 2, 3, 1, 1], [3, 3, 1, 1, 2]
, [3, 3, 1, 2, 1], [3, 3, 2, 1, 1]]
, [[1, 1, 2, 2, 3], [1, 1, 2, 3, 2], [1, 1, 3, 2, 2], [1, 2, 1, 2, 3]
, [1, 2, 1, 3, 2], [1, 2, 2, 1, 3], [1, 2, 2, 3, 1], [1, 2, 3, 1, 2]
, [1, 2, 3, 2, 1], [1, 3, 1, 2, 2], [1, 3, 2, 1, 2], [1, 3, 2, 2, 1]
, [2, 1, 1, 2, 3], [2, 1, 1, 3, 2], [2, 1, 2, 1, 3], [2, 1, 2, 3, 1]
, [2, 1, 3, 1, 2], [2, 1, 3, 2, 1], [2, 2, 1, 1, 3], [2, 2, 1, 3, 1]
, [2, 2, 3, 1, 1], [2, 3, 1, 1, 2], [2, 3, 1, 2, 1], [2, 3, 2, 1, 1]
, [3, 1, 1, 2, 2], [3, 1, 2, 1, 2], [3, 1, 2, 2, 1], [3, 2, 1, 1, 2]
, [3, 2, 1, 2, 1], [3, 2, 2, 1, 1]]
, [[1, 1, 1, 2, 3], [1, 1, 1, 3, 2], [1, 1, 2, 1, 3], [1, 1, 2, 3, 1]
, [1, 1, 3, 1, 2], [1, 1, 3, 2, 1], [1, 2, 1, 1, 3], [1, 2, 1, 3, 1]
, [1, 2, 3, 1, 1], [1, 3, 1, 1, 2], [1, 3, 1, 2, 1], [1, 3, 2, 1, 1]
, [2, 1, 1, 1, 3], [2, 1, 1, 3, 1], [2, 1, 3, 1, 1], [2, 3, 1, 1, 1]
, [3, 1, 1, 1, 2], [3, 1, 1, 2, 1], [3, 1, 2, 1, 1], [3, 2, 1, 1, 1]]]
"""

recursive removal of elements in array

Given an array of n elements, remove any adjacent pair of elements which are equal. Repeat this operation until there are no more adjacent pairs to remove; that will be the final array.
For e.g 1 2 2 3 4 should return the array 1 3 4.
please note array need not to be sorted.
check this test case also: 1,2,2,3,4,4,3,5 o/p should be 1,5.
(2,2) and (4,4) gets removed, then (3,3) which became adjacent after the removal of (4,4)

Any time you remove a pair of elements, you also need to see if you generated another pair that you want to remove.
The algorithm should follow naturally from that observation.

In Python:
>>> l=[1,2,2,3,4,4,3,5]
>>> [x for x in l if not l.count(x) > 1]
[1, 5]
This removes all integers that occur more than once in the list. This is a correct result for your example but I think that you are really trying to state something different. I think you are saying:
list:=(an unsorted list of integers)
while adjacent_pairs(list) is True:
remove_adjacent_pairs(list)
Once again, in Python:
#!/usr/bin/env python
def dedupe_adjacent(l):
for i in xrange(len(l) - 1, 0, -1):
if l[i] == l[i-1]:
del l[i-1:i+1]
return True
return False
def process_list(l):
print "input list: ",l
i=1
while(dedupe_adjacent(l)):
print " loop ",i,":",l
i+=1
print "processed list=",l
print
process_list([1,2,2,3,4,4,3,5])
process_list([1,2,2,3,4,4,6,3,5])
Output:
input list: [1, 2, 2, 3, 4, 4, 3, 5]
loop 1 : [1, 2, 2, 3, 3, 5]
loop 2 : [1, 2, 2, 5]
loop 3 : [1, 5]
processed list= [1, 5]
input list: [1, 2, 2, 3, 4, 4, 6, 3, 5]
loop 1 : [1, 2, 2, 3, 6, 3, 5]
loop 2 : [1, 3, 6, 3, 5]
processed list= [1, 3, 6, 3, 5]

The following:
function compress(arr) {
var prev, res = [];
for (var i in arr) {
if (i == 0 || (arr[i] != arr[i - 1]))
res.push(arr[i]);
}
return res;
}
compress([1, 2, 2, 3, 3, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]);
Returns:
[1, 2, 3, 4, 3, 5, 6, 7, 8]
Also (JavaScript 1.6 solution):
[1, 2, 2, 3, 3, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8].filter(function(el, i, arr) {
return i == 0 || (el != arr[i - 1]);
})
Edit: Removing any item that appears in the array more than once requires a different solution:
function dedup(arr) {
var res = [], seen = {};
for (var i in arr)
seen[arr[i]] = seen[arr[i]] ? ++seen[arr[i]] : 1;
for (var j in arr) {
if (seen[arr[j]] == 1)
res.push(arr[j]);
}
return res;
}
The following:
dedup([1, 2, 2, 3, 4, 4, 3, 5]);
Produces:
[1, 5]

I have a solution to this in Java. You need to use replaceAll method in String class in Java. You can use regular expession to remove such adjacent redundant characters:
public class MyString {
public static void main(String[] args) {
String str = "12234435";
while(!str.replaceAll("(\\w)\\1+", "").equalsIgnoreCase(str))
str = str.replaceAll("(\\w)\\1+", "");
System.out.println(str);
}
}
You can find how to give a regular expression here

I would:
Sort the array.
From the start of the array, until you are at the last element of the array do:
`count` = count the number of array[i] elements.
remove the first `count` elements of the array if `count` > 1.

The following Python 3 code will remove duplicates from a list (array). It does this by scanning the array from start towards end and compares the target element with the element one larger. If they are the same they are removed. If the element pointer is not pointing at 0, then it is reduced by 1 in order to catch nested pairs. If the two compared elements are different then the pointer is incremented.
I'm sure there's a more pythonic way to remove two adjacent elements from a list, but I'm new to Python and haven't figured that out yet. Also, you'll want to get rid of the print(indx, SampleArray) statement--I left it in there to let you follow the progress in the output listing below.
# Algorithm to remove duplicates in a semi-sorted list
def CompressArray(SampleArray):
indx=0
while(indx < len(SampleArray)-1):
print(indx, SampleArray)
if(SampleArray[indx]==SampleArray[indx+1]):
del(SampleArray[indx])
del(SampleArray[indx])
if(indx>0):
indx-=1
else:
indx+=1
return SampleArray
Here are sample runs for:
[1, 2, 2, 3, 4]
[1, 2, 2, 3, 4, 4, 3, 5]
[1, 2, 2, 3, 3, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]
[1, 2, 2, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
[1, 1, 2, 3, 3, 2, 4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
================================
0 [1, 2, 2, 3, 4]
1 [1, 2, 2, 3, 4]
0 [1, 3, 4]
1 [1, 3, 4]
[1, 3, 4]
================================
0 [1, 2, 2, 3, 4, 4, 3, 5]
1 [1, 2, 2, 3, 4, 4, 3, 5]
0 [1, 3, 4, 4, 3, 5]
1 [1, 3, 4, 4, 3, 5]
2 [1, 3, 4, 4, 3, 5]
1 [1, 3, 3, 5]
0 [1, 5]
[1, 5]
================================
0 [1, 2, 2, 3, 3, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]
1 [1, 2, 2, 3, 3, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]
0 [1, 3, 3, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]
1 [1, 3, 3, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]
0 [1, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]
1 [1, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]
0 [1, 4, 3, 3, 5, 6, 7, 8, 8]
1 [1, 4, 3, 3, 5, 6, 7, 8, 8]
2 [1, 4, 3, 3, 5, 6, 7, 8, 8]
1 [1, 4, 5, 6, 7, 8, 8]
2 [1, 4, 5, 6, 7, 8, 8]
3 [1, 4, 5, 6, 7, 8, 8]
4 [1, 4, 5, 6, 7, 8, 8]
5 [1, 4, 5, 6, 7, 8, 8]
[1, 4, 5, 6, 7]
================================
0 [1, 2, 2, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
1 [1, 2, 2, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
0 [1, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
1 [1, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
2 [1, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
3 [1, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
4 [1, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
3 [1, 3, 4, 6, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
2 [1, 3, 4, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
1 [1, 3, 3, 8, 8, 5, 9, 10, 10, 9, 11]
0 [1, 8, 8, 5, 9, 10, 10, 9, 11]
1 [1, 8, 8, 5, 9, 10, 10, 9, 11]
0 [1, 5, 9, 10, 10, 9, 11]
1 [1, 5, 9, 10, 10, 9, 11]
2 [1, 5, 9, 10, 10, 9, 11]
3 [1, 5, 9, 10, 10, 9, 11]
2 [1, 5, 9, 9, 11]
1 [1, 5, 11]
[1, 5, 11]
================================
0 [1, 1, 2, 3, 3, 2, 4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
0 [2, 3, 3, 2, 4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
1 [2, 3, 3, 2, 4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
0 [2, 2, 4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
0 [4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
1 [4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
2 [4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
1 [4, 5, 5, 7, 8, 8, 7, 4, 9]
0 [4, 7, 8, 8, 7, 4, 9]
1 [4, 7, 8, 8, 7, 4, 9]
2 [4, 7, 8, 8, 7, 4, 9]
1 [4, 7, 7, 4, 9]
0 [4, 4, 9]
[9]
================================

I love Java, but functional solutions should get more time on this site.
In Haskell, doing things the way the question asks:
compress lst = if (length lst == length b) then lst else (compress b) where
b = helper lst
helper [] = []
helper [x] = [x]
helper (x:y:xs) = if (x == y) then (helper xs) else (x:helper (y:xs))
You can solve this problem in O(n) time, although it is a bit more complicated
compress' lst = reverse (helper [] lst) where
helper xs [] = xs
helper [] (x:xs) = helper [x] xs
helper (a:as) (x:xs)
| a == x = helper as xs
| otherwise = helper (x:a:as) xs

I think we could use a stack to check adjacent duplicated elements.
Scan the array. For each new element, if it is equal to the top element in the stack, drop it and pop the top element from the stack. Otherwise, push it into the stack.

Here is the stack based algorithm based upon the edited question.
// pseudo code, not tested
void RemoveDupp(vector<int> & vin, vector<int> & vout)
{
int i = 0, int j = -1;
vout.resize(vin.size());
while (i < vin.size())
{
if (j == -1 || vout[j] != vin[i])
vout[++j] = vin[i++]; //push
else
j--, i++; //pop
}
vout.resize(j + 1);
}

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