Develop Reference

Has this snippet O(log^2(n)) complexity?

If not, witch complexity it would be? Thanks:
public static int f(int n, int x) {
for (int i = n; i > 0; i /= 2) {
for (int j = 0; j < i; j++) {
x += j; // Assume, this operation costs 1.
}
}
return x;
}

This is an interesting one. The assumption of log^2(n) is wrong. Henry gave a good reductio ad absurdum why it cannot be log^2(n) in the comments:
We can see that, O(log^2(n)) ⊊ O(n).
the first iteration of the inner loop takes O(n).
Since O(log^2(n)) ⊊ O(n), the assumption must be wrong because the first iteration alone is ∈ O(n).
This also provides us with a lower bound for the algorithm: Since the first iteration of the algorithm is ∈ O(n), then whole algorithm takes at least Ω(n).
Now let us get to estimating the execution time. Normally, the first approach is to estimate the inner and outer loop separately and multiplying them together. Clearly, the outer loop has complexity log(n). Estimating the inner loop, however, is not trivial. So we can estimate it with n (which is an overestimation) and get a result of n log(n). This is an upper bound.
To get a more precise estimation, let us make two observations:
The inner loop basically adds up all values of outer loop variable i
Loop variable i follows the pattern of n, n/2, n/4, ..., 1, 0
Let us assume that n = 2^k, k ∈ ℕ, k > 0, i.e. n is a power of 2. Then n/2 = 2^(k-1), n/4 = 2^(k-2), ... To generalize from this assumtion, if n is not a power of 2, we set it to the next smaller power of 2. This is, in fact, an exact estimation. I leave the reasoning as to why as an exercise for the reader.
It is a well-known fact that 2^k + 2^(k-1) + 2^(k-2) + ... + 1 (+ 0) =sum_(i=0)^k 2^i = 2^(k+1) - 1. Since our input is n = 2^k, we know that 2^(k+1)= 2 * 2^k = 2 * n ∈ O(n). The algorithm's runtime complexity is, in fact, Θ(n), i.e. this is an upper and a lower bound. It is also a lower bound since the estimation we made is exact. Alternatively, we can use our observation of the Algorithm being ∈ Ω(n) and thus arrive this way at Θ(n).

First of all, look at the outer loop. You can see it iterates until i < 1 or i = 0. So, outer loop executes for values for i = N, N/2, N/4 … N/2^k (executing k number of times)
N/2^k < 1
N<2^k
k>log(N)
so, outer loop executes logN times.
Now, looking at inner loop. First of all, it executes for N times, then N/2 times then N/4 times until it reaches 1. Basically, executing logN times.
So, time complexity will be N + N/2 + … logN terms.
By Geometric progression:
a=N, r= 1/2, n= logn (Remember logn has base 2)
Also, using a^logb = b^loga and log2 is 1.
N((1- (1/2)^logN)/(1-1/2)) = 2N(1-(1^logN)/(N^log2)) = 2N(1-1/N)=2(N-1) = 2*N = O(N)
So, time complexity is O(N)

Linear O(n)
Total cost = n (first outer loop iteration) + n/2 (second outer loop iteration) + n/4 (third) + ... etc to a total of log(n) iterations. This sum is bounded by 2n (sum of a geometric series with a = n, r = 1/2).

Maximizing GCD with some condition

This is a problem given in HackWithInfy2019 in hackerrank.
I am stuck with this problem since yesterday.
Question:
You are given array of N integers.You have to find a pair (i,j)
which maximizes the value of GCD(a[i],a[j])+(j - i)
and 1<=i< j<=n
Constraints are:
2<= N <= 10^5
1<= a[i] <= 10^5
I've tried this problem using python

Here is an approach that could work:
result = 0
min_i = array[1 ... 100000] initialized to 0
for j in [1, 2, ..., n]
for d in divisors of a[j]
let i = min_i[d]
if i > 0
result = max(result, d + j - i)
else
min_i[d] = j
Here, min_i[d] for each d is the smallest i such that a[i] % d == 0. We use this in the inner loop to, for each d, find the first element in the array whose GCD with a[j] is at least d. When j is one of the possible values for which gcd(a[i], a[j]) + j - i is maximal, when the inner loop runs with d equal to the required GCD, result will be set to the correct answer.
The maximum possible number of divisors for a natural number less than or equal to 100,000 is 128 (see here). Therefore the inner loop runs at most 128 * 100,000 = 12.8 million times. I imagine this could pass with some optimizations (although maybe not in Python).
(To iterate over divisors, use a sieve to precompute the smallest nontrivial divisor for each integer from 1 to 100000.)

Here is one way of doing it.
Create a mutable class MinMax for storing the min. and max. index.
Create a Map<Integer, MinMax> for storing the min. and max. index for a particular divisor.
For each value in a, find all divisors for a[i], and update the map accordingly, such that the MinMax object stores the min. and max. i of the number with that particular divisor.
When done, iterate the map and find the entry with largest result of calculating key + value.max - value.min.
The min. and max. values of that entry is your answer.

Algorithm to efficiently determine the [n][n] element in a matrix

This is a question regarding a piece of coursework so would rather you didn't fully answer the question but rather give tips to improve the run time complexity of my current algorithm.
I have been given the following information:
A function g(n) is given by g(n) = f(n,n) where f may be defined recursively by
I have implemented this algorithm recursively with the following code:
public static double f(int i, int j)
{
if (i == 0 && j == 0) {
return 0;
}
if (i ==0 || j == 0) {
return 1;
}
return ((f(i-1, j)) + (f(i-1, j-1)) + (f(i, j-1)))/3;
}
This algorithm gives the results I am looking for, but it is extremely inefficient and I am now tasked to improve the run time complexity.
I wrote an algorithm to create an n*n matrix and it then computes every element up to the [n][n] element in which it then returns the [n][n] element, for example f(1,1) would return 0.6 recurring. The [n][n] element is 0.6 recurring because it is the result of (1+0+1)/3.
I have also created a spreadsheet of the result from f(0,0) to f(7,7) which can be seen below:
Now although this is much faster than my recursive algorithm, it has a huge overhead of creating a n*n matrix.
Any suggestions to how I can improve this algorithm will be greatly appreciated!
I can now see that is it possible to make the algorithm O(n) complexity, but is it possible to work out the result without creating a [n][n] 2D array?
I have created a solution in Java that runs in O(n) time and O(n) space and will post the solution after I have handed in my coursework to stop any plagiarism.

This is another one of those questions where it's better to examine it, before diving in and writing code.
The first thing i'd say you should do is look at a grid of the numbers, and to not represent them as decimals, but fractions instead.
The first thing that should be obvious is that the total number of you have is just a measure of the distance from the origin, .
If you look at a grid in this way, you can get all of the denominators:
Note that the first row and column are not all 1s - they've been chosen to follow the pattern, and the general formula which works for all of the other squares.
The numerators are a little bit more tricky, but still doable. As with most problems like this, the answer is related to combinations, factorials, and then some more complicated things. Typical entries here include Catalan numbers, Stirling's numbers, Pascal's triangle, and you will nearly always see Hypergeometric functions used.
Unless you do a lot of maths, it's unlikely you're familiar with all of these, and there is a hell of a lot of literature. So I have an easier way to find out the relations you need, which nearly always works. It goes like this:
Write a naive, inefficient algorithm to get the sequence you want.
Copy a reasonably large amount of the numbers into google.
Hope that a result from the Online Encyclopedia of Integer Sequences pops up.
3.b. If one doesn't, then look at some differences in your sequence, or some other sequence related to your data.
Use the information you find to implement said sequence.
So, following this logic, here are the numerators:
Now, unfortunately, googling those yielded nothing. However, there are a few things you can notice about them, the main being that the first row/column are just powers of 3, and that the second row/column are one less than powers of three. This kind boundary is exactly the same as Pascal's triangle, and a lot of related sequences.
Here is the matrix of differences between the numerators and denominators:
Where we've decided that the f(0,0) element shall just follow the same pattern. These numbers already look much simpler. Also note though - rather interestingly, that these numbers follow the same rules as the initial numbers - except the that the first number is one (and they are offset by a column and a row). T(i,j) = T(i-1,j) + T(i,j-1) + 3*T(i-1,j-1):
1
1 1
1 5 1
1 9 9 1
1 13 33 13 1
1 17 73 73 17 1
1 21 129 245 192 21 1
1 25 201 593 593 201 25 1
This looks more like the sequences you see a lot in combinatorics.
If you google numbers from this matrix, you do get a hit.
And then if you cut off the link to the raw data, you get sequence A081578, which is described as a "Pascal-(1,3,1) array", which exactly makes sense - if you rotate the matrix, so that the 0,0 element is at the top, and the elements form a triangle, then you take 1* the left element, 3* the above element, and 1* the right element.
The question now is implementing the formulae used to generate the numbers.
Unfortunately, this is often easier said than done. For example, the formula given on the page:
T(n,k)=sum{j=0..n, C(k,j-k)*C(n+k-j,k)*3^(j-k)}
is wrong, and it takes a fair bit of reading the paper (linked on the page) to work out the correct formula. The sections you want are proposition 26, corollary 28. The sequence is mentioned in Table 2 after proposition 13. Note that r=4
The correct formula is given in proposition 26, but there is also a typo there :/. The k=0 in the sum should be a j=0:
Where T is the triangular matrix containing the coefficients.
The OEIS page does give a couple of implementations to calculate the numbers, but neither of them are in java, and neither of them can be easily transcribed to java:
There is a mathematica example:
Table[ Hypergeometric2F1[-k, k-n, 1, 4], {n, 0, 10}, {k, 0, n}] // Flatten
which, as always, is ridiculously succinct. And there is also a Haskell version, which is equally terse:
a081578 n k = a081578_tabl !! n !! k
a081578_row n = a081578_tabl !! n
a081578_tabl = map fst $ iterate
(\(us, vs) -> (vs, zipWith (+) (map (* 3) ([0] ++ us ++ [0])) $
zipWith (+) ([0] ++ vs) (vs ++ [0]))) ([1], [1, 1])
I know you're doing this in java, but i could not be bothered to transcribe my answer to java (sorry). Here's a python implementation:
from __future__ import division
import math
#
# Helper functions
#
def cache(function):
cachedResults = {}
def wrapper(*args):
if args in cachedResults:
return cachedResults[args]
else:
result = function(*args)
cachedResults[args] = result
return result
return wrapper
#cache
def fact(n):
return math.factorial(n)
#cache
def binomial(n,k):
if n < k: return 0
return fact(n) / ( fact(k) * fact(n-k) )
def numerator(i,j):
"""
Naive way to calculate numerator
"""
if i == j == 0:
return 0
elif i == 0 or j == 0:
return 3**(max(i,j)-1)
else:
return numerator(i-1,j) + numerator(i,j-1) + 3*numerator(i-1,j-1)
def denominator(i,j):
return 3**(i+j-1)
def A081578(n,k):
"""
http://oeis.org/A081578
"""
total = 0
for j in range(n-k+1):
total += binomial(k, j) * binomial(n-k, j) * 4**(j)
return int(total)
def diff(i,j):
"""
Difference between the numerator, and the denominator.
Answer will then be 1-diff/denom.
"""
if i == j == 0:
return 1/3
elif i==0 or j==0:
return 0
else:
return A081578(j+i-2,i-1)
def answer(i,j):
return 1 - diff(i,j) / denominator(i,j)
# And a little bit at the end to demonstrate it works.
N, M = 10,10
for i in range(N):
row = "%10.5f"*M % tuple([numerator(i,j)/denominator(i,j) for j in range(M)])
print row
print ""
for i in range(N):
row = "%10.5f"*M % tuple([answer(i,j) for j in range(M)])
print row
So, for a closed form:
Where the are just binomial coefficients.
Here's the result:
One final addition, if you are looking to do this for large numbers, then you're going to need to compute the binomial coefficients a different way, as you'll overflow the integers. Your answers are lal floating point though, and since you're apparently interested in large f(n) = T(n,n) then I guess you could use Stirling's approximation or something.

Well for starters here are some things to keep in mind:
This condition can only occur once, yet you test it every time through every loop.
if (x == 0 && y == 0) {
matrix[x][y] = 0;
}
You should instead: matrix[0][0] = 0; right before you enter your first loop and set x to 1. Since you know x will never be 0 you can remove the first part of your second condition x == 0 :
for(int x = 1; x <= i; x++)
{
for(int y = 0; y <= j; y++)
{
if (y == 0) {
matrix[x][y] = 1;
}
else
matrix[x][y] = (matrix[x-1][y] + matrix[x-1][y-1] + matrix[x][y-1])/3;
}
}
No point in declaring row and column since you only use it once. double[][] matrix = new double[i+1][j+1];

This algorithm has a minimum complexity of Ω(n) because you just need to multiply the values in the first column and row of the matrix with some factors and then add them up. The factors stem from unwinding the recursion n times.
However you therefore need to do the unwinding of the recursion. That itself has a complexity of O(n^2). But by balancing unwinding and evaluation of recursion, you should be able to reduce complexity to O(n^x) where 1 <= x <= 2. This is some kind of similiar to algorithms for matrix-matrix multiplication, where the naive case has a complexity of O(n^3) but Strassens's algorithm is for example O(n^2.807).
Another point is the fact that the original formula uses a factor of 1/3. Since this is not accurately representable by fixed point numbers or ieee 754 floating points, the error increases when evaluating the recursion successively. Therefore unwinding the recursion could give you higher accuracy as a nice side effect.
For example when you unwind the recursion sqr(n) times then you have complexity O((sqr(n))^2+(n/sqr(n))^2). The first part is for unwinding and the second part is for evaluating a new matrix of size n/sqr(n). That new complexity actually can be simplified to O(n).

To describe time complexity we usually use a big O notation. It is important to remember that it only describes the growth given the input. O(n) is linear time complexity, but it doesn't say how quickly (or slowly) the time grows when we increase input. For example:
n=3 -> 30 seconds
n=4 -> 40 seconds
n=5 -> 50 seconds
This is O(n), we can clearly see that every increase of n increases the time by 10 seconds.
n=3 -> 60 seconds
n=4 -> 80 seconds
n=5 -> 100 seconds
This is also O(n), even though for every n we need twice that much time, and the raise is 20 seconds for every increase of n, the time complexity grows linearly.
So if you have O(n*n) time complexity and you will half the number of operations you perform, you will get O(0.5*n*n) which is equal to O(n*n) - i.e. your time complexity won't change.
This is theory, in practice the number of operations sometimes makes a difference. Because you have a grid n by n, you need to fill n*n cells, so the best time complexity you can achieve is O(n*n), but there are a few optimizations you can do:
Cells on the edges of the grid could be filled in separate loops. Currently in majority of the cases you have two unnecessary conditions for i and j equal to 0.
You grid has a line of symmetry, you could utilize it to calculate only half of it and then copy the results onto the other half. For every i and j grid[i][j] = grid[j][i]
On final note, the clarity and readability of the code is much more important than performance - if you can read and understand the code, you can change it, but if the code is so ugly that you cannot understand it, you cannot optimize it. That's why I would do only first optimization (it also increases readability), but wouldn't do the second one - it would make the code much more difficult to understand.
As a rule of thumb, don't optimize the code, unless the performance is really causing problems. As William Wulf said:
More computing sins are committed in the name of efficiency (without necessarily achieving it) than for any other single reason - including blind stupidity.
EDIT:
I think it may be possible to implement this function with O(1) complexity. Although it gives no benefits when you need to fill entire grid, with O(1) time complexity you can instantly get any value without having a grid at all.
A few observations:
denominator is equal to 3 ^ (i + j - 1)
if i = 2 or j = 2, numerator is one less than denominator
EDIT 2:
The numerator can be expressed with the following function:
public static int n(int i, int j) {
if (i == 1 || j == 1) {
return 1;
} else {
return 3 * n(i - 1, j - 1) + n(i - 1, j) + n(i, j - 1);
}
}
Very similar to original problem, but no division and all numbers are integers.

If the question is about how to output all values of the function for 0<=i<N, 0<=j<N, here is a solution in time O(N²) and space O(N). The time behavior is optimal.
Use a temporary array T of N numbers and set it to all ones, except for the first element.
Then row by row,
use a temporary element TT and set it to 1,
then column by column, assign simultaneously T[I-1], TT = TT, (TT + T[I-1] + T[I])/3.

Thanks to will's (first) answer, I had this idea:
Consider that any positive solution comes only from the 1's along the x and y axes. Each of the recursive calls to f divides each component of the solution by 3, which means we can sum, combinatorially, how many ways each 1 features as a component of the solution, and consider it's "distance" (measured as how many calls of f it is from the target) as a negative power of 3.
JavaScript code:
function f(n){
var result = 0;
for (var d=n; d<2*n; d++){
var temp = 0;
for (var NE=0; NE<2*n-d; NE++){
temp += choose(n,NE);
}
result += choose(d - 1,d - n) * temp / Math.pow(3,d);
}
return 2 * result;
}
function choose(n,k){
if (k == 0 || n == k){
return 1;
}
var product = n;
for (var i=2; i<=k; i++){
product *= (n + 1 - i) / i
}
return product;
}
Output:
for (var i=1; i<8; i++){
console.log("F(" + i + "," + i + ") = " + f(i));
}
F(1,1) = 0.6666666666666666
F(2,2) = 0.8148148148148148
F(3,3) = 0.8641975308641975
F(4,4) = 0.8879743941472337
F(5,5) = 0.9024030889600163
F(6,6) = 0.9123609205913732
F(7,7) = 0.9197747256986194

What does someArray[--n] mean?

I tried googling this but Google doesn't handle "--n" well. I saw this in my professor's code:
f[--n];
f[n++];
where f is an array of double values.
My guess is that it returns the value of f[n] before reducing (or adding) to n.

f[--n]; means :
n = n -1;
f[n];
f[n++]; means :
f[n];
n = n + 1;

It's actually a type of operator called a pre-decrement, and it's part of a family of 4 operators (see table of java operators)
For an integer-type variable called n:
post-increment n++ is the equivalent of n = n + 1, the 'post' part means that if you see it in a line of code (ex. foo(n++);) then the line of code will be called Before n is incremented.
pre-increment ++n is also the same as n = n + 1 but it occurs Before the line of code it belongs in has been run.
post-decrement n-- is the equivalent of n = n - 1 and occurs After the current line of code has been run
pre-decrement --n is the equivalent of n = n - 1 and occurs Before the current line of code has been run
Example of post vs pre decrement:
int n = 5;
System.out.println(n--); //This prints 5
System.out.println(n); //This prints 4
System.out.println(--n); //This prints 3
System.out.println(n); //this prints 3

you can look it up under predecrement (--n) or postincrement (n++).
It works like this:
f[--n]: first n is reduced by 1 then the value of f[n] (this is already the reduced n) is returned
f[n++]: first the value of f[n] is returned then n is increased by 1
Example:
f {1,3,5}
n=1;
f[--n] returns 1
f[n++] returns 3

The code
f[--n];
f[n++];
Is the same as
n--;
f[n];
f[n];
n++;

Algorithm for selecting k element out of n element array whose sum is zero [duplicate]

recently I became interested in the subset-sum problem which is finding a zero-sum subset in a superset. I found some solutions on SO, in addition, I came across a particular solution which uses the dynamic programming approach. I translated his solution in python based on his qualitative descriptions. I'm trying to optimize this for larger lists which eats up a lot of my memory. Can someone recommend optimizations or other techniques to solve this particular problem? Here's my attempt in python:
import random
from time import time
from itertools import product
time0 = time()
# create a zero matrix of size a (row), b(col)
def create_zero_matrix(a,b):
return [[0]*b for x in xrange(a)]
# generate a list of size num with random integers with an upper and lower bound
def random_ints(num, lower=-1000, upper=1000):
return [random.randrange(lower,upper+1) for i in range(num)]
# split a list up into N and P where N be the sum of the negative values and P the sum of the positive values.
# 0 does not count because of additive identity
def split_sum(A):
N_list = []
P_list = []
for x in A:
if x < 0:
N_list.append(x)
elif x > 0:
P_list.append(x)
return [sum(N_list), sum(P_list)]
# since the column indexes are in the range from 0 to P - N
# we would like to retrieve them based on the index in the range N to P
# n := row, m := col
def get_element(table, n, m, N):
if n < 0:
return 0
try:
return table[n][m - N]
except:
return 0
# same definition as above
def set_element(table, n, m, N, value):
table[n][m - N] = value
# input array
#A = [1, -3, 2, 4]
A = random_ints(200)
[N, P] = split_sum(A)
# create a zero matrix of size m (row) by n (col)
#
# m := the number of elements in A
# n := P - N + 1 (by definition N <= s <= P)
#
# each element in the matrix will be a value of either 0 (false) or 1 (true)
m = len(A)
n = P - N + 1;
table = create_zero_matrix(m, n)
# set first element in index (0, A[0]) to be true
# Definition: Q(1,s) := (x1 == s). Note that index starts at 0 instead of 1.
set_element(table, 0, A[0], N, 1)
# iterate through each table element
#for i in xrange(1, m): #row
# for s in xrange(N, P + 1): #col
for i, s in product(xrange(1, m), xrange(N, P + 1)):
if get_element(table, i - 1, s, N) or A[i] == s or get_element(table, i - 1, s - A[i], N):
#set_element(table, i, s, N, 1)
table[i][s - N] = 1
# find zero-sum subset solution
s = 0
solution = []
for i in reversed(xrange(0, m)):
if get_element(table, i - 1, s, N) == 0 and get_element(table, i, s, N) == 1:
s = s - A[i]
solution.append(A[i])
print "Solution: ",solution
time1 = time()
print "Time execution: ", time1 - time0

I'm not quite sure if your solution is exact or a PTA (poly-time approximation).
But, as someone pointed out, this problem is indeed NP-Complete.
Meaning, every known (exact) algorithm has an exponential time behavior on the size of the input.
Meaning, if you can process 1 operation in .01 nanosecond then, for a list of 59 elements it'll take:
2^59 ops --> 2^59 seconds --> 2^26 years --> 1 year
-------------- ---------------
10.000.000.000 3600 x 24 x 365
You can find heuristics, which give you just a CHANCE of finding an exact solution in polynomial time.
On the other side, if you restrict the problem (to another) using bounds for the values of the numbers in the set, then the problem complexity reduces to polynomial time. But even then the memory space consumed will be a polynomial of VERY High Order.
The memory consumed will be much larger than the few gigabytes you have in memory.
And even much larger than the few tera-bytes on your hard drive.
( That's for small values of the bound for the value of the elements in the set )
May be this is the case of your Dynamic programing algorithm.
It seemed to me that you were using a bound of 1000 when building your initialization matrix.
You can try a smaller bound. That is... if your input is consistently consist of small values.
Good Luck!

Someone on Hacker News came up with the following solution to the problem, which I quite liked. It just happens to be in python :):
def subset_summing_to_zero (activities):
subsets = {0: []}
for (activity, cost) in activities.iteritems():
old_subsets = subsets
subsets = {}
for (prev_sum, subset) in old_subsets.iteritems():
subsets[prev_sum] = subset
new_sum = prev_sum + cost
new_subset = subset + [activity]
if 0 == new_sum:
new_subset.sort()
return new_subset
else:
subsets[new_sum] = new_subset
return []
I spent a few minutes with it and it worked very well.

An interesting article on optimizing python code is available here. Basically the main result is that you should inline your frequent loops, so in your case this would mean instead of calling get_element twice per loop, put the actual code of that function inside the loop in order to avoid the function call overhead.
Hope that helps! Cheers

, 1st eye catch
def split_sum(A):
N_list = 0
P_list = 0
for x in A:
if x < 0:
N_list+=x
elif x > 0:
P_list+=x
return [N_list, P_list]
Some advices:
Try to use 1D list and use bitarray to reduce memory footprint at minimum (http://pypi.python.org/pypi/bitarray) so you will just change get / set functon. This should reduce your memory footprint by at lest 64 (integer in list is pointer to integer whit type so it can be factor 3*32)
Avoid using try - catch, but figure out proper ranges at beginning, you might found out that you will gain huge speed.

The following code works for Python 3.3+ , I have used the itertools module in Python that has some great methods to use.
from itertools import chain, combinations
def powerset(iterable):
s = list(iterable)
return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))
nums = input("Enter the Elements").strip().split()
inputSum = int(input("Enter the Sum You want"))
for i, combo in enumerate(powerset(nums), 1):
sum = 0
for num in combo:
sum += int(num)
if sum == inputSum:
print(combo)
The Input Output is as Follows:
Enter the Elements 1 2 3 4
Enter the Sum You want 5
('1', '4')
('2', '3')

Just change the values in your set w and correspondingly make an array x as big as the len of w then pass the last value in the subsetsum function as the sum for which u want subsets and you wl bw done (if u want to check by giving your own values).
def subsetsum(cs,k,r,x,w,d):
x[k]=1
if(cs+w[k]==d):
for i in range(0,k+1):
if x[i]==1:
print (w[i],end=" ")
print()
elif cs+w[k]+w[k+1]<=d :
subsetsum(cs+w[k],k+1,r-w[k],x,w,d)
if((cs +r-w[k]>=d) and (cs+w[k]<=d)) :
x[k]=0
subsetsum(cs,k+1,r-w[k],x,w,d)
#driver for the above code
w=[2,3,4,5,0]
x=[0,0,0,0,0]
subsetsum(0,0,sum(w),x,w,7)