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Checking whether there is a subset of size k of an array which has a sum multiple of n

Good evening, I have an array in java with n integer numbers. I want to check if there is a subset of size k of the entries that satisfies the condition:
The sum of those k entries is a multiple of m.
How may I do this as efficiently as possible? There are n!/k!(n-k)! subsets that I need to check.

You can use dynamic programming. The state is (prefix length, sum modulo m, number of elements in a subset). Transitions are obvious: we either add one more number(increasing the number of elements in a subset and computing new sum modulo m), or we just increase prefix lenght(not adding the current number). If you just need a yes/no answer, you can store only the last layer of values and apply bit optimizations to compute transitions faster. The time complexity is O(n * m * k), or about n * m * k / 64 operations with bit optimizations. The space complexity is O(m * k). It looks feasible for a few thousands of elements. By bit optimizations I mean using things like bitset in C++ that can perform an operation on a group of bits at the same time using bitwise operations.

I don't like this solution, but it may work for your needs
public boolean containsSubset( int[] a , int currentIndex, int currentSum, int depth, int divsor, int maxDepth){
//you could make a, maxDepth, and divisor static as well
//If maxDepthis equal to depth, then our subset has k elements, in addition the sum of
//elements must be divisible by out divsor, m
//If this condition is satisafied, then there exists a subset of size k whose sum is divisible by m
if(depth==maxDepth&&currentSum%divsor==0)
return true;
//If the depth is greater than or equal maxDepth, our subset has more than k elements, thus
//adding more elements can not satisfy the necessary conditions
//additionally we know that if it contains k elements and is divisible by m, it would've satisafied the above condition.
if(depth>=maxdepth)
return false;
//boolean to be returned, initialized to false because we have not found any sets yet
boolean ret = false;
//iterate through all remaining elements of our array
for (int i = currentIndex+1; i < a.length; i++){
//this may be an optimization or this line
//for (int i = currentIndex+1; i < a.length-maxDepth+depth; i++){
//by recursing, we add a[i] to our set we then use an or operation on all our subsets that could
//be constructed from the numbers we have so far so that if any of them satisfy our condition (return true)
//then the value of the variable ret will be true
ret |= containsSubset(a,i,currentSum+a[i],depth+1,divisor, maxDepth);
} //end for
//return the variable storing whether any sets of numbers that could be constructed from the numbers so far.
return ret;
}
Then invoke this method as such
//this invokes our method with "no numbers added to our subset so far" so it will try adding
// all combinations of other elements to determine if the condition is satisfied.
boolean answer = containsSubset(myArray,-1,0,0,m,k);
EDIT:
You could probably optimize this by taking everything modulo (%) m and deleting repeats. For examples with large values of n and/or k, but small values of m, this could be a pretty big optimization.
EDIT 2:
The above optimization I listed isn't helpful. You may need the repeats to get the correct information. My bad.
Happy Coding! Let me know if you have any questions!

If numbers have lower and upper bounds, it might be better to:
Iterate all multiples of n where lower_bound * k < multiple < upper_bound * k
Check if there is a subset with sum multiple in the array (see Subset Sum problem) using dynamic programming.
Complexity is O(k^2 * (lower_bound + upper_bound)^2). This approach can be optimized further, I believe with careful thinking.
Otherwise you can find all subsets of size k. Complexity is O(n!). Using backtracking (pseudocode-ish):
function find_subsets(array, k, index, current_subset):
if current_subset.size = k:
add current_subset to your solutions list
return
if index = array.size:
return
number := array[index]
add number to current_subset
find_subsets(array, k, index + 1, current_subset)
remove number from current_subset
find_subsets(array, k, index + 1, current_subset)

repeated element in Array

I have an array of N elements and contain 1 to (N-1) integers-a sequence of integers starting from 1 to the max number N-1-, meaning that there is only one number is repeated, and I want to write an algorithm that return this repeated element, I have found a solution but it only could work if the array is sorted, which is may not be the case.
?
int i=0;
while(i<A[i])
{
i++
}
int rep = A[i];

I do not know why RC removed his comment but his idea was good.
With the knowledge of N you easy can calculate that the sum of [1:N-1]. then sum up all elementes in your array and subtract the above sum and you have your number.
This comes at the cost of O(n) and is not beatable.
However this only works with the preconditions you mentioned.
A more generic approach would be to sort the array and then simply walk through it. This would be O(n log(n)) and still better than your O(n²).
I you know the maximum number you may create a lookup table and init it with all zeros, walk through the array and check for one and mark the entries with one. The complexity is also just O(n) but at the expense of memory.
if the value range is unknown a simiar approach can be used but instead of using a lookup table a hashset canbe used.

Linear search will help you with complexity O(n):
final int n = ...;
final int a[] = createInput(n); // Expect each a[i] < n && a[i] >= 0
final int b[] = new int[n];
for (int i = 0; i < n; i++)
b[i]++;
for (int i = 0; i < n; i++)
if (b[i] >= 2)
return a[i];
throw new IllegalArgumentException("No duplicates found");

A possible solution is to sum all elements in the array and then to compute the sym of the integers up to N-1. After that subtract the two values and voila - you found your number. This is the solution proposed by vlad_tepesch and it is good, but has a drawback - you may overflow the integer type. To avoid this you can use 64 bit integer.
However I want to propose a slight modification - compute the xor sum of the integers up to N-1(that is compute 1^2^3^...(N-1)) and compute the xor sum of your array(i.e. a0^a1^...aN-1). After that xor the two values and the result will be the repeated element.

How to efficiently generate a set of unique random numbers with a predefined distribution?

I have a map of items with some probability distribution:
Map<SingleObjectiveItem, Double> itemsDistribution;
Given a certain m I have to generate a Set of m elements sampled from the above distribution.
As of now I was using the naive way of doing it:
while(mySet.size < m)
mySet.add(getNextSample(itemsDistribution));
The getNextSample(...) method fetches an object from the distribution as per its probability. Now, as m increases the performance severely suffers. For m = 500 and itemsDistribution.size() = 1000 elements, there is too much thrashing and the function remains in the while loop for too long. Generate 1000 such sets and you have an application that crawls.
Is there a more efficient way to generate a unique set of random numbers with a "predefined" distribution? Most collection shuffling techniques and the like are uniformly random. What would be a good way to address this?
UPDATE: The loop will call getNextSample(...) "at least" 1 + 2 + 3 + ... + m = m(m+1)/2 times. That is in the first run we'll definitely get a sample for the set. The 2nd iteration, it may be called at least twice and so on. If getNextSample is sequential in nature, i.e., goes through the entire cumulative distribution to find the sample, then the run time complexity of the loop is at least: n*m(m+1)/2, 'n' is the number of elements in the distribution. If m = cn; 0<c<=1 then the loop is at least Sigma(n^3). And that too is the lower bound!
If we replace sequential search by binary search, the complexity would be at least Sigma(log n * n^2). Efficient but may not be by a large margin.
Also, removing from the distribution is not possible since I call the above loop k times, to generate k such sets. These sets are part of a randomized 'schedule' of items. Hence a 'set' of items.

Start out by generating a number of random points in two dimentions.
Then apply your distribution
Now find all entries within the distribution and pick the x coordinates, and you have your random numbers with the requested distribution like this:

The problem is unlikely to be the loop you show:
Let n be the size of the distribution, and I be the number of invocations to getNextSample. We have I = sum_i(C_i), where C_i is the number of invocations to getNextSample while the set has size i. To find E[C_i], observe that C_i is the inter-arrival time of a poisson process with λ = 1 - i / n, and therefore exponentially distributed with λ. Therefore, E[C_i] = 1 / λ = therefore E[C_i] = 1 / (1 - i / n) <= 1 / (1 - m / n). Therefore, E[I] < m / (1 - m / n).
That is, sampling a set of size m = n/2 will take, on average, less than 2m = n invocations of getNextSample. If that is "slow" and "crawls", it is likely because getNextSample is slow. This is actually unsurprising, given the unsuitable way the distrubution is passed to the method (because the method will, of necessity, have to iterate over the entire distribution to find a random element).
The following should be faster (if m < 0.8 n)
class Distribution<T> {
private double[] cummulativeWeight;
private T[] item;
private double totalWeight;
Distribution(Map<T, Double> probabilityMap) {
int i = 0;
cummulativeWeight = new double[probabilityMap.size()];
item = (T[]) new Object[probabilityMap.size()];
for (Map.Entry<T, Double> entry : probabilityMap.entrySet()) {
item[i] = entry.getKey();
totalWeight += entry.getValue();
cummulativeWeight[i] = totalWeight;
i++;
}
}
T randomItem() {
double weight = Math.random() * totalWeight;
int index = Arrays.binarySearch(cummulativeWeight, weight);
if (index < 0) {
index = -index - 1;
}
return item[index];
}
Set<T> randomSubset(int size) {
Set<T> set = new HashSet<>();
while(set.size() < size) {
set.add(randomItem());
}
return set;
}
}
public class Test {
public static void main(String[] args) {
int max = 1_000_000;
HashMap<Integer, Double> probabilities = new HashMap<>();
for (int i = 0; i < max; i++) {
probabilities.put(i, (double) i);
}
Distribution<Integer> d = new Distribution<>(probabilities);
Set<Integer> set = d.randomSubset(max / 2);
//System.out.println(set);
}
}
The expected runtime is O(m / (1 - m / n) * log n). On my computer, a subset of size 500_000 of a set of 1_000_000 is computed in about 3 seconds.
As we can see, the expected runtime approaches infinity as m approaches n. If that is a problem (i.e. m > 0.9 n), the following more complex approach should work better:
Set<T> randomSubset(int size) {
Set<T> set = new HashSet<>();
while(set.size() < size) {
T randomItem = randomItem();
remove(randomItem); // removes the item from the distribution
set.add(randomItem);
}
return set;
}
To efficiently implement remove requires a different representation for the distribution, for instance a binary tree where each node stores the total weight of the subtree whose root it is.
But that is rather complicated, so I wouldn't go that route if m is known to be significantly smaller than n.

If you are not concerning with randomness properties too much then I do it like this:
create buffer for pseudo-random numbers
double buff[MAX]; // [edit1] double pseudo random numbers
MAX is size should be big enough ... 1024*128 for example
type can be any (float,int,DWORD...)
fill buffer with numbers
you have range of numbers x = < x0,x1 > and probability function probability(x) defined by your probability distribution so do this:
for (i=0,x=x0;x<=x1;x+=stepx)
for (j=0,n=probability(x)*MAX,q=0.1*stepx/n;j<n;j++,i++) // [edit1] unique pseudo-random numbers
buff[i]=x+(double(i)*q); // [edit1] ...
The stepx is your accuracy for items (for integral types = 1) now the buff[] array has the same distribution as you need but it is not pseudo-random. Also you should add check if j is not >= MAX to avoid array overruns and also at the end the real size of buff[] is j (can be less than MAX due to rounding)
shuffle buff[]
do just few loops of swap buff[i] and buff[j] where i is the loop variable and j is pseudo-random <0-MAX)
write your pseudo-random function
it just return number from the buffer. At first call returns the buff[0] at second buff[1] and so on ... For standard generators When you hit the end of buff[] then shuffle buff[] again and start from buff[0] again. But as you need unique numbers then you can not reach the end of buffer so so set MAX to be big enough for your task otherwise uniqueness will not be assured.
[Notes]
MAX should be big enough to store the whole distribution you want. If it is not big enough then items with low probability can be missing completely.
[edit1] - tweaked answer a little to match the question needs (pointed by meriton thanks)
PS. complexity of initialization is O(N) and for get number is O(1).

You should implement your own random number generator (using a MonteCarlo methode or any good uniform generator like mersen twister) and basing on the inversion method (here).
For example : exponential law: generate a uniform random number u in [0,1] then your random variable of the exponential law would be : ln(1-u)/(-lambda) lambda being the exponential law parameter and ln the natural logarithm.
Hope it'll help ;).

I think you have two problems:
Your itemDistribution doesn't know you need a set, so when the set you are building gets
large you will pick a lot of elements that are already in the set. If you start with the
set all full and remove elements you will run into the same problem for very small sets.
Is there a reason why you don't remove the element from the itemDistribution after you
picked it? Then you wouldn't pick the same element twice?
The choice of datastructure for itemDistribution looks suspicious to me. You want the
getNextSample operation to be fast. Doesn't the map from values to probability force you
to iterate through large parts of the map for each getNextSample. I'm no good at
statistics but couldn't you represent the itemDistribution the other way, like a map from
probability, or maybe the sum of all smaller probabilities + probability to a element
of the set?

Your performance depends on how your getNextSample function works. If you have to iterate over all probabilities when you pick the next item, it might be slow.
A good way to pick several unique random items from a list is to first shuffle the list and then pop items off the list. You can shuffle the list once with the given distribution. From then on, picking your m items ist just popping the list.
Here's an implementation of a probabilistic shuffle:
List<Item> prob_shuffle(Map<Item, int> dist)
{
int n = dist.length;
List<Item> a = dist.keys();
int psum = 0;
int i, j;
for (i in dist) psum += dist[i];
for (i = 0; i < n; i++) {
int ip = rand(psum); // 0 <= ip < psum
int jp = 0;
for (j = i; j < n; j++) {
jp += dist[a[j]];
if (ip < jp) break;
}
psum -= dist[a[j]];
Item tmp = a[i];
a[i] = a[j];
a[j] = tmp;
}
return a;
}
This in not Java, but pseudocude after an implementation in C, so please take it with a grain of salt. The idea is to append items to the shuffled area by continuously picking items from the unshuffled area.
Here, I used integer probabilities. (The proabilities don't have to add to a special value, it's just "bigger is better".) You can use floating-point numbers but because of inaccuracies, you might end up going beyond the array when picking an item. You should use item n - 1 then. If you add that saftey net, you could even have items with zero probability that always get picked last.
There might be a method to speed up the picking loop, but I don't really see how. The swapping renders any precalculations useless.

Accumulate your probabilities in a table
Probability
Item Actual Accumulated
Item1 0.10 0.10
Item2 0.30 0.40
Item3 0.15 0.55
Item4 0.20 0.75
Item5 0.25 1.00
Make a random number between 0.0 and 1.0 and do a binary search for the first item with a sum that is greater than your generated number. This item would have been chosen with the desired probability.

Ebbe's method is called rejection sampling.
I sometimes use a simple method, using an inverse cumulative distribution function, which is a function that maps a number X between 0 and 1 onto the Y axis.
Then you just generate a uniformly distributed random number between 0 and 1, and apply the function to it.
That function is also called the "quantile function".
For example, suppose you want to generate a normally distributed random number.
It's cumulative distribution function is called Phi.
The inverse of that is called probit.
There are many ways to generate normal variates, and this is just one example.
You can easily construct an approximate cumulative distribution function for any univariate distribution you like, in the form of a table.
Then you can just invert it by table-lookup and interpolation.

Efficiently determine the parity of a permutation

I have an int[] array of length N containing the values 0, 1, 2, .... (N-1), i.e. it represents a permutation of integer indexes.
What's the most efficient way to determine if the permutation has odd or even parity?
(I'm particularly keen to avoid allocating objects for temporary working space if possible....)

I think you can do this in O(n) time and O(n) space by simply computing the cycle decomposition.
You can compute the cycle decomposition in O(n) by simply starting with the first element and following the path until you return to the start. This gives you the first cycle. Mark each node as visited as you follow the path.
Then repeat for the next unvisited node until all nodes are marked as visited.
The parity of a cycle of length k is (k-1)%2, so you can simply add up the parities of all the cycles you have discovered to find the parity of the overall permutation.
Saving space
One way of marking the nodes as visited would be to add N to each value in the array when it is visited. You would then be able to do a final tidying O(n) pass to turn all the numbers back to the original values.

I selected the answer by Peter de Rivaz as the correct answer as this was the algorithmic approach I ended up using.
However I used a couple of extra optimisations so I thought I would share them:
Examine the size of data first
If it is greater than 64, use a java.util.BitSet to store the visited elements
If it is less than or equal to 64, use a long with bitwise operations to store the visited elements. This makes it O(1) space for many applications that only use small permutations.
Actually return the swap count rather than the parity. This gives you the parity if you need it, but is potentially useful for other purposes, and is no more expensive to compute.
Code below:
public int swapCount() {
if (length()<=64) {
return swapCountSmall();
} else {
return swapCountLong();
}
}
private int swapCountLong() {
int n=length();
int swaps=0;
BitSet seen=new BitSet(n);
for (int i=0; i<n; i++) {
if (seen.get(i)) continue;
seen.set(i);
for(int j=data[i]; !seen.get(j); j=data[j]) {
seen.set(j);
swaps++;
}
}
return swaps;
}
private int swapCountSmall() {
int n=length();
int swaps=0;
long seen=0;
for (int i=0; i<n; i++) {
long mask=(1L<<i);
if ((seen&mask)!=0) continue;
seen|=mask;
for(int j=data[i]; (seen&(1L<<j))==0; j=data[j]) {
seen|=(1L<<j);
swaps++;
}
}
return swaps;
}

You want the parity of the number of inversions. You can do this in O(n * log n) time using merge sort, but either you lose the initial array, or you need extra memory on the order of O(n).
A simple algorithm that uses O(n) extra space and is O(n * log n):
inv = 0
mergesort A into a copy B
for i from 1 to length(A):
binary search for position j of A[i] in B
remove B[j] from B
inv = inv + (j - 1)
That said, I don't think it's possible to do it in sublinear memory. See also:
https://cs.stackexchange.com/questions/3200/counting-inversion-pairs
https://mathoverflow.net/questions/72669/finding-the-parity-of-a-permutation-in-little-space

Consider this approach...
From the permutation, get the inverse permutation, by swapping the rows and
sorting according to the top row order. This is O(nlogn)
Then, simulate performing the inverse permutation and count the swaps, for O(n). This should give the parity of the permutation, according to this
An even permutation can be obtained as the composition of an even
number and only an even number of exchanges (called transpositions) of
two elements, while an odd permutation be obtained by (only) an odd
number of transpositions.
from Wikipedia.
Here's some code I had lying around, which performs an inverse permutation, I just modified it a bit to count swaps, you can just remove all mention of a, p contains the inverse permutation.
size_t
permute_inverse (std::vector<int> &a, std::vector<size_t> &p) {
size_t cnt = 0
for (size_t i = 0; i < a.size(); ++i) {
while (i != p[i]) {
++cnt;
std::swap (a[i], a[p[i]]);
std::swap (p[i], p[p[i]]);
}
}
return cnt;
}

UVa 10134: Is Bigger Smarter? (Dynamic programming and longest increasing subsequence)

private void findLDS() {
Integer[] array = Arrays.copyOf(elephants.iq, elephants.iq.length);
Hashtable<Integer, Integer> eq = elephants.elephantiqs;
Integer[] lds = new Integer[array.length];
Integer[] prev= new Integer[array.length];
lds[0] = 0;
prev[0] = 0;
int maxlds = 1, ending=0;
for(int i = 0; i < array.length; ++i) {
lds[i] = 1;
prev[i] = -1;
for (int j = i; j >= 0; --j) {
if(lds[j] + 1 > lds[i] && array[j] > array[i] && eq.get(array[j]) < eq.get(array[i])) {
lds[i] = lds[j]+1;
prev[i] = j;
}
}
if(lds[i] > maxlds) {
ending = i;
maxlds = lds[i];
}
}
System.out.println(maxlds);
for(int i = ending; i >= 0; --i) {
if(prev[i] != -1) {
System.out.println(eq.get(array[prev[i]]));
}
}
I have based this algorithm on this SO question. This code is trying to find longest decreasing subsequence instead of increasing. array[] is sorted in descending order, and I also have a hashtable with the elephants IQ's as keys for their weights.
I'm having a hard time properly understanding DP, and I need some help.
My algorithm seems to work fine besides tracking the chosen sequence in prev[], where it always misses one element. Does anyone know how to do this?

A few ways to approach this one:
Sort by weight in decreasing order, then find the longest increasing subsequence.
Sort by IQ in decreasing order, then find the longest increasing subsequence of weights.
and 4. are just (1) and (2), switching the words "increasing" and "decreasing"
If you don't understand the DP for longest increasing subsequence O(N^2), it's basically this:
Since the list has to be strictly increasing/decreasing anyway, you can just eliminate some elephants beforehand to make the set unique.
Create an array, which I will call llis standing for "Longest Increasing Subsequence", of length N, the number of elephants there now are. Create another array called last with the same length. I will assume the sorted list of elephants is called array as it is in your problem statement.
Assuming that you've already sorted the elephants in decreasing order, you will want to find the longest increasing subsequence of IQs.
Tell yourself that the element in the array llis at index n (this is a different "n") < N will be the length of the longest increasing subsequence for the sub-array of array from index 0 to n, inclusive. Also say that the element in the next array at index n will be the next index in array in the longest increasing subsequence.
Therefore, finding the length of the longest increasing subsequence in the "sub-array" of 0 to N - 1 inclusive, which is also the whole array, would only require you to find the N - 1 th element in the array llis after the DP calculations, and finding the actual subsequence would simplify to following the indices in the next array.
Now that you know what you're looking for, you can proceed with the algorithm. At index n in the array, how do you know what the longest increasing subsequence is? Well, if you've calculated the length of the longest increasing subsequence and the last value in the subsequences for every k < n, you can try adding the elephant at index n to the longest increasing subsequence ending at k if the IQ of the elephant n is higher than the IQ of the elephant at k. In this case, the length of the longest increasing subsequence ending at elephant n would be llis[k] + 1. (Also, remember to set next[k] to be n, since the next elephant in the increasing subsequence will be the one at n.)
We've found the DP relation that llis[n] = max(llis[n], llis[k] + 1), after going through all k s that come strictly before n. Just process the n s in the right order (linearly) and you should get the correct result.
Procedure/warnings: 1) Process n in order from 0 to N - 1. 2) For every n, process k in order from n - 1 to 0 because you want to minimize the k that you choose. 3) After you're done processing, make sure to find the maximum number in the array llis to get your final result.
Since this is tagged as homework, I won't explicitly say how to modify this to find the longest decreasing subsequence, but I hope my explanation has helped with your understanding of DP. It should be easy to figure out the decreasing version on your own, if you choose to use it. (Note that this problem can be solved using the increasing version, as described in approaches 1 or 2.)