I need to merge k (1 <= k <= 16) sorted arrays into one sorted array. This is for a homework assignment and the Professor requires that this be done using an O(n) algorithm. Merging 2 arrays is no problem and I can do it easily using an O(n) algorithm. I feel that what my professor is asking is undoable for n arrays with an O(n) algorithm.
I am using the below algorithm to split the array indices and running InsertionSort on each partition. I could save these start and end indices into a 2D array. I just don't see how the merging can be done using O(n) because this is going to require more than one loop. If it is possible, does anyone have any hints. I'm not looking for actual code, just a hint as to where I should start/
int chunkSize = round(float(arraySize) / numThreads);
for (int i = 0; i < numThreads; i++) {
int start = i * chunkSize;
int end = start + chunkSize - 1;
if (i == numThreads - 1) {
end = arraySize - 1;
}
InsertionSort(&array[start], end - start + 1);
}
EDIT: The requirement is that the algorithm be O(n) where n is the number of elements in the array. Also, I need to solve this without using a min heap.
EDIT #2: Here is an algorithm I came up with. The problem here is that I'm not storing the result of each iteration back into the original array. I could just copy all of it back in for a loop but that would be expensive. Is there any way I can do this, other than using something memcpy? In the below code, indices is a 2D array [numThreads][2] where array[i][0] is the start index and array[i][1] is the end index of the ith array.
void mergeArrays(int array[], int indices[][2], int threads, int result[]) {
for (int i = 0; i < threads - 1; i++) {
int resPos = 0;
int lhsPos = 0;
int lhsEnd = indices[i][1];
int rhsPos = indices[i+1][0];
int rhsEnd = indices[i+1][1];
while (lhsPos <= lhsEnd && rhsPos <= rhsEnd) {
if (array[lhsPos] <= array[rhsPos]) {
result[resPos] = array[lhsPos];
lhsPos++;
} else {
result[resPos] = array[rhsPos];
rhsPos++;
}
resPos++;
}
while (lhsPos <= lhsEnd) {
result[resPos] = array[lhsPos];
lhsPos++;
resPos++;
}
while (rhsPos <= rhsEnd) {
result[resPos] = array[rhsPos];
rhsPos++;
resPos++;
}
}
}
You can merge K sorted arrays in one sorted array with O(N*log(K)) algorithm, using priority queue with K entries, where N is overall number of elements in all arrays.
If K is considered as constant value (it is limited by 16 in your case), then complexity is O(N).
Note again: N is number of elements in my post, not number of arrays.
It is impossible to merge arrays in O(K) - simple copy takes O(N)
Using the facts you provided:
(1) n is the number of arrays to to merge;
(2) the arrays to be merged are already sorted;
(3) the merge needs to be of order n, that is linear in the number of arrays
(and NOT linear in the number of elements in each array, as you might mistakenly think at first sight).
Use the analogy of merging 4 sorted piles of cards, low to high, face up. You would pick the card with the lowest face value from one of the piles and put it (face down) on the merged deck, until all piles are exhausted.
For your program: keep a counter for each array for the number of elements you have already transferred to the output. This is at the same time an index to the next element in each array NOT merged in the output. Pick the smallest element that you find at one of these locations. You have to lookup the first waiting element in all the arrays for that, so that is of order n.
Also, I don't understand why the answer from MoB got up-votes, it does not answer the question.
Here is one way to do it (pseudocode)
input array[k][n]
init indices[k] = { 0, 0, 0, ... }
init queue = { empty priority queue }
for i in 0..k:
insert i into queue with priority (array[i][0])
while queue is not empty:
let x = pop queue
output array[x, indices[x]]
increment indices[x]
insert x into queue with priority (array[x][indices[x]])
This can probably be simplified further in C. You would have to find a suitable queue implementation to use though as there are none in libc.
Complexity for this operation:
"while queue is not empty" => O(n)
"insert x into queue ..." => O(log k)
=> O(n log k)
Which, if you consider k = constant, is O(n).
After sorting the k sub-arrays (the method doesn't matter), the code does a k-way merge. The simplest implementation does k-1 compares to determine the smallest leading element of each of the k arrays, then moves that element from it's sub-array to the output array and gets the next element from that array. When the end of an array is reached, the algorithm drops down to a (k-1) way merge, then (k-2) way merge, finally there's just one sub-array left and it's copied. This will be O(n) time since k-1 is a constant.
The k-1 compares can be sped up by using a minimum heap (which is how some priority queues are implemented), but it's still O(n), with just a smaller constant. The heap needs to be initialized at the start, then updated each time an element is removed and a new one added.
Related
I have been attempting to solve the following problem:
You are given an array of n+1 integers where all the elements lies in [1,n]. You are also given that one of the elements is duplicated a certain number of times, whilst the others are distinct. Develop an algorithm to find both the duplicated number and the number of times it is duplicated.
Here is my solution where I let k = number of duplications:
struct LatticePoint{ // to hold duplicate and k
int a;
int b;
LatticePoint(int a_, int b_) : a(a_), b(b_) {}
}
LatticePoint findDuplicateAndK(const std::vector<int>& A){
int n = A.size() - 1;
std::vector<int> Numbers (n);
for(int i = 0; i < n + 1; ++i){
++Numbers[A[i] - 1]; // A[i] in range [1,n] so no out-of-access
}
int i = 0;
while(i < n){
if(Numbers[i] > 1) {
int duplicate = i + 1;
int k = Numbers[i] - 1;
LatticePoint result{duplicate, k};
return LatticePoint;
}
So, the basic idea is this: we go along the array and each time we see the number A[i] we increment the value of Numbers[A[i]]. Since only the duplicate appears more than once, the index of the entry of Numbers with value greater than 1 must be the duplicate number with the value of the entry the number of duplications - 1. This algorithm of O(n) in time complexity and O(n) in space.
I was wondering if someone had a solution that is better in time and/or space? (or indeed if there are any errors in my solution...)
You can reduce the scratch space to n bits instead of n ints, provided you either have or are willing to write a bitset with run-time specified size (see boost::dynamic_bitset).
You don't need to collect duplicate counts until you know which element is duplicated, and then you only need to keep that count. So all you need to track is whether you have previously seen the value (hence, n bits). Once you find the duplicated value, set count to 2 and run through the rest of the vector, incrementing count each time you hit an instance of the value. (You initialise count to 2, since by the time you get there, you will have seen exactly two of them.)
That's still O(n) space, but the constant factor is a lot smaller.
The idea of your code works.
But, thanks to the n+1 elements, we can achieve other tradeoffs of time and space.
If we have some number of buckets we're dividing numbers between, putting n+1 numbers in means that some bucket has to wind up with more than expected. This is a variant on the well-known pigeonhole principle.
So we use 2 buckets, one for the range 1..floor(n/2) and one for floor(n/2)+1..n. After one pass through the array, we know which half the answer is in. We then divide that half into halves, make another pass, and so on. This leads to a binary search which will get the answer with O(1) data, and with ceil(log_2(n)) passes, each taking time O(n). Therefore we get the answer in time O(n log(n)).
Now we don't need to use 2 buckets. If we used 3, we'd take ceil(log_3(n)) passes. So as we increased the fixed number of buckets, we take more space and save time. Are there other tradeoffs?
Well you showed how to do it in 1 pass with n buckets. How many buckets do you need to do it in 2 passes? The answer turns out to be at least sqrt(n) bucekts. And 3 passes is possible with the cube root. And so on.
So you get a whole family of tradeoffs where the more buckets you have, the more space you need, but the fewer passes. And your solution is merely at the extreme end, taking the most spaces and the least time.
Here's a cheekier algorithm, which requires only constant space but rearranges the input vector. (It only reorders; all the original elements are still present at the end.)
It's still O(n) time, although that might not be completely obvious.
The idea is to try to rearrange the array so that A[i] is i, until we find the duplicate. The duplicate will show up when we try to put an element at the right index and it turns out that that index already holds that element. With that, we've found the duplicate; we have a value we want to move to A[j] but the same value is already at A[j]. We then scan through the rest of the array, incrementing the count every time we find another instance.
#include <utility>
#include <vector>
std::pair<int, int> count_dup(std::vector<int> A) {
/* Try to put each element in its "home" position (that is,
* where the value is the same as the index). Since the
* values start at 1, A[0] isn't home to anyone, so we start
* the loop at 1.
*/
int n = A.size();
for (int i = 1; i < n; ++i) {
while (A[i] != i) {
int j = A[i];
if (A[j] == j) {
/* j is the duplicate. Now we need to count them.
* We have one at i. There's one at j, too, but we only
* need to add it if we're not going to run into it in
* the scan. And there might be one at position 0. After that,
* we just scan through the rest of the array.
*/
int count = 1;
if (A[0] == j) ++count;
if (j < i) ++count;
for (++i; i < n; ++i) {
if (A[i] == j) ++count;
}
return std::make_pair(j, count);
}
/* This swap can only happen once per element. */
std::swap(A[i], A[j]);
}
}
/* If we get here, every element from 1 to n is at home.
* So the duplicate must be A[0], and the duplicate count
* must be 2.
*/
return std::make_pair(A[0], 2);
}
A parallel solution with O(1) complexity is possible.
Introduce an array of atomic booleans and two atomic integers called duplicate and count. First set count to 1. Then access the array in parallel at the index positions of the numbers and perform a test-and-set operation on the boolean. If a boolean is set already, assign the number to duplicate and increment count.
This solution may not always perform better than the suggested sequential alternatives. Certainly not if all numbers are duplicates. Still, it has constant complexity in theory. Or maybe linear complexity in the number of duplicates. I am not quite sure. However, it should perform well when using many cores and especially if the test-and-set and increment operations are lock-free.
Given an Array of integers, Find the smallest Lexical subsequence with size k.
EX: Array : [3,1,5,3,5,9,2] k =4
Expected Soultion : 1 3 5 2
The problem can be solved in O(n) by maintaining a double ended queue(deque). We iterate the element from left to right and ensure that the deque always holds the smallest lexicographic sequence upto that point. We should only pop off element if the current element is smaller than the elements in deque and the total elements in deque plus remaining to be processed are at least k.
vector<int> smallestLexo(vector<int> s, int k) {
deque<int> dq;
for(int i = 0; i < s.size(); i++) {
while(!dq.empty() && s[i] < dq.back() && (dq.size() + (s.size() - i - 1)) >= k) {
dq.pop_back();
}
dq.push_back(s[i]);
}
return vector<int> (dq.begin(), dq.end());
}
Here is a greedy algorithm that should work:
Choose Next Number ( lastChoosenIndex, k ) {
minNum = Find out what is the smallest number from lastChoosenIndex to ArraySize-k
//Now we know this number is the best possible candidate to be the next number.
lastChoosenIndex = earliest possible occurance of minNum after lastChoosenIndex
//do the same process for k-1
Choose Next Number ( lastChoosenIndex, k-1 )
}
Algorithm above is high complexity.
But we can pre-sort all the array elements paired with their array index and do the same process greedily using a single loop.
Since we used sorting complexity still will be n*log(n)
Ankit Joshi's answer works. But I think it can be done with just a vector itself, not using a deque as all the operations done are available in vector too. Also in Ankit Joshi's answer, the deque can contain extra elements, we have to manually pop off those elements before returning. Add these lines before returning.
while(dq.size() > k)
{
dq.pop_back();
}
It can be done with RMQ in O(n) + Klog(n).
Construct an RMQ in O(n).
Now find the sequence where every ith element will be the smallest no. from pos [x(i-1)+1 to n-(K-i)] (for i [1 to K] , where x0 = 0, xi is the position of the ith smallest element in the given array)
If I've understood the question right, here's a DP Algorithm that should work but it takes O(NK) time.
//k is the given size and n is the size of the array
create an array dp[k+1][n+1]
initialize the first column with the maximum integer value (we'll need it later)
and the first row with 0's (keep element dp[0][0] = 0)
now run the loop while building the solution
for(int i=1; i<=k; i++) {
for(int j=1; j<=n; j++) {
//if the number of elements in the array is less than the size required (K)
//initialize it with the maximum integer value
if( j < i ) {
dp[i][j] = MAX_INT_VALUE;
}else {
//last minimum of size k-1 with present element or last minimum of size k
dp[i][j] = minimun (dp[i-1][j-1] + arr[j-1], dp[i][j-1]);
}
}
}
//it consists the solution
return dp[k][n];
The last element of the array contains the solution.
I suggest you can try use modified merge sort. The place for
modified is merge part, discard the duplicate value.
select the smallest four
The complexity is o(n logn)
Still thinking whether complexity can be o(n)
So, you have n sorted arrays (not necessarily of equal length), and you are to return the kth smallest element in the combined array (i.e the combined array formed by merging all the n sorted arrays)
I have been trying it and its other variants for quite a while now, and till now I only feel comfortable in the case where there are two arrays of equal length, both sorted and one has to return the median of these two.
This has logarithmic time complexity.
After this I tried to generalize it to finding kth smallest among two sorted arrays. Here is the question on SO.
Even here the solution given is not obvious to me. But even if I somehow manage to convince myself of this solution, I am still curious as to how to solve the absolute general case (which is my question)
Can somebody explain me a step by step solution (which again in my opinion should take logarithmic time i.e O( log(n1) + log(n2) ... + log(nN) where n1, n2...nN are the lengths of the n arrays) which starts from the more specific cases and moves on to the more general one?
I know similar questions for more specific cases are there all over the internet, but I haven't found a convincing and clear answer.
Here is a link to a question (and its answer) on SO which deals with 5 sorted arrays and finding the median of the combined array. The answer just gets too complicated for me to able to generalize it.
Even clean approaches for the more specific cases (as I mentioned during the post) are welcome.
PS: Do you think this can be further generalized to the case of unsorted arrays?
PPS: It's not a homework problem, I am just preparing for interviews.
This doesn't generalize the links, but does solve the problem:
Go through all the arrays and if any have length > k, truncate to length k (this is silly, but we'll mess with k later, so do it anyway)
Identify the largest remaining array A. If more than one, pick one.
Pick the middle element M of the largest array A.
Use a binary search on the remaining arrays to find the same element (or the largest element <= M).
Based on the indexes of the various elements, calculate the total number of elements <= M and > M. This should give you two numbers: L, the number <= M and G, the number > M
If k < L, truncate all the arrays at the split points you've found and iterate on the smaller arrays (use the bottom halves).
If k > L, truncate all the arrays at the split points you've found and iterate on the smaller arrays (use the top halves, and search for element (k-L).
When you get to the point where you only have one element per array (or 0), make a new array of size n with those data, sort, and pick the kth element.
Because you're always guaranteed to remove at least half of one array, in N iterations, you'll get rid of half the elements. That means there are N log k iterations. Each iteration is of order N log k (due to the binary searches), so the whole thing is N^2 (log k)^2 That's all, of course, worst case, based on the assumption that you only get rid of half of the largest array, not of the other arrays. In practice, I imagine the typical performance would be quite a bit better than the worst case.
It can not be done in less than O(n) time. Proof Sketch If it did, it would have to completely not look at at least one array. Obviously, one array can arbitrarily change the value of the kth element.
I have a relatively simple O(n*log(n)*log(m)) where m is the length of the longest array. I'm sure it is possible to be slightly faster, but not a lot faster.
Consider the simple case where you have n arrays each of length 1. Obviously, this is isomorphic to finding the kth element in an unsorted list of length n. It is possible to find this in O(n), see Median of Medians algorithm, originally by Blum, Floyd, Pratt, Rivest and Tarjan, and no (asymptotically) faster algorithms are possible.
Now the problem is how to expand this to longer sorted arrays. Here is the algorithm: Find the median of each array. Sort the list of tuples (median,length of array/2) and sort it by median. Walk through keeping a sum of the lengths, until you reach a sum greater than k. You now have a pair of medians, such that you know the kth element is between them. Now for each median, we know if the kth is greater or less than it, so we can throw away half of each array. Repeat. Once the arrays are all one element long (or less), we use the selection algorithm.
Implementing this will reveal additional complexities and edge conditions, but nothing that increases the asymptotic complexity. Each step
Finds the medians or the arrays, O(1) each, so O(n) total
Sorts the medians O(n log n)
Walks through the sorted list O(n)
Slices the arrays O(1) each so, O(n) total
that is O(n) + O(n log n) + O(n) + O(n) = O(n log n). And, we must perform this untill the longest array is length 1, which will take log m steps for a total of O(n*log(n)*log(m))
You ask if this can be generalized to the case of unsorted arrays. Sadly, the answer is no. Consider the case where we only have one array, then the best algorithm will have to compare at least once with each element for a total of O(m). If there were a faster solution for n unsorted arrays, then we could implement selection by splitting our single array into n parts. Since we just proved selection is O(m), we are stuck.
You could look at my recent answer on the related question here. The same idea can be generalized to multiple arrays instead of 2. In each iteration you could reject the second half of the array with the largest middle element if k is less than sum of mid indexes of all arrays. Alternately, you could reject the first half of the array with the smallest middle element if k is greater than sum of mid indexes of all arrays, adjust k. Keep doing this until you have all but one array reduced to 0 in length. The answer is kth element of the last array which wasn't stripped to 0 elements.
Run-time analysis:
You get rid of half of one array in each iteration. But to determine which array is going to be reduced, you spend time linear to the number of arrays. Assume each array is of the same length, the run time is going to be cclog(n), where c is the number of arrays and n is the length of each array.
There exist an generalization that solves the problem in O(N log k) time, see the question here.
Old question, but none of the answers were good enough. So I am posting the solution using sliding window technique and heap:
class Node {
int elementIndex;
int arrayIndex;
public Node(int elementIndex, int arrayIndex) {
super();
this.elementIndex = elementIndex;
this.arrayIndex = arrayIndex;
}
}
public class KthSmallestInMSortedArrays {
public int findKthSmallest(List<Integer[]> lists, int k) {
int ans = 0;
PriorityQueue<Node> pq = new PriorityQueue<>((a, b) -> {
return lists.get(a.arrayIndex)[a.elementIndex] -
lists.get(b.arrayIndex)[b.elementIndex];
});
for (int i = 0; i < lists.size(); i++) {
Integer[] arr = lists.get(i);
if (arr != null) {
Node n = new Node(0, i);
pq.add(n);
}
}
int count = 0;
while (!pq.isEmpty()) {
Node curr = pq.poll();
ans = lists.get(curr.arrayIndex)[curr.elementIndex];
if (++count == k) {
break;
}
curr.elementIndex++;
pq.offer(curr);
}
return ans;
}
}
The maximum number of elements that we need to access here is O(K) and there are M arrays. So the effective time complexity will be O(K*log(M)).
This would be the code. O(k*log(m))
public int findKSmallest(int[][] A, int k) {
PriorityQueue<int[]> queue = new PriorityQueue<>(Comparator.comparingInt(x -> A[x[0]][x[1]]));
for (int i = 0; i < A.length; i++)
queue.offer(new int[] { i, 0 });
int ans = 0;
while (!queue.isEmpty() && --k >= 0) {
int[] el = queue.poll();
ans = A[el[0]][el[1]];
if (el[1] < A[el[0]].length - 1) {
el[1]++;
queue.offer(el);
}
}
return ans;
}
If the k is not that huge, we can maintain a priority min queue. then loop for every head of the sorted array to get the smallest element and en-queue. when the size of the queue is k. we get the first k smallest .
maybe we can regard the n sorted array as buckets then try the bucket sort method.
This could be considered the second half of a merge sort. We could simply merge all the sorted lists into a single list...but only keep k elements in the combined lists from merge to merge. This has the advantage of only using O(k) space, but something slightly better than merge sort's O(n log n) complexity. That is, it should in practice operate slightly faster than a merge sort. Choosing the kth smallest from the final combined list is O(1). This is kind of complexity is not so bad.
It can be done by doing binary search in each array, while calculating the number of smaller elements.
I used the bisect_left and bisect_right to make it work for non-unique numbers as well,
from bisect import bisect_left
from bisect import bisect_right
def kthOfPiles(givenPiles, k, count):
'''
Perform binary search for kth element in multiple sorted list
parameters
==========
givenPiles are list of sorted list
count is the total number of
k is the target index in range [0..count-1]
'''
begins = [0 for pile in givenPiles]
ends = [len(pile) for pile in givenPiles]
#print('finding k=', k, 'count=', count)
for pileidx,pivotpile in enumerate(givenPiles):
while begins[pileidx] < ends[pileidx]:
mid = (begins[pileidx]+ends[pileidx])>>1
midval = pivotpile[mid]
smaller_count = 0
smaller_right_count = 0
for pile in givenPiles:
smaller_count += bisect_left(pile,midval)
smaller_right_count += bisect_right(pile,midval)
#print('check midval', midval,smaller_count,k,smaller_right_count)
if smaller_count <= k and k < smaller_right_count:
return midval
elif smaller_count > k:
ends[pileidx] = mid
else:
begins[pileidx] = mid+1
return -1
Please find the below C# code to Find the k-th Smallest Element in the Union of Two Sorted Arrays. Time Complexity : O(logk)
public int findKthElement(int k, int[] array1, int start1, int end1, int[] array2, int start2, int end2)
{
// if (k>m+n) exception
if (k == 0)
{
return Math.Min(array1[start1], array2[start2]);
}
if (start1 == end1)
{
return array2[k];
}
if (start2 == end2)
{
return array1[k];
}
int mid = k / 2;
int sub1 = Math.Min(mid, end1 - start1);
int sub2 = Math.Min(mid, end2 - start2);
if (array1[start1 + sub1] < array2[start2 + sub2])
{
return findKthElement(k - mid, array1, start1 + sub1, end1, array2, start2, end2);
}
else
{
return findKthElement(k - mid, array1, start1, end1, array2, start2 + sub2, end2);
}
}
Based on a this logic given as an answer on SO to a different(similar) question, to remove repeated numbers in a array in O(N) time complexity, I implemented that logic in C, as shown below. But the result of my code does not return unique numbers. I tried debugging but could not get the logic behind it to fix this.
int remove_repeat(int *a, int n)
{
int i, k;
k = 0;
for (i = 1; i < n; i++)
{
if (a[k] != a[i])
{
a[k+1] = a[i];
k++;
}
}
return (k+1);
}
main()
{
int a[] = {1, 4, 1, 2, 3, 3, 3, 1, 5};
int n;
int i;
n = remove_repeat(a, 9);
for (i = 0; i < n; i++)
printf("a[%d] = %d\n", i, a[i]);
}
1] What is incorrect in above code to remove duplicates.
2] Any other O(N) or O(NlogN) solution for this problem. Its logic?
Heap sort in O(n log n) time.
Iterate through in O(n) time replacing repeating elements with a sentinel value (such as INT_MAX).
Heap sort again in O(n log n) to distil out the repeating elements.
Still bounded by O(n log n).
Your code only checks whether an item in the array is the same as its immediate predecessor.
If your array starts out sorted, that will work, because all instances of a particular number will be contiguous.
If your array isn't sorted to start with, that won't work because instances of a particular number may not be contiguous, so you have to look through all the preceding numbers to determine whether one has been seen yet.
To do the job in O(N log N) time, you can sort the array, then use the logic you already have to remove duplicates from the sorted array. Obviously enough, this is only useful if you're all right with rearranging the numbers.
If you want to retain the original order, you can use something like a hash table or bit set to track whether a number has been seen yet or not, and only copy each number to the output when/if it has not yet been seen. To do this, we change your current:
if (a[k] != a[i])
a[k+1] = a[i];
to something like:
if (!hash_find(hash_table, a[i])) {
hash_insert(hash_table, a[i]);
a[k+1] = a[i];
}
If your numbers all fall within fairly narrow bounds or you expect the values to be dense (i.e., most values are present) you might want to use a bit-set instead of a hash table. This would be just an array of bits, set to zero or one to indicate whether a particular number has been seen yet.
On the other hand, if you're more concerned with the upper bound on complexity than the average case, you could use a balanced tree-based collection instead of a hash table. This will typically use more memory and run more slowly, but its expected complexity and worst case complexity are essentially identical (O(N log N)). A typical hash table degenerates from constant complexity to linear complexity in the worst case, which will change your overall complexity from O(N) to O(N2).
Your code would appear to require that the input is sorted. With unsorted inputs as you are testing with, your code will not remove all duplicates (only adjacent ones).
You are able to get O(N) solution if the number of integers is known up front and smaller than the amount of memory you have :). Make one pass to determine the unique integers you have using auxillary storage, then another to output the unique values.
Code below is in Java, but hopefully you get the idea.
int[] removeRepeats(int[] a) {
// Assume these are the integers between 0 and 1000
Boolean[] v = new Boolean[1000]; // A lazy way of getting a tri-state var (false, true, null)
for (int i=0;i<a.length;++i) {
v[a[i]] = Boolean.TRUE;
}
// v[i] = null => number not seen
// v[i] = true => number seen
int[] out = new int[a.length];
int ptr = 0;
for (int i=0;i<a.length;++i) {
if (v[a[i]] != null && v[a[i]].equals(Boolean.TRUE)) {
out[ptr++] = a[i];
v[a[i]] = Boolean.FALSE;
}
}
// Out now doesn't contain duplicates, order is preserved and ptr represents how
// many elements are set.
return out;
}
You are going to need two loops, one to go through the source and one to check each item in the destination array.
You are not going to get O(N).
[EDIT]
The article you linked to suggests a sorted output array which means the search for duplicates in the output array can be a binary search...which is O(LogN).
Your logic just wrong, so the code is wrong too. Do your logic by yourself before coding it.
I suggest a O(NlnN) way with a modification of heapsort.
With heapsort, we join from a[i] to a[n], find the minimum and replace it with a[i], right?
So now is the modification, if the minimum is the same with a[i-1] then swap minimum and a[n], reduce your array item's number by 1.
It should do the trick in O(NlnN) way.
Your code will work only on particular cases. Clearly, you're checking adjacent values but duplicate values can occur any where in array. Hence, it's totally wrong.
I want to write a program to find the n-th smallest element without using any sorting technique..
Can we do it recursively, divide and conquer style like quick-sort?
If not, how?
You can find information about that problem here: Selection algorithm.
What you are referring to is the Selection Algorithm, as previously noted. Specifically, your reference to quicksort suggests you are thinking of the partition based selection.
Here's how it works:
Like in Quicksort, you start by picking a good
pivot: something that you think is nearly
half-way through your list. Then you
go through your entire list of items
swapping things back and forth until
all the items less than your pivot
are in the beginning of the list, and
all things greater than your pivot
are at the end. Your pivot goes into the leftover spot in the middle.
Normally in a quicksort you'd recurse
on both sides of the pivot, but for
the Selection Algorithm you'll only
recurse on the side that contains the
index you are interested in. So, if
you want to find the 3rd lowest
value, recurse on whichever side
contains index 2 (because index 0 is
the 1st lowest value).
You can stop recursing when you've
narrowed the region to just the one
index. At the end, you'll have one
unsorted list of the "m-1" smallest
objects, and another unsorted list of the "n-m" largest
objects. The "m"th object will be inbetween.
This algorithm is also good for finding a sorted list of the highest m elements... just select the m'th largest element, and sort the list above it. Or, for an algorithm that is a little bit faster, do the Quicksort algorithm, but decline to recurse into regions not overlapping the region for which you want to find the sorted values.
The really neat thing about this is that it normally runs in O(n) time. The first time through, it sees the entire list. On the first recursion, it sees about half, then one quarter, etc. So, it looks at about 2n elements, therefore it runs in O(n) time. Unfortunately, as in quicksort, if you consistently pick a bad pivot, you'll be running in O(n2) time.
This task is quite possible to complete within roughly O(n) time (n being the length of the list) by using a heap structure (specifically, a priority queue based on a Fibonacci heap), which gives O(1) insertion time and O(log n) removal time).
Consider the task of retrieving the m-th smallest element from the list. By simply looping over the list and adding each item to the priority queue (of size m), you can effectively create a queue of each of the items in the list in O(n) time (or possibly fewer using some optimisations, though I'm not sure this is exceedingly helpful). Then, it is a straightforward matter of removing the element with lowest priority in the queue (highest priority being the smallest item), which only takes O(log m) time in total, and you're finished.
So overall, the time complexity of the algorithm would be O(n + log n), but since log n << n (i.e. n grows a lot faster than log n), this reduces to simply O(n). I don't think you'll be able to get anything significantly more efficient than this in the general case.
You can use Binary heap, if u dont want to use fibonacci heap.
Algo:
Contruct the min binary heap from the array this operation will take O(n) time.
Since this is a min binary heap, the element at the root is the minimum value.
So keep on removing element frm root, till u get ur kth minimum value. o(1) operation
Make sure after every remove you re-store the heap kO(logn) operation.
So running time here is O(klogn) + O(n)............so it is O(klogn)...
Two stacks can be used like this to locate the Nth smallest number in one pass.
Start with empty Stack-A and Stack-B
PUSH the first number into Stack-A
The next number onwards, choose to PUSH into Stack-A only if the number is smaller than its top
When you have to PUSH into Stack-A, run through these steps
While TOP of Stack-A is larger than new number, POP TOP of Stack-A and push it into Stack-B
When Stack-A goes empty or its TOP is smaller than new number, PUSH in the new number and restore the contents of Stack-B over it
At this point you have inserted the new number to its correct (sorted) place in Stack-A and Stack-B is empty again
If Stack-A depth is now sufficient you have reached the end of your search
I generally agree to Noldorins' optimization analysis.
This stack solution is towards a simple scheme that will work (with relatively more data movement -- across the two stacks). The heap scheme reduces the fetch for Nth smallest number to a tree traversal (log m).
If your target is an optimal solution (say for a large set of numbers or maybe for a programming assignment, where optimization and the demonstration of it are critical) you should use the heap technique.
The stack solution can be compressed in space requirements by implementing the two stacks within the same space of K elements (where K is the size of your data set). So, the downside is just extra stack movement as you insert.
Here is the Ans to find Kth smallest element from an array:
#include<stdio.h>
#include<conio.h>
#include<iostream>
using namespace std;
int Nthmin=0,j=0,i;
int GetNthSmall(int numbers[],int NoOfElements,int Nthsmall);
int main()
{
int size;
cout<<"Enter Size of array\n";
cin>>size;
int *arr=(int*)malloc(sizeof(int)*size);
cout<<"\nEnter array elements\n";
for(i=0;i<size;i++)
cin>>*(arr+i);
cout<<"\n";
for(i=0;i<size;i++)
cout<<*(arr+i)<<" ";
cout<<"\n";
int n=sizeof(arr)/sizeof(int);
int result=GetNthSmall(arr,size,3);
printf("Result = %d",result);
getch();
return 0;
}
int GetNthSmall(int numbers[],int NoOfElements,int Nthsmall)
{
int min=numbers[0];
while(j<Nthsmall)
{
Nthmin=numbers[0];
for(i=1;i<NoOfElements;i++)
{
if(j==0)
{
if(numbers[i]<min)
{
min=numbers[i];
}
Nthmin=min;
}
else
{
if(numbers[i]<Nthmin && numbers[i]>min)
Nthmin=numbers[i];
}
}
min=Nthmin;
j++;
}
return Nthmin;
}
The simplest way to find the nth largest element in an array without using any sorting methods.
public static void kthLargestElement() {
int[] a = { 5, 4, 3, 2, 1, 9, 8 };
int n = 3;
int max = a[0], min = a[0];
for (int i = 0; i < a.length; i++) {
if (a[i] < min) {
min = a[i];
}
if (a[i] > max) {
max = a[i];
}
}
int max1 = max, c = 0;
for (int i = 0; i < a.length; i++) {
for (int j = 0; j < a.length; j++) {
if (a[j] > min && a[j] < max) {
max = a[j];
}
}
min = max;
max = max1;
c++;
if (c == (a.length - n)) {
System.out.println(min);
}
}
}