You are given an array of integers, arr
, of size n
, which is analogous to a continuous stream of integers input. Your task is to find k
largest elements from a given stream of numbers.
By definition, we don’t know the size of the input stream. Hence, produce k
largest elements seen so far, at any given time. For repeated numbers, return them only once.
If there are less than k
distinct elements in arr
, return all of them.
arr
.{
"arr": [1, 5, 4, 4, 2],
"k": 2
}
Output:
[4, 5]
{
"arr": [1, 5, 1, 5, 1],
"k": 3
}
Output:
[5, 1]
k
largest elements.Constraints:
n
<= 105k
<= 105arr
may contain duplicate numbersarr
may or may not be sortedWe have provided one solution for this problem.
We need to preserve the order of elements in a sorted manner. If we can do that, we can obtain top k
elements. Also, if an element is smaller than the last element in top k
, then that element can be dropped as we are not deleting elements.
We can maintain a balanced BST or a sorted set collection. Keep adding new elements to the sorted set and if the size of the tree increases more than k
, remove the smallest element.
O(n * log(k)).
O(k).
O(n + k).
/*
* Asymptotic complexity in terms of size of `arr` `n` and `k`:
* Time: O(n * log(k)).
* Auxiliary space: O(k).
* Total space: O(n + k).
*/
static ArrayList<Integer> top_k(ArrayList<Integer> arr, int k) {
// TreeSet will maintain set of elements in a sorted fashion
TreeSet<Integer> tree = new TreeSet<Integer>();
/*
We will add all elements to the sorted set and when size of the set increases over
required size k, we will remove the smallest element.
*/
for (int x : arr) {
tree.add(x);
if (tree.size() > k) {
tree.pollFirst();
}
}
ArrayList<Integer> ans= new ArrayList<Integer>();
for (int x : tree) {
ans.add(x);
}
return ans;
}
We hope that these solutions to top Kth problem have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.
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