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### How to Nail your next Technical Interview

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## You may be missing out on a 66.5% salary hike*

### Nick Camilleri

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## FREE course on 'Sorting Algorithms' by Omkar Deshpande (Stanford PhD, Head of Curriculum, IK)

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# Kth Largest In A Stream Problem

Given an initial list along with another list of numbers to be appended with the initial list and an integer K, return an array consisting of the Kth largest element after adding each element from the first list to the second list.

Example

Input: [ 2, [4, 6], [5, 2, 20] ]

Output: [5, 5, 6]

Notes

Constraints:

• 1 <= length of both lists <= 10^5.
• 1 <= K <= length of initial list + 1.
• 0 <= any value in the list <= 10^9.
• The stream can contain duplicates.

#### Solutions

We have provided two solutions.

We will start with a brute-force approach that solves the problem in polynomial time complexity, later we present an optimized linear time complexity solution. We will refer to the initial list as intial_stream and the second list as append_stream.

### 1) brute_force_solution.cpp

The idea is to follow the exact same steps described in the problem statement.

for i = 1 to size_of_append_stream:

-  Sort initial_stream in non-decreasing order.

-  The Kth element from the end will be the Kth largest element.

Time Complexity:

O(|append_stream| * (|append_stream| + |initial_stream|) * log(|append_stream| + |initial_stream|)).

To sort the initial stream = O(|initial_stream| * log(|initial_stream|)).

To add an element from append_stream to the sorted stream = O((|append_stream| + |initial_stream|) * log(|append_stream| + |initial_stream|)).

We add |append_stream| number of elements to the sorted stream.

Auxiliary Space Used:

O(|append_stream|).

Memory used to add elements from append_stream to initial_stream = O(|append_stream|).

Space Complexity:

O(|initial_stream| + |append_stream|).

Memory used for input = O(|initial_stream| + |append_stream|).

Memory used for output = O(|append_stream|).

Auxiliary space = O(|append_stream|).

Total space complexity = O(|initial_stream| + |append_stream|).

``````
// -------- START --------

vector kth_largest(int k, vector initial_stream, vector append_stream)
{
vector result;

for(int i = 0; i < append_stream.size(); i++) {
initial_stream.push_back(append_stream[i]);

sort(initial_stream.begin(), initial_stream.end());

// kth element from the end will be the kth largest element.
result.push_back(initial_stream[initial_stream.size() - k]);
}

return result;
}

// -------- END --------
``````

### 2) heap_solution.cpp

The idea is to keep track of only the K largest elements of the stream. Create a min-heap storing the K largest elements from initial_stream. To create such a heap, we can sort the inital_stream and push K largest elements in the heap but this will require O(|inital_stream| * log(|inital_stream|)). Instead, we will directly push elements from inital_stream to min-heap to achieve a complexity of O(|inital_stream| * log(K)). While pushing the elements we will keep a check such that the size of min-heap does not exceed K which can be achieved by popping the top element of the heap when its size becomes (K + 1).

#### How to process a new element of the append_stream?

- If the new element is smaller than the top element of the heap, ignore it

- else remove the topmost element of the heap and insert the new element in the heap. To remove or insert a new element, the time complexity is O(log(K)).

The top element of the heap is always the Kth largest element of the current stream.

Note: It is possible to get a solution using a max-heap also instead of a min-heap, but would require some extra handling i.e. maintaining a max-heap of size (number_of_elements_at_time - K + 1) to get the Kth largest element.

Time Complexity:

O( log(K) * ( |initial_stream| + |append_stream| )).

To maintain a heap of size K = O(log(K)).

We push and pop every element of the initial and append stream at most once.

Auxiliary Space Used:

O(K).

Memory used to maintain a heap of size K = O(K).

Space Complexity:

O(|initial_stream| + |append_stream| + K).

Memory used for input = O(|initial_stream| + |append_stream|).

Memory used for output = O(|append_stream|).

Auxiliary space = O(K).

Total space complexity = O(|initial_stream| + |append_stream| + K).

``````
// -------- START --------

vector kth_largest(int k, vector initial_stream, vector append_stream)
{
// Priority queue's internal implementation is the same as a binary heap.
priority_queue, greater> min_heap;

for(int i = 0; i < initial_stream.size(); i++) {
min_heap.push(initial_stream[i]);

// Make sure that the heap size does not exceed K.
if(min_heap.size() > k)
min_heap.pop(); // Remove the element smaller than the K largest elements.
}

vector result;

for(int i = 0; i < append_stream.size(); i++) {
min_heap.push(append_stream[i]);

// Make sure that the heap size does not exceed K.
if(min_heap.size() > k)
min_heap.pop(); // Remove the element smaller than the K largest elements.

// Adding current Kth largest element.
result.push_back(min_heap.top());
}

return result;
}

// -------- END --------
``````

### Try yourself in the Editor

Note: Input and Output will already be taken care of.

# Kth Largest In A Stream Problem

Given an initial list along with another list of numbers to be appended with the initial list and an integer K, return an array consisting of the Kth largest element after adding each element from the first list to the second list.

Example

Input: [ 2, [4, 6], [5, 2, 20] ]

Output: [5, 5, 6]

Notes

Constraints:

• 1 <= length of both lists <= 10^5.
• 1 <= K <= length of initial list + 1.
• 0 <= any value in the list <= 10^9.
• The stream can contain duplicates.

#### Solutions

We have provided two solutions.

We will start with a brute-force approach that solves the problem in polynomial time complexity, later we present an optimized linear time complexity solution. We will refer to the initial list as intial_stream and the second list as append_stream.

### 1) brute_force_solution.cpp

The idea is to follow the exact same steps described in the problem statement.

for i = 1 to size_of_append_stream:

-  Sort initial_stream in non-decreasing order.

-  The Kth element from the end will be the Kth largest element.

Time Complexity:

O(|append_stream| * (|append_stream| + |initial_stream|) * log(|append_stream| + |initial_stream|)).

To sort the initial stream = O(|initial_stream| * log(|initial_stream|)).

To add an element from append_stream to the sorted stream = O((|append_stream| + |initial_stream|) * log(|append_stream| + |initial_stream|)).

We add |append_stream| number of elements to the sorted stream.

Auxiliary Space Used:

O(|append_stream|).

Memory used to add elements from append_stream to initial_stream = O(|append_stream|).

Space Complexity:

O(|initial_stream| + |append_stream|).

Memory used for input = O(|initial_stream| + |append_stream|).

Memory used for output = O(|append_stream|).

Auxiliary space = O(|append_stream|).

Total space complexity = O(|initial_stream| + |append_stream|).

``````
// -------- START --------

vector kth_largest(int k, vector initial_stream, vector append_stream)
{
vector result;

for(int i = 0; i < append_stream.size(); i++) {
initial_stream.push_back(append_stream[i]);

sort(initial_stream.begin(), initial_stream.end());

// kth element from the end will be the kth largest element.
result.push_back(initial_stream[initial_stream.size() - k]);
}

return result;
}

// -------- END --------
``````

### 2) heap_solution.cpp

The idea is to keep track of only the K largest elements of the stream. Create a min-heap storing the K largest elements from initial_stream. To create such a heap, we can sort the inital_stream and push K largest elements in the heap but this will require O(|inital_stream| * log(|inital_stream|)). Instead, we will directly push elements from inital_stream to min-heap to achieve a complexity of O(|inital_stream| * log(K)). While pushing the elements we will keep a check such that the size of min-heap does not exceed K which can be achieved by popping the top element of the heap when its size becomes (K + 1).

#### How to process a new element of the append_stream?

- If the new element is smaller than the top element of the heap, ignore it

- else remove the topmost element of the heap and insert the new element in the heap. To remove or insert a new element, the time complexity is O(log(K)).

The top element of the heap is always the Kth largest element of the current stream.

Note: It is possible to get a solution using a max-heap also instead of a min-heap, but would require some extra handling i.e. maintaining a max-heap of size (number_of_elements_at_time - K + 1) to get the Kth largest element.

Time Complexity:

O( log(K) * ( |initial_stream| + |append_stream| )).

To maintain a heap of size K = O(log(K)).

We push and pop every element of the initial and append stream at most once.

Auxiliary Space Used:

O(K).

Memory used to maintain a heap of size K = O(K).

Space Complexity:

O(|initial_stream| + |append_stream| + K).

Memory used for input = O(|initial_stream| + |append_stream|).

Memory used for output = O(|append_stream|).

Auxiliary space = O(K).

Total space complexity = O(|initial_stream| + |append_stream| + K).

``````
// -------- START --------

vector kth_largest(int k, vector initial_stream, vector append_stream)
{
// Priority queue's internal implementation is the same as a binary heap.
priority_queue, greater> min_heap;

for(int i = 0; i < initial_stream.size(); i++) {
min_heap.push(initial_stream[i]);

// Make sure that the heap size does not exceed K.
if(min_heap.size() > k)
min_heap.pop(); // Remove the element smaller than the K largest elements.
}

vector result;

for(int i = 0; i < append_stream.size(); i++) {
min_heap.push(append_stream[i]);

// Make sure that the heap size does not exceed K.
if(min_heap.size() > k)
min_heap.pop(); // Remove the element smaller than the K largest elements.

// Adding current Kth largest element.
result.push_back(min_heap.top());
}

return result;
}

// -------- END --------
``````

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