No. 33 - Maximums in Sliding Windows

No. 33 - Maximums in Sliding Windows

Question: Given an array of numbers and a sliding window size, how to get the maximal numbers in all sliding windows?

For example, if the input array is {2, 3, 4, 2, 6, 2, 5, 1} and the size of sliding windows is 3, the output of maximums are {4, 4, 6, 6, 6, 5}, as illustrated in Table1.

Sliding Windows in an Array

Maximums in Sliding Windows

[2, 3, 4], 2, 6, 2, 5, 1

4

2, [3, 4, 2], 6, 2, 5, 1

4

2, 3, [4, 2, 6], 2, 5, 1

6

2, 3, 4, [2, 6, 2], 5, 1

6

2, 3, 4, 2, [6, 2, 5], 1

6

2, 3, 4, 2, 6, [2, 5, 1]

5

Table 1: Maximums of all sliding windows with size 3 in an array {2, 3, 4, 2, 6, 2, 5, 1}. A pair of brackets indicates a sliding window.

Analysis: It is not difficult to get a solution with brute force: Scan numbers in every sliding window to get its maximal value. The overall time complexity is O(nk) if the length of array is n and the size of sliding windows is k.

The naïve solution is not the best solution. Let us explore better solutions.

Solution 1: Maximal value in a queue

A window can be viewed as a queue. When it slides, a number is pushed into its back, and its front is popped off. Therefore, the problem is solved if we can get the maximal value of a queue.

There are no straightforward approaches to getting the maximal value of a queue. However, there are solutions to get the maximal value of a stack, which is similar to the solution introduced in the blog “Stack with Function min()”. Additionally, a queue can also be implemented with two stacks (details are discussed in another blog “Queue implemented with Two Stacks”). 

If a new type of queue is implemented with two stacks, in which a function max() is defined to get the maximal value, the maximal value in a queue is the greater number of the two maximal numbers in two stacks.

This solution is workable. However, we may not have enough time to write all code to implement our own queue and stack data structures during interviews. Let us continue exploring a more concise solution.

Solution 2: Saving the maximal value into the front of a queue

Instead of pushing every numbers inside a sliding window into a queue, we try to push the candidates of maximum only into a queue. Let us take the array {2, 3, 4, 2, 6, 2, 5, 1} as an example to analyze the solution step by step.

The first number in the array is 2, we push it into a queue. The second number is 3, which is greater than the previous number 2. The number 2 should be popped off, because it is less than 3 and it has no chance to be the maximal value. There is only one number left in the queue when we pop 2 at the back and push 3 at the back. The operations are similar when we push the next number 4. There is only a number 4 remaining in the queue. Now the sliding window already has three elements, we can get the maximum value at the front of the queue.

We continue to push the fourth number.  It is pushed at the back of queue, because it is less than the previous number 4 and it might be a maximal number in the future when the previous numbers are popped off. There are two numbers, 4 and 2, in the queue, and 4 is the maximum.

The next number to be pushed is 6. Since it is greater than the existing numbers, 4 and 2, these two numbers can be popped off because they have no chance to be the maximum. Now there is only one number in the queue, which is 6, after the current number is pushed. Of course, the maximum is 6.

The next number is 2, which is pushed into the back of the queue because it is less than the previous number 6. There are two numbers in the queue, 6 and 2, and the number 6 at the front of the queue is the maximal value.

It is time to push the number 5. Because it is greater than the number 2 at the back of the queue, 2 is popped off and then 5 is pushed. There are two numbers in the queue, 6 and 5, and the number 6 at the front of the queue is the maximal value.

Now let us push the last number 1. It can be pushed into the queue. It is noticeable that the number at the front is beyond the scope the current sliding window, and it should be popped off.  How do we know whether the number at the front of the queue is out of sliding window? Rather than storing numbers in the queue directly, we can store indices instead. If the distance between the index at the front of queue and the index of the current number to be pushed is greater than or equal to the window size, the number corresponding to be the index at the font of queue is out of sliding window.

The analysis process above is summarized in Table 2.

Step

Number to Be Pushed

Numbers in Sliding Window

Indices in queue

Maximum in Window

1

2

2

0(2)

2

3

2, 3

1(3)

3

4

2, 3, 4

2(4)

4

4

2

3, 4, 2

2(4), 3(2)

4

5

6

4, 2, 6

4(6)

6

6

2

2, 6, 2

4(6), 5(2)

6

7

5

6, 2, 5

4(6), 6(5)

6

8

1

2, 5, 1

6(5), 7(1)

5

Table 2: The process to get the maximal number in all sliding windows with window size 3 in the array {2, 3, 4, 2, 6, 2, 5, 1}. In the column “Indices in queue”, the number inside a pair of parentheses is the number indexed by the number before it in the array.

We can implement a solution based on the analysis above. Some sample code in C++ is shown below, which utilizes the type deque of STL.

vector<int> maxInWindows(const vector<int>& numbers, int windowSize)

{

    vector<int> maxInSlidingWindows;

    if(numbers.size() >= windowSize && windowSize > 1)

    {

        deque<int> indices;

        for(int i = 0; i < windowSize; ++i)

        {

            while(!indices.empty() && numbers[i] >= numbers[indices.back()])

                indices.pop_back();

            indices.push_back(i);

        }

        for(int i = windowSize; i < numbers.size(); ++i)

        {

            maxInSlidingWindows.push_back(numbers[indices.front()]);

            while(!indices.empty() && numbers[i] >= numbers[indices.back()])

                indices.pop_back();

            if(!indices.empty() && indices.front() <= i - windowSize)

                indices.pop_front();

            indices.push_back(i);

        }

        maxInSlidingWindows.push_back(numbers[indices.front()]);

    }

    return maxInSlidingWindows;

}

Extension: Another solution to get the maximum of a queue

As we mentioned before, a sliding window can be viewed as a queue. Therefore, we can implement a new solution to get the maximal value of a queue based on the second solution to get the maximums of sliding windows.

The following is the sample code:

template<typename T> class QueueWithMax

{

public:

    QueueWithMax(): currentIndex(0)

    {

    }

    void push_back(T number)

    {

        while(!maximums.empty() && number >= maximums.back().number)

            maximums.pop_back();

        InternalData internalData = {number, currentIndex};

        data.push_back(internalData);

        maximums.push_back(internalData);

        ++currentIndex;

    }

    void pop_front()

    {

        if(maximums.empty())

            throw new exception("queue is empty");

        if(maximums.front().index == data.front().index)

            maximums.pop_front();

        data.pop_front();

    }

    T max() const

    {

        if(maximums.empty())

            throw new exception("queue is empty");

        return maximums.front().number;

    }

private:

    struct InternalData

    {

        T number;

        int index;

    };

    deque<InternalData> data;

    deque<InternalData> maximums;

    int currentIndex;

};

Since this solution is similar to the second solution to get maximums of sliding windows, we won’t analyze the process step by step, and leave it as an exercise if you are interested.

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