No. 27 - Area of Rectangles

Problem: Please implement a function which gets area of a set of rectangles, whose edges are parallel to X-axis or Y-axis.

Rectangles are defined as the following:

/* === Assumming: left <= right; top <= bottom === */

struct Rect

{

    int left;

    int right;

    int top;

    int bottom;

};

Analysis: The merged shape of overlapping rectangles is no longer a rectangle, and there is no equation to calculate the merged area directly. Therefore, we should split it into a set of rectangles.

A solution to split overlapping rectangles is based on x-values of all left and right edges. For example, there are three rectangles in Figure 1.

Figure 1: Three rectangles

We get all x-values of the three rectangles, which are X1, X2, …, X6 in Figure 2, and then split the merged shape into five rectangles based on x-values.

We should merge y-values of rectangles in ranges of x-value. For instance, both Rect1 and Rect2 fill into the x-value range between X2 and X3. Therefore, we should merge the y-value range of Rect1 and Rect2 to calculate the rectangle piece between X2 and X3.

Figure 2: Split rectangles into pieces based on x-values

Since ranges of x-values and y-values are necessary information to calculate area, we can define a class Range:

struct Range

{

    int less;

    int greater;

    Range(int l, int g)

    {

        less = (l < g) ? l : g;

        greater = (l + g) - less;

    }

    bool IsOverlapping(const Range& other)

    {

        return !(less > other.greater || other.less > greater);

    }

    void Merge(const Range& other)

    {

        if(IsOverlapping(other))

        {

            less = (less < other.less) ? less : other.less;

            greater = (greater > other.greater) ? greater : other.greater;

        }

    }

};

The overall algorithm to calculate the area of overlapping rectangles is: We firstly get all x-values of all rectangles, and then check whether every rectangle fills into each range formed by two neighboring x-values. If there are multiple rectangles fill into an x-value range, we merge their ranges of y-value before we calculate the area of a piece of merged rectangle.

The following function GetArea implements this algorithm. For optimization purpose, we sort all input rectangles according to their x-values of right edges, and sort a vector of all x-values.

int GetArea(vector<Rect>& rects)

{

    // sort rectangles according to x-value of right edges

    sort(rects.begin(), rects.end());

    vector<int> xes;

    GetAllX(rects, xes);

    sort(xes.begin(), xes.end());

    int area = 0;

    vector<int>::iterator iterX1 = xes.begin();

    vector<Rect>::const_iterator iterRect = rects.begin();

    for(; iterX1 != xes.end() && iterX1 != xes.end() - 1; ++ iterX1)

    {

        vector<int>::iterator iterX2 = iterX1 + 1;

        // filter out duplicated X-es

        if(*iterX1 < *iterX2)

        {

            Range rangeX(*iterX1, *iterX2);

            while(iterRect->right < *iterX1)

                ++ iterRect;

            list<Range> rangesOfY;

            GetRangesOfY(rects, iterRect, rangeX, rangesOfY);

            area += GetRectArea(rangeX, rangesOfY);

        }

    }

    return area;

}

bool operator < (const Rect& rect1, const Rect& rect2)

{

    return (rect1.right < rect2.right);

}

The following function GetAllX gets all x-values of all input rectangles.

void GetAllX(const vector<Rect>& rects, vector<int>& xes)

{

    vector<Rect>::const_iterator iter = rects.begin();

    for(; iter != rects.end(); ++ iter)

    {

        xes.push_back(iter->left);

        xes.push_back(iter->right);

    }

}

The function GetRangesOfY gets the ranges of y-values which fill into a range of x-value. It should be noticed that there might be multiple non-overlapping rectangles fill into an x-value range. For example, Rect1 and Rect2 in Figure do not overlap with each other. Therefore, we cannot merge their ranges of y-values. That is why it has a list of ranges of y-values for a range of x-value in the function GetRangesOfY.

Figure 3: Another example of three overlaping rectangles

void GetRangesOfY(const vector<Rect>& rects, vector<Rect>::const_iterator iterRect, const Range& rangeX, list<Range>& rangesOfY)

{

    for(; iterRect != rects.end(); ++ iterRect)

    {

        if(rangeX.less < iterRect->right && rangeX.greater > iterRect->left)

            InsertRangeY(rangesOfY, Range(iterRect->top, iterRect->bottom));

    }

}

The function InsertRangeY inserts a range of y-value into a list. When the range overlaps with another range existing in the list, we should remove the existing range and then insert a new merged range.

void InsertRangeY(list<Range>& rangesOfY, Range& rangeY)

{

    list<Range>::iterator iter = rangesOfY.begin();

    while(iter != rangesOfY.end())

    {

        if(rangeY.IsOverlapping(*iter))

        {

            rangeY.Merge(*iter);

            list<Range>::iterator iterCopy = iter;

            ++ iter;

            rangesOfY.erase(iterCopy);

        }

        else

            ++ iter;

    }

    rangesOfY.push_back(rangeY);

}

The function GetRectArea below gets the total area of all rectangle pieces in a range of x-value.

int GetRectArea(const Range& rangeX, const list<Range>& rangesOfY)

{

    int width = rangeX.greater - rangeX.less;

   

    list<Range>::const_iterator iter = rangesOfY.begin();

    int area = 0;

    for(; iter != rangesOfY.end(); ++ iter)

    {

        int height = iter->greater - iter->less;

        area += width * height;

    }

    return area;

}

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