Find largest rectangle containing only zeros in an N×N binary matrix

Question

Given an NxN binary matrix (containing only 0\'s or 1\'s), how can we go about finding largest rectangle containing all 0\'s?

Example:

      I
    0          


        

Answer 1

To be complete, here's the C# version which outputs the rectangle coordinates. It's based on dmarra's answer but without any other dependencies. There's only the function bool GetPixel(int x, int y), which returns true when a pixel is set at the coordinates x,y.

    public struct INTRECT
    {
        public int Left, Right, Top, Bottom;

        public INTRECT(int aLeft, int aTop, int aRight, int aBottom)
        {
            Left = aLeft;
            Top = aTop;
            Right = aRight;
            Bottom = aBottom;
        }

        public int Width { get { return (Right - Left + 1); } }

        public int Height { get { return (Bottom - Top + 1); } }

        public bool IsEmpty { get { return Left == 0 && Right == 0 && Top == 0 && Bottom == 0; } }

        public static bool operator ==(INTRECT lhs, INTRECT rhs)
        {
            return lhs.Left == rhs.Left && lhs.Top == rhs.Top && lhs.Right == rhs.Right && lhs.Bottom == rhs.Bottom;
        }

        public static bool operator !=(INTRECT lhs, INTRECT rhs)
        {
            return !(lhs == rhs);
        }

        public override bool Equals(Object obj)
        {
            return obj is INTRECT && this == (INTRECT)obj;
        }

        public bool Equals(INTRECT obj)
        {
            return this == obj;
        }

        public override int GetHashCode()
        {
            return Left.GetHashCode() ^ Right.GetHashCode() ^ Top.GetHashCode() ^ Bottom.GetHashCode();
        }
    }

    public INTRECT GetMaximumFreeRectangle()
    {
        int XEnd = 0;
        int YStart = 0;
        int MaxRectTop = 0;
        INTRECT MaxRect = new INTRECT();
        // STEP 1:
        // build a seed histogram using the first row of grid points
        // example: [true, true, false, true] = [1,1,0,1]
        int[] hist = new int[Height];
        for (int y = 0; y < Height; y++)
        {
            if (!GetPixel(0, y))
            {
                hist[y] = 1;
            }
        }

        // STEP 2:
        // get a starting max area from the seed histogram we created above.
        // using the example from above, this value would be [1, 1], as the only valid area is a single point.
        // another example for [0,0,0,1,0,0] would be [1, 3], because the largest area of contiguous free space is 3.
        // Note that at this step, the heigh fo the found rectangle will always be 1 because we are operating on
        // a single row of data.
        Tuple<int, int> maxSize = MaxRectSize(hist, out YStart);
        int maxArea = (int)(maxSize.Item1 * maxSize.Item2);
        MaxRectTop = YStart;
        // STEP 3:
        // build histograms for each additional row, re-testing for new possible max rectangluar areas
        for (int x = 1; x < Width; x++)
        {
            // build a new histogram for this row. the values of this row are
            // 0 if the current grid point is occupied; otherwise, it is 1 + the value
            // of the previously found historgram value for the previous position. 
            // What this does is effectly keep track of the height of continous avilable spaces.
            // EXAMPLE:
            //      Given the following grid data (where 1 means occupied, and 0 means free; for clairty):
            //          INPUT:        OUTPUT:
            //      1.) [0,0,1,0]   = [1,1,0,1]
            //      2.) [0,0,1,0]   = [2,2,0,2]
            //      3.) [1,1,0,1]   = [0,0,1,0]
            //
            //  As such, you'll notice position 1,0 (row 1, column 0) is 2, because this is the height of contiguous
            //  free space.
            for (int y = 0; y < Height; y++)
            {
                if (!GetPixel(x, y))
                {
                    hist[y]++;
                }
                else
                {
                    hist[y] = 0;
                }
            }

            // find the maximum size of the current histogram. If it happens to be larger
            // that the currently recorded max size, then it is the new max size.
            Tuple<int, int> maxSizeTemp = MaxRectSize(hist, out YStart);
            int tempArea = (int)(maxSizeTemp.Item1 * maxSizeTemp.Item2);
            if (tempArea > maxArea)
            {
                maxSize = maxSizeTemp;
                maxArea = tempArea;
                MaxRectTop = YStart;
                XEnd = x;
            }
        }
        MaxRect.Left = XEnd - maxSize.Item1 + 1;
        MaxRect.Top = MaxRectTop;
        MaxRect.Right = XEnd;
        MaxRect.Bottom = MaxRectTop + maxSize.Item2 - 1;

        // at this point, we know the max size
        return MaxRect;
    }

    private Tuple<int, int> MaxRectSize(int[] histogram, out int YStart)
    {
        Tuple<int, int> maxSize = new Tuple<int, int>(0, 0);
        int maxArea = 0;
        Stack<Tuple<int, int>> stack = new Stack<Tuple<int, int>>();
        int x = 0;
        YStart = 0;
        for (x = 0; x < histogram.Length; x++)
        {
            int start = x;
            int height = histogram[x];
            while (true)
            {
                if (stack.Count == 0 || height > stack.Peek().Item2)
                {
                    stack.Push(new Tuple<int, int>(start, height));
                }
                else if (height < stack.Peek().Item2)
                {
                    int tempArea = (int)(stack.Peek().Item2 * (x - stack.Peek().Item1));
                    if (tempArea > maxArea)
                    {
                        YStart = stack.Peek().Item1;
                        maxSize = new Tuple<int, int>(stack.Peek().Item2, (x - stack.Peek().Item1));
                        maxArea = tempArea;
                    }
                    Tuple<int, int> popped = stack.Pop();
                    start = (int)popped.Item1;
                    continue;
                }
                break;
            }
        }

        foreach (Tuple<int, int> data in stack)
        {
            int tempArea = (int)(data.Item2 * (x - data.Item1));
            if (tempArea > maxArea)
            {
                YStart = data.Item1;
                maxSize = new Tuple<int, int>(data.Item2, (x - data.Item1));
                maxArea = tempArea;
            }
        }

        return maxSize;
    }

Answer 2

Solution with space complexity O(columns) [Can be modified to O(rows) also] and time complexity O(rows*columns)

public int maximalRectangle(char[][] matrix) {
    int m = matrix.length;
    if (m == 0)
        return 0;
    int n = matrix[0].length;
    int maxArea = 0;
    int[] aux = new int[n];
    for (int i = 0; i < n; i++) {
        aux[i] = 0;
    }
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            aux[j] = matrix[i][j] - '0' + aux[j];
            maxArea = Math.max(maxArea, maxAreaHist(aux));
        }
    }
    return maxArea;
}

public int maxAreaHist(int[] heights) {
    int n = heights.length;
    Stack<Integer> stack = new Stack<Integer>();
    stack.push(0);
    int maxRect = heights[0];
    int top = 0;
    int leftSideArea = 0;
    int rightSideArea = heights[0];
    for (int i = 1; i < n; i++) {
        if (stack.isEmpty() || heights[i] >= heights[stack.peek()]) {
            stack.push(i);
        } else {
            while (!stack.isEmpty() && heights[stack.peek()] > heights[i]) {
                top = stack.pop();
                rightSideArea = heights[top] * (i - top);
                leftSideArea = 0;
                if (!stack.isEmpty()) {
                    leftSideArea = heights[top] * (top - stack.peek() - 1);
                } else {
                    leftSideArea = heights[top] * top;
                }
                maxRect = Math.max(maxRect, leftSideArea + rightSideArea);
            }
            stack.push(i);
        }
    }
    while (!stack.isEmpty()) {
        top = stack.pop();
        rightSideArea = heights[top] * (n - top);
        leftSideArea = 0;
        if (!stack.isEmpty()) {
            leftSideArea = heights[top] * (top - stack.peek() - 1);
        } else {
            leftSideArea = heights[top] * top;
        }
        maxRect = Math.max(maxRect, leftSideArea + rightSideArea);
    }
    return maxRect;
}

But I get Time Limite exceeded excpetion when I try this on LeetCode. Is there any less complex solution?