Maximal Rectangle

Stack, Dynamic Programming, Array, Hash Table

Hard

Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing only 1's and return its area.

Example:

Input:
[
  ["1","0","1","0","0"],
  ["1","0","1","1","1"],
  ["1","1","1","1","1"],
  ["1","0","0","1","0"]
]
Output: 6

Solution & Analysis

Via @wangyushawn's, @calis's answer

This question is similar as [Largest Rectangle in Histogram]:

You can maintain a row length of Integer array H recorded its height of '1's, and scan and update row by row to find out the largest rectangle of each row.

For each row, if matrix[row][i] == '1'. H[i] +=1, or reset the H[i] to zero.
and accroding the algorithm of [Largest Rectangle in Histogram], to update the maximum area.

class Solution {
    public int maximalRectangle(char[][] matrix) {
       int rLen = matrix.length, cLen = rLen == 0 ? 0 : matrix[0].length, max = 0;
        int[] h = new int[cLen+1];

        for (int row = 0; row < rLen; row++) {
            Stack<Integer> s = new Stack<Integer>();
            s.push(-1);
            for (int i = 0; i <= cLen ;i++) {
                if(i < cLen && matrix[row][i] == '1')
                    h[i] += 1;
                else h[i] = 0;

                while(s.peek() != -1 && h[i] < h[s.peek()]) {
                    int height = h[s.pop()];
                    int width = i - s.peek() - 1;
                    max = Math.max(max, height * width);
                }
                s.push(i);
            }
        }
        return max;
    }
}

See also: Solution based on Maximum Rectangle in Histogram

Two Pass

by @agritsik

class Solution {

    public int maximalRectangle(char[][] matrix) {
        if(matrix.length==0) return 0;

        int[][] dp = new int[matrix.length][matrix[0].length];
        for (int i = 0; i < dp.length; i++) {
            for (int j = 0; j < dp[0].length; j++) {
                dp[i][j] = matrix[i][j]-'0';
                if (dp[i][j] > 0 && i>0) dp[i][j] += dp[i - 1][j];
            }
        }

        int max = 0;
        for (int[] a : dp) max=Math.max(largestRectangleArea(a), max);

        return max;
    }

    // copied "Largest Rectangle in Histogram" solution
    public int largestRectangleArea(int[] a) {
        LinkedList<Integer> stack = new LinkedList<>();
        int max = 0;

        for (int i = 0; i <= a.length; i++) {
            while (!stack.isEmpty() && (i == a.length || a[stack.peek()] > a[i])) {
                int height = a[stack.pop()];
                int width = (!stack.isEmpty()) ? i - stack.peek() - 1 : i;
                max = Math.max(max, height * width);
            }

            stack.push(i);

        }

        return max;
    }
}

Dynamic Programming

Via @morrischen2008's answer:

The DP solution proceeds row by row, starting from the first row. Let the maximal rectangle area at row i and column j be computed by [right(i,j) - left(i,j)]*height(i,j).

All the 3 variables left, right, and height can be determined by the information from previous row, and also information from the current row. So it can be regarded as a DP solution. The transition equations are:

left(i,j) = max(left(i-1,j), cur_left), cur_left can be determined from the current row

right(i,j) = min(right(i-1,j), cur_right), cur_right can be determined from the current row

height(i,j) = height(i-1,j) + 1, if matrix[i][j]=='1';

height(i,j) = 0, if matrix[i][j]=='0'

-------@wahcheung--------

[
   ["1","0","1","0","0"],
   ["1","0","1","1","1"],
   ["1","1","1","1","1"],
   ["1","0","0","1","0"]
 ]

策略: 把matrix看成多个直方图, 每一行及其上方的数据都构成一个直方图, 需要考察matrix.size()个直方图

  • 对于每个点(row, col), 我们最后都计算以这个点上方的连续的'1'往left, right方向延申可以得到的最大的矩形的面积
  • 通过这种方法获取的矩形一定会把最大的矩形包含在内
  • height[row][col]记录的是(row, col)这个坐标为底座的直方图柱子的高度, 如果这个点是'0', 那么高度当然是0了
  • left[row][col]记录的是(row, col)这个坐标点对应的height可以延申到的最左边的位置
  • right[row][col]记录的是(row, col)这个坐标点对应的height可以延申到的最右边的位置+1

以上面的matrix为例,

  • 对于(row=2, col=1)这个点, left=0, right=5, height=1
  • 对于(row=2, col=2)这个点, left=2, right=3, height=3
  • (2,2)这个点与(2,1)紧挨着,left和right却已经变化如此之大了, 这是因为left和right除了受左右两边的'1'影响, 还受到了其上方连续的'1'的制约
  • 由于点(2,2)上有height=3个'1', 这几个'1'的left的最大值作为当前点的left, 这几个'1'的right的最小值作为当前点的right

因此, 实际上, 我们是要找以height对应的这条线段往左右两边移动(只能往全是'1'的地方移动), 可以扫过的最大面积。

当height与目标最大矩形区域的最短的height重合时, 最大矩形的面积就找到了, 如上面的例子, 就是点(2,3)或(2,4)对应的height

--

Code by @Self_Learner

/* we start from the first row, and move downward;
 * height[i] record the current number of countinous '1' in column i;
 * left[i] record the left most index j which satisfies that for any index k from j to  i, height[k] >= height[i];
 * right[i] record the right most index j which satifies that for any index k from i to  j, height[k] >= height[i];
 * by understanding the definition, we can easily figure out we need to update maxArea with value (height[i] * (right[i] - left[i] + 1));
 * 
 * Then the question is how to update the array of height, left, and right
 * =================================
 * for updating height, it is easy
 * for (int j = 0; j < n; j++) {
 *    if (matrix[i][j] == '1') height[j]++;
 *    else height[j] = 0;
 * }
 * =============================================================================
 * It is a little bit tricky for initializing and updating left and right array
 * for initialization: 
 * we know initially, height array contains all 0, so according to the definition of left and right array, left array should contains all 0, and right array should contain all n - 1
 * for updating left and right, it is kind of tricky to understand...
 *     ==============================================================
 *     Here is the code for updating left array, we scan from left to right
 *     int lB = 0;  //lB is indicating the left boundry for the current row of the matrix (for cells with char "1"), not the height array...
 *     for (int j = 0; j < n; j++) {
 *          if (matrix[i][j] == '1') {
 *              left[j] = Math.max(left[j], lB); // this means the current boundry should satisfy two conditions: within the boundry of the previous height array, and within the boundry of the current row...
 *          } else {
 *              left[j] = 0; // when matrix[i][j] = 0, height[j] will get update to 0, so left[j] becomes 0, since all height in between 0 - j satisfies the condition of height[k] >= height[j];
 *              lB = j + 1; //and since current position is '0', so the left most boundry for next "1" cell is at least j + 1;
 *          }
 *     }
 *     ==============================================================
 *     the idea for updating the right boundary is similar, we just need to iterate from right to left
 *     int rB = n - 1;
 *     for (int j = n - 1; j >= 0; j--) {
 *         if (matrix[i][j] == '1') {
 *              right[j] = Math.min(right[j], rB);
 *         } else {
 *              right[j] = n - 1;
 *              rB = j - 1;
 *      }
 */
class Solution {
    public int maximalRectangle(char[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0] == null || matrix[0].length == 0) return 0;
        int m = matrix.length, n = matrix[0].length, maxArea = 0;
        int[] left = new int[n];
        int[] right = new int[n];
        int[] height = new int[n];
        Arrays.fill(right, n - 1);
        for (int i = 0; i < m; i++) {
            int rB = n - 1;
            for (int j = n - 1; j >= 0; j--) {
                if (matrix[i][j] == '1') {
                    right[j] = Math.min(right[j], rB);
                } else {
                    right[j] = n - 1;
                    rB = j - 1;
                }
            }
            int lB = 0;
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == '1') {
                    left[j] = Math.max(left[j], lB);
                    height[j]++;
                    maxArea = Math.max(maxArea, height[j] * (right[j] - left[j] + 1));
                } else {
                    height[j] = 0;
                    left[j] = 0;
                    lB = j + 1;
                }
            }
        }
        return maxArea;
    }
}

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