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package stack;

import java.util.Stack;

/**

* Created by gouthamvidyapradhan on 29/11/2017.

*

* <p>Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing only

* 1's and return its area.

*

* <p>For example, given the following matrix:

*

* <p>1 0 1 0 0 1 0 1 1 1 1 1 1 1 1 1 0 0 1 0 Return 6.

*

* <p>Solution O(n * m): This problem is similar to LargestRectangleInHistogram. Run the largest

* rectangle in histogram algorithm for each row.

*/

public class MaximalRectangle {

/**

* Main method

*

* @param args

* @throws Exception

*/

public static void main(String[] args) throws Exception {

char[][] matrix = {

{'1', '0', '1', '0', '0'},

{'1', '0', '1', '1', '1'},

{'1', '1', '1', '1', '1'},

{'1', '0', '0', '1', '0'}

};

System.out.println(new MaximalRectangle().maximalRectangle(matrix));

}

public int maximalRectangle(char[][] matrix) {

if (matrix.length == 0 || matrix[0].length == 0) return 0;

int[] A = new int[matrix[0].length];

int max = Integer.MIN_VALUE;

for (int i = 0; i < matrix.length; i++) {

for (int j = 0; j < matrix[0].length; j++) {

if (matrix[i][j] == '1') {

if (i > 0 && matrix[i - 1][j] == '1') {

A[j] = A[j] + 1;

} else {

A[j] = 1;

}

} else {

A[j] = 0;

}

}

// calculate max rectangle for this row

max = Math.max(max, getMaxRectangle(A));

}

return max;

}

/**

* Get max rectangle algorithm similar to max rectangle in histogram

*

* @param heights

* @return

*/

private int getMaxRectangle(int[] heights) {

int maxArea = Integer.MIN_VALUE;

Stack<Integer> stack = new Stack<>();

int i = 0;

for (; i < heights.length; i++) {

while (!stack.isEmpty() && heights[stack.peek()] >= heights[i]) {

int top = stack.pop();

int base = stack.isEmpty() ? i : i - stack.peek() - 1;

maxArea = Math.max(maxArea, base * heights[top]);

}

stack.push(i);

}

while (!stack.isEmpty()) {

int top = stack.pop();

int base = stack.isEmpty() ? i : i - stack.peek() - 1;

maxArea = Math.max(maxArea, base * heights[top]);

}

return maxArea;

}

}