Maximal Rectangle : Accepted

Author: gouthampradhan

README.md

@@ -170,6 +170,7 @@ My accepted leetcode solutions to some of the common interview problems.
 - [Valid Parentheses](problems/src/stack/ValidParentheses.java) (Easy)
 - [Largest Rectangle In Histogram](problems/src/stack/LargestRectangleInHistogram.java) (Hard)
 - [Implement Queue using Stacks](problems/src/stack/MyQueue.java) (Easy)
+- [Maximal Rectangle](problems/src/stack/MaximalRectangle.java) (Hard)
 
 
 #### [String](problems/src/string)

problems/src/stack/MaximalRectangle.java

@@ -0,0 +1,80 @@
+package stack;
+
+import java.util.Stack;
+
+/**
+ * Created by gouthamvidyapradhan on 29/11/2017.
+ *
+ * Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing only 1's and return its area.
+
+ For example, given the following matrix:
+
+ 1 0 1 0 0
+ 1 0 1 1 1
+ 1 1 1 1 1
+ 1 0 0 1 0
+ Return 6.
+
+ 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;
+    }
+}