package com.freetymekiyan.algorithms.level.hard;
import java.util.Arrays;
/**
* 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>
* Company Tags: Facebook
* Tags: Array, Hash Table, Stack, Dynamic Programming
* Similar Problems: (H) Largest Rectangle in Histogram, (M) Maximal Square
*/
public class MaximalRectangle {
/**
* DP.
* Three matrices: left, right, and height.
* Maximal rectangle area at row i and column j is: [right(i,j) - left(i,j)] * height(i,j).
* left is the left boundary of current row and previous row.
* left(i,j) = max(left(i-1,j), curLeft)
* curLeft is the left boundary candidate of current row.
* <p>
* right is the right boundary of current row. Actually is index + 1 for easy implementation.
* right(i,j) = min(right(i-1,j), curRight)
* curRight is the right boundary candidate of current row.
* <p>
* height is how many 1s, including current 1, from top to current position.
* height(i,j) = height(i-1,j) + 1, if matrix[i][j]=='1'.
* height(i,j) = 0, if matrix[i][j]=='0'.
* <p>
* Implementation:
* Initialize left array and height as all zeroes, right array as the column length.
* For each row in the matrix, update height, left, right arrays.
* Then compute the max area.
* Stop when all rows are looped through.
* https://discuss.leetcode.com/topic/6650/share-my-dp-solution
*/
public int maximalRectangle(char[][] matrix) {
int m = matrix.length;
int n = matrix[0].length;
int[] left = new int[n];
int[] right = new int[n];
int[] height = new int[n];
// Arrays.fill(left, 0);
Arrays.fill(right, n);
// Arrays.fill(height, 0);
int maxA = 0;
for (int i = 0; i < m; i++) {
// Compute height (can do this from either side).
for (int j = 0; j < n; j++) {
if (matrix[i][j] == '1') {
height[j]++;
} else {
height[j] = 0;
}
}
// Compute left (must from left to right).
int curLeft = 0; // Index of leftmost 1 of current row.
for (int j = 0; j < n; j++) {
if (matrix[i][j] == '1') {
left[j] = Math.max(left[j], curLeft);
} else {
left[j] = 0;
curLeft = j + 1;
}
}
// Compute right (must from right to left).
int curRight = n; // Index+1 of rightmost 1 of current row.
for (int j = n - 1; j >= 0; j--) {
if (matrix[i][j] == '1') {
right[j] = Math.min(right[j], curRight);
} else {
right[j] = n; // Like reset. Make sure right[j] >= curRight.
curRight = j;
}
}
// Compute the area of rectangle (can do this from either side).
for (int j = 0; j < n; j++) {
maxA = Math.max(maxA, (right[j] - left[j]) * height[j]);
}
}
return maxA;
}
}