# Puzzle: Find largest rectangle (maximal rectangle problem)

###### https://stackoverflow.com/questions/7245

### Full question

- math - algorithm - language-agnostic - geometry |
- |

### Question

What's the most efficient algorithm to find the rectangle with the largest area which will fit in the empty space?

Let's say the screen looks like this ('#' represents filled area):

```
....................
..............######
##..................
.................###
.................###
#####...............
#####...............
#####...............
```

A probable solution is:

```
....................
..............######
##...++++++++++++...
.....++++++++++++###
.....++++++++++++###
#####++++++++++++...
#####++++++++++++...
#####++++++++++++...
```

Normally I'd enjoy figuring out a solution. Although this time I'd like to avoid wasting time fumbling around on my own since this has a practical use for a project I'm working on. Is there a well-known solution?

**Shog9** wrote:

Is your input an array (as implied by the other responses), or a list of occlusions in the form of arbitrarily sized, positioned rectangles (as might be the case in a windowing system when dealing with window positions)?

Yes, I have a structure which keeps track of a set of windows placed on the screen. I also have a grid which keeps track of all the areas between each edge, whether they are empty or filled, and the pixel position of their left or top edge. I think there is some modified form which would take advantage of this property. Do you know of any?

### OTHER TIPS

Here's a page that has some code and some analysis.

Your particular problem begins a bit down on the page, search the page for the text *maximal rectangle problem*.

http://www.seas.gwu.edu/~simhaweb/cs151/lectures/module6/module6.html

@lassevk

```
// 4. Outer double-for-loop to consider all possible positions
// for topleft corner.
for (int i=0; i < M; i++) {
for (int j=0; j < N; j++) {
// 2.1 With (i,j) as topleft, consider all possible bottom-right corners.
for (int a=i; a < M; a++) {
for (int b=j; b < N; b++) {
```

HAHA... O(m2 n2).. That's probably what I would have come up with.

I see they go on to develop optmizations... looks good, I'll have a read.

I implemented the solution of Dobbs in Java.

No warranty for anything.

```
package com.test;
import java.util.Stack;
public class Test {
public static void main(String[] args) {
boolean[][] test2 = new boolean[][] { new boolean[] { false, true, true, false },
new boolean[] { false, true, true, false }, new boolean[] { false, true, true, false },
new boolean[] { false, true, false, false } };
solution(test2);
}
private static class Point {
public Point(int x, int y) {
this.x = x;
this.y = y;
}
public int x;
public int y;
}
public static int[] updateCache(int[] cache, boolean[] matrixRow, int MaxX) {
for (int m = 0; m < MaxX; m++) {
if (!matrixRow[m]) {
cache[m] = 0;
} else {
cache[m]++;
}
}
return cache;
}
public static void solution(boolean[][] matrix) {
Point best_ll = new Point(0, 0);
Point best_ur = new Point(-1, -1);
int best_area = 0;
final int MaxX = matrix[0].length;
final int MaxY = matrix.length;
Stack<Point> stack = new Stack<Point>();
int[] cache = new int[MaxX + 1];
for (int m = 0; m != MaxX + 1; m++) {
cache[m] = 0;
}
for (int n = 0; n != MaxY; n++) {
int openWidth = 0;
cache = updateCache(cache, matrix[n], MaxX);
for (int m = 0; m != MaxX + 1; m++) {
if (cache[m] > openWidth) {
stack.push(new Point(m, openWidth));
openWidth = cache[m];
} else if (cache[m] < openWidth) {
int area;
Point p;
do {
p = stack.pop();
area = openWidth * (m - p.x);
if (area > best_area) {
best_area = area;
best_ll.x = p.x;
best_ll.y = n;
best_ur.x = m - 1;
best_ur.y = n - openWidth + 1;
}
openWidth = p.y;
} while (cache[m] < openWidth);
openWidth = cache[m];
if (openWidth != 0) {
stack.push(p);
}
}
}
}
System.out.printf("The maximal rectangle has area %d.\n", best_area);
System.out.printf("Location: [col=%d, row=%d] to [col=%d, row=%d]\n", best_ll.x + 1, best_ll.y + 1,
best_ur.x + 1, best_ur.y + 1);
}
}
```

After struggling so much I've wrote this algorithm...Just see the code...

You understand that.This code speaks !!

```
import java.util.Scanner;
import java.util.Stack;
/**
* Created by BK on 05-08-2017.
*/
public class LargestRectangleUnderHistogram {
public static void main(String... args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int[] input = new int[n];
for (int j = 0; j < n; j++) {
input[j] = scanner.nextInt();
}
/*
* This is the procedure used for solving :
*
* Travel from first element to last element of the array
*
* If stack is empty add current element to stack
*
* If stack is not empty check for the top element of stack if
* it is smaller than the current element push into stack
*
* If it is larger than the current element pop the stack until we get an
* element smaller than the current element or until stack becomes empty
*
* After popping each element check if the stack is empty or not..
*
* If stack is empty it means that this is the smallest element encountered till now
*
* So we can multiply i with this element to get a big rectangle which is contributed by
*
* this
*
* If stack is not empty then check for individual areas(Not just one bar individual area means largest rectangle by this) (i-top)*input[top]
*
*
* */
/*
* Initializing the maxarea as we check each area with maxarea
*/
int maxarea = -1;
int i = 0;
Stack<Integer> stack = new Stack<>();
for (i = 0; i < input.length; i++) {
/*
* Pushing the element if stack is empty
* */
if (stack.isEmpty()) {
stack.push(i);
} else {
/*
* If stack top element is less than current element push
* */
if (input[stack.peek()] < input[i]) {
stack.push(i);
} else {
/*
* Else pop until we encounter an element smaller than this in stack or stack becomes empty
*
* */
while (!stack.isEmpty() && input[stack.peek()] > input[i]) {
int top = stack.pop();
/*
* If stack is empty means this is the smallest element encountered so far
*
* So we can multiply this with i
* */
if (stack.isEmpty()) {
maxarea = maxarea < (input[top] * i) ? (input[top] * i) : maxarea;
}
/*
* If stack is not empty we find areas of each individual rectangle
* Remember we are in while loop
*/
else {
maxarea = maxarea < (input[top] * (i - top)) ? (input[top] * (i - top)) : maxarea;
}
}
/*
* Finally pushing the current element to stack
* */
stack.push(i);
}
}
}
/*
* This is for checking if stack is not empty after looping the last element of input
*
* This happens if input is like this 4 5 6 1 2 3 4 5
*
* Here 2 3 4 5 remains in stack as they are always increasing and we never got
*
* a chance to pop them from stack by above process
*
* */
while (!stack.isEmpty()) {
int top = stack.pop();
maxarea = maxarea < (i - top) * input[top] ? (i - top) * input[top] : maxarea;
}
System.out.println(maxarea);
}
}
```

I am the author of the Maximal Rectangle Solution on LeetCode, which is what this answer is based on.

Since the stack-based solution has already been discussed in the other answers, I would like to present an optimal `O(NM)`

dynamic programming solution which originates from user morrischen2008.

**Intuition**

Imagine an algorithm where for each point we computed a rectangle by doing the following:

Finding the maximum height of the rectangle by iterating upwards until a filled area is reached

Finding the maximum width of the rectangle by iterating outwards left and right until a height that doesn't accommodate the maximum height of the rectangle

For example finding the rectangle defined by the yellow point:

We know that the maximal rectangle must be one of the rectangles constructed in this manner (the max rectangle must have a point on its base where the next filled square is *height* above that point).

For each point we define some variables:

`h`

- the height of the rectangle defined by that point

`l`

- the left bound of the rectangle defined by that point

`r`

- the right bound of the rectangle defined by that point

These three variables uniquely define the rectangle at that point. We can compute the area of this rectangle with `h * (r - l)`

. The global maximum of all these areas is our result.

Using dynamic programming, we can use the `h`

, `l`

, and `r`

of each point in the previous row to compute the `h`

, `l`

, and `r`

for every point in the next row in linear time.

**Algorithm**

Given row `matrix[i]`

, we keep track of the `h`

, `l`

, and `r`

of each point in the row by defining three arrays - `height`

, `left`

, and `right`

.

`height[j]`

will correspond to the height of `matrix[i][j]`

, and so on and so forth with the other arrays.

The question now becomes how to update each array.

`height`

`h`

is defined as the number of continuous unfilled spaces in a line from our point. We increment if there is a new space, and set it to zero if the space is filled (we are using '1' to indicate an empty space and '0' as a filled one).

```
new_height[j] = old_height[j] + 1 if row[j] == '1' else 0
```

`left`

:

Consider what causes changes to the left bound of our rectangle. Since all instances of filled spaces occurring in the row above the current one have already been factored into the current version of `left`

, the only thing that affects our `left`

is if we encounter a filled space in our current row.

As a result we can define:

```
new_left[j] = max(old_left[j], cur_left)
```

`cur_left`

is one greater than rightmost filled space we have encountered. When we "expand" the rectangle to the left, we know it can't expand past that point, otherwise it'll run into the filled space.

`right`

:

Here we can reuse our reasoning in `left`

and define:

```
new_right[j] = min(old_right[j], cur_right)
```

`cur_right`

is the leftmost occurrence of a filled space we have encountered.

**Implementation**

```
def maximalRectangle(matrix):
if not matrix: return 0
m = len(matrix)
n = len(matrix[0])
left = [0] * n # initialize left as the leftmost boundary possible
right = [n] * n # initialize right as the rightmost boundary possible
height = [0] * n
maxarea = 0
for i in range(m):
cur_left, cur_right = 0, n
# update height
for j in range(n):
if matrix[i][j] == '1': height[j] += 1
else: height[j] = 0
# update left
for j in range(n):
if matrix[i][j] == '1': left[j] = max(left[j], cur_left)
else:
left[j] = 0
cur_left = j + 1
# update right
for j in range(n-1, -1, -1):
if matrix[i][j] == '1': right[j] = min(right[j], cur_right)
else:
right[j] = n
cur_right = j
# update the area
for j in range(n):
maxarea = max(maxarea, height[j] * (right[j] - left[j]))
return maxarea
```