Puzzle: Find largest rectangle (maximal rectangle problem)

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?

Answer 1

I'm the author of that Dr. Dobb's article and get occasionally asked about an implementation. Here is a simple one in C:

#include <assert.h> #include <stdio.h> #include <stdlib.h>  typedef struct {   int one;   int two; } Pair;  Pair best_ll = { 0, 0 }; Pair best_ur = { -1, -1 }; int best_area = 0;  int *c; /* Cache */ Pair *s; /* Stack */ int top = 0; /* Top of stack */  void push(int a, int b) {   s[top].one = a;   s[top].two = b;   ++top; }  void pop(int *a, int *b) {   --top;   *a = s[top].one;   *b = s[top].two; }   int M, N; /* Dimension of input; M is length of a row. */  void update_cache() {   int m;   char b;   for (m = 0; m!=M; ++m) {     scanf(" %c", &b);     fprintf(stderr, " %c", b);     if (b=='0') {       c[m] = 0;     } else { ++c[m]; }   }   fprintf(stderr, "\n"); }   int main() {   int m, n;   scanf("%d %d", &M, &N);   fprintf(stderr, "Reading %dx%d array (1 row == %d elements)\n", M, N, M);   c = (int*)malloc((M+1)*sizeof(int));   s = (Pair*)malloc((M+1)*sizeof(Pair));   for (m = 0; m!=M+1; ++m) { c[m] = s[m].one = s[m].two = 0; }   /* Main algorithm: */   for (n = 0; n!=N; ++n) {     int open_width = 0;     update_cache();     for (m = 0; m!=M+1; ++m) {       if (c[m]>open_width) { /* Open new rectangle? */         push(m, open_width);         open_width = c[m];       } else /* "else" optional here */       if (c[m]<open_width) { /* Close rectangle(s)? */         int m0, w0, area;         do {           pop(&m0, &w0);           area = open_width*(m-m0);           if (area>best_area) {             best_area = area;             best_ll.one = m0; best_ll.two = n;             best_ur.one = m-1; best_ur.two = n-open_width+1;           }           open_width = w0;         } while (c[m]<open_width);         open_width = c[m];         if (open_width!=0) {           push(m0, w0);         }       }     }   }   fprintf(stderr, "The maximal rectangle has area %d.\n", best_area);   fprintf(stderr, "Location: [col=%d, row=%d] to [col=%d, row=%d]\n",                   best_ll.one+1, best_ll.two+1, best_ur.one+1, best_ur.two+1);   return 0; } 

It takes its input from the console. You could e.g. pipe this file to it:

16 12 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 1 0 0 0 1 1 0 1 0 0 0 0 1 1 0 1 1 1 0 1 1 1 0 1 0 0 0 0 0 1 1 * * * * * * 0 0 1 0 0 0 0 0 0 0 * * * * * * 0 0 1 0 0 0 0 0 0 0 1 1 0 1 1 1 1 1 1 0 0 0 1 0 0 0 0 1 0 0 1 1 1 0 1 0  0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0  0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 

And after printing its input, it will output:

The maximal rectangle has area 12. Location: [col=7, row=6] to [col=12, row=5] 

The implementation above is nothing fancy of course, but it's very close to the explanation in the Dr. Dobb's article and should be easy to translate to whatever is needed.