I have a 2D bitmap with various shapes (each pixel is either 1 or 0).

How can I find the largest axis-aligned rectangle that covers only set pixels?

  • $\begingroup$ What do you mean by largest? In terms of area? $\endgroup$ Dec 3, 2021 at 19:58
  • $\begingroup$ @YuvalFilmus yes, largest area $\endgroup$ Dec 3, 2021 at 19:59
  • $\begingroup$ @PålGD 1D is easy, you can just do it in a single O(n) swipe left to right $\endgroup$ Dec 3, 2021 at 20:00
  • $\begingroup$ Are you ok with an approximation? $\endgroup$ Dec 3, 2021 at 20:04
  • 1
    $\begingroup$ You can in linear time compute the number of zeros in the rectangle (0,0) to (i,j). You can use this to query whether (up,left),(down,right) defines a valid rectangle in constant time. $\endgroup$
    – Pål GD
    Dec 3, 2021 at 20:15

1 Answer 1


It can be done in $O(n^3)$ time, for a $n\times n$ bitmap.

Let $A$ denote the image, so $A[i,j]$ is the pixel at coordinates $(i,j)$.

For each pixel $(i,j)$, let $H[i,j]$ denote the height of the column of all-1 pixels below it, i.e., the largest $h$ such that $A[i',j]=1$ for all $i' \in [i,i+h-1]$.

For each pixel $(i,j)$ and each height $h$, let $W[i,j,h]$ denote the width of the rectangle of height $h$ with upper-left corner at $(i,j)$ that has covers only set pixels, i.e., the largest $w$ such that $H[i,j'] \ge h$ for all $j' \in [j,j+w-1]$.

You can compute each element of $H$ in $O(1)$ time using dynamic programming (fill them in from the bottom up). Also, once you have $H$, you can compute each element of $W$ in $O(1)$ time using dynamic programming (fill them in the right to left).

Finally, once you have $W$, it is trivial to find the largest rectangle, by scanning all of its elements.

In other words, if you have a $n\times n$ bitmap, you can fill in $H$ in $O(n^2)$ time and fill in $W$ in $O(n^3)$ time, and then you can find the largest rectangle in $O(n^3)$ time. So, this gives an $O(n^3)$ time algorithm. That is an improvement over Pål GD's algorithm, which takes $O(n^4)$ time.

If the bitmap is rectangular rather than square, you can use the same algorithm. I suggest you transpose it first if it is taller than it is wide; that way you will have fewer values of $h$ to iterate over. In this way, if you have a $m\times n$ bitmap, the running time will be $O(\min(mn^2,m^2n))$.


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