algorithm - Finding maximum size sub-matrix of all 1's in a matrix having 1's and 0's

Question

Suppose you are given an mXn bitmap, represented by an array M[1..m,1.. n] whose entries are all 0 or 1. A all-one block is a subarray of the form M[i .. i0, j .. j0] in which every bit is equal to 1. Describe and analyze an efficient algorithm to find an all-one block in M with maximum area

I am trying to make a dynamic programming solution. But my recursive algorithm runs in O(n^n) time, and even after memoization I cannot think of bringing it down below O(n^4). Can someone help me find a more efficient solution?

Answer 1

An O(N) (number of elements) solution:

A
1 1 0 0 1 0
0 1 1 1 1 1
1 1 1 1 1 0
0 0 1 1 0 0 

Generate an array C where each element represents the number of 1s above and including it, up until the first 0.

C
1 1 0 0 1 0
0 2 1 1 2 1
1 3 2 2 3 0
0 0 3 3 0 0 

We want to find the row R, and left, right indices l , r that maximizes (r-l+1)*min(C[R][l..r]). Here is an algorithm to inspect each row in O(cols) time:

Maintain a stack of pairs (h, i), where C[R][i-1] < h ≤ C[R][i]. At any position cur, we should have h=min(C[R][i..cur]) for all pairs (h, i) on the stack.

For each element:

• If h_cur>h_top
• • Push (h, i).
• Else:
• • While h_cur<h_top:
• • • Pop the top of the stack.
• • • Check whether it would make a new best, i.e. (i_cur-i_pop)*h_pop > best.
• • If h_cur>h_top
• • • Push (h, i_lastpopped).

An example of this in execution for the third row in our example:

  i =0      1      2      3      4      5
C[i]=1      3      2      2      3      0
                                 (3, 4)
 S=         (3, 1) (2, 1) (2, 1) (2, 1)
     (1, 0) (1, 0) (1, 0) (1, 0) (1, 0)    
     (0,-1) (0,-1) (0,-1) (0,-1) (0,-1) (0,-1) 

i=0, C[i]=1) Push (1, 0).
i=1, C[i]=3) Push (3, 1).
i=2, C[i]=2) Pop (3, 1). Check whether (2-1)*3=3 is a new best.
????????The last i popped was 1, so push (2, 1).
i=3, C[i]=2) h_cur=h_top so do nothing.
i=4, C[i]=3) Push (3, 4).
i=5, C[i]=0) Pop (3, 4). Check whether (5-4)*3=3 is a new best.
????????Pop (2, 1). Check whether (5-1)*2=8 is a new best.
????????Pop (1, 0). Check whether (5-0)*1=5 is a new best.
????????End. (Okay, we should probably add an extra term C[cols]=0 on the end for good measure).