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 whi
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:
h_cur>h_top(h, i).h_cur: (i_cur-i_pop)*h_pop > best. h_cur>h_top (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).