The following algorithm finds such occurences in O(n) time given any two strings.
One assumption worth noting for the O(n) time bound is that the size of the alphabet is limited to some constant size.

Nested loops in this algorithm scream O(n^2)!! However, those loops are bounded by a constant that is proportional to the size of the alphabet, squared. Since the alphabet is bounded by a constant (by my assumption above), these nested loops will be bounded by a constant — so they don’t take us out of O(n) time.

__author__ = "Kris Olhovsky"
__date__ = "$Dec 29, 2009 6:54:11 PM$"

class Char:
    def __init__(self, char):
        self.char = char
        self.paths = {} # Key is length of paths, value is the # of paths.

    ''' Return the number of paths to this char of length len. '''
    def retrieve_matches(self, len):
        count = 0
        for c in self.paths:
            if len in self.paths[c]:
                count += self.paths[c][len]
                self.paths[c][len] = 0 # Delete recorded matches.
        return count

    ''' Copies paths from char c to this char, adding 1 to the
        length of every path. Paths of length greater than len are ignored. '''
    def add_paths(self, c, len):
        temp = {} # Avoid altering map during iteration.
        for char in c.paths:
            for length in c.paths[char]:
                if length + 1 <= len:
                    if c not in temp:
                        temp[c] = {}
                    if length + 1 not in temp[c]:
                        temp[c][length+1] = 0
                    temp[c][length+1] += c.paths[char][length]

        for k in temp: # Apply changes to map.
            for j in temp[k]:
                if k not in self.paths:
                    self.paths[c] = {}
                if j not in self.paths[k]:
                    self.paths[k][j] = 0
                self.paths[k][j] += temp[k][j]

''' Returns the number of occurences of string s1 in string s2. '''
def subseq(s1, s2):
    if s1 == "": # Special case defined by the problem.
        return 1

    prev = {}   # Maps a char to a dict of chars that appear one index to
                # the left of that char anywhere in s1.

    chars = {}  # All of the chars that appear in s1.

    for c in s1: # Initialize maps.
        prev[c] = {}
        chars[c] = Char(c)

    i = 1
    while i < len(s1): # Populate maps with data from s1.
        prev[s1[i]][s1[i-1]] = chars[s1[i-1]]
        i += 1

    start = Char('start')
    chars[s1[0]].paths[start] = {}
    chars[s1[0]].paths[start][0] = 0
    occurences = 0 # Record number of occurences of s1 in s2.
    for char in s2:

        if char in chars: # A char in s2 matches a char in s1.
            if char in prev[char]: # If same exists, prioritize this.
                chars[char].add_paths(prev[char][char], len(s1) - 1)
            for c in prev[char]:
                if c != char:
                    chars[char].add_paths(prev[char][c], len(s1) - 1)
            if char == s1[0]: # First char in s1 always starts a new path.
                chars[s1[0]].paths[start][0] += 1

            if char == s1[-1]: # Update occurences when last char of s1 seen.
                occurences += chars[char].retrieve_matches(len(s1) - 1)

    return occurences

Here are some example usages:

if __name__ == "__main__":
    s1 = "ABC"
    s2 = "ABBC"
    print subseq(s1, s2)

    s1 = "ABCB"
    s2 = "ABBCBBCB"
    print subseq(s1, s2)

    s1 = "ABCB"
    s2 = "ABCB"
    print subseq(s1, s2)

    s1 = "ABBB"
    s2 = "ABBBBB"
    print subseq(s1, s2)

    s1 = "ABBB"
    s2 = "ABABB"
    print subseq(s1, s2)

This runs in O(n log n) time, and uses O(n) memory.

''' An item in the final sequence, used to form a linked list. '''
class SeqItem():
    val = 0      # This item's value.
    prev = None  # The value before this one.
    def __init__(self, val, prev):
        self.val = val
        self.prev = prev

''' Extract longest non-decreasing subsequence from sequence seq.'''
def extract_sorted(seq):
    subseqs = [SeqItem(seq[0], None)] # Track decreasing subsequences in seq.
    result_list = [subseqs[0]]
    for i in range(1, len(seq)):
        result = search_insert(subseqs, seq[i], 0, len(subseqs))

    # Build Python list from custom linked list:
    final_list = []
    result = subseqs[-1] # Longest nondecreasing subsequence is found by
                         # traversing the linked list backwards starting from
                         # the final smallest value in the last nonincreasing
                         # subsequence found.
    while(result != None and result.val != None):
        final_list.append(result.val)
        result = result.prev # Walk backwards through longest sequence.

    final_list.reverse()
    return final_list

''' Seq tracks the smallest value of each nonincreasing subsequence constructed.
Find smallest item in seq that is greater than search_val.
If such a value does not exist, append search_val to seq, creating the beginning
of a new nonincreasing subsequence.
If such a value does exist, replace the value in seq at that position, and
search_val will be considered the new candidate for the longest subseq if
a value in the following nonincreasing subsequence is added.
Seq is guaranteed to be in increasing sorted order.
Returns the index of the element in seq that should be added to results. '''
def search_insert(seq, search_val, start, end):
    median = (start + end)/2

    if end - start < 2: # End of the search.
        if seq[start].val > search_val:
            if start > 0:
                new_item = SeqItem(search_val, seq[start - 1])
            else:
                new_item = SeqItem(search_val, None)

            seq[start] = new_item
            return new_item
        else: # seq[start].val <= search_val
            if start + 1 < len(seq):
                new_item = SeqItem(search_val, seq[start])
                seq[start + 1] = new_item
                return new_item
            else:
                new_item = SeqItem(search_val, seq[start])
                seq.append(new_item)
                return new_item

    if search_val < seq[median].val: # Search left side
        return search_insert(seq, search_val, start, median)
    else: #search_val >= seq[median].val: # Search right side
        return search_insert(seq, search_val, median, end)

And here is an example usage:

import random
if __name__ == '__main__':
    seq = []
    for i in range(100000):
        seq.append(int(random.random() * 1000))

    print extract_sorted(seq)

'''
2D CDF 9/7 Wavelet Forward and Inverse Transform (lifting implementation)

This code is provided "as is" and is given for educational purposes.
2008 - Kris Olhovsky - code.inquiries@olhovsky.com
'''

from PIL import Image # Part of the standard Python Library

''' Example matrix as a list of lists: '''
mat4x4 = [
         [0,   1,  2,  3], # Row 1
         [4,   5,  6,  7], # Row 2
         [8,   9, 10, 11], # Row 3
         [12, 13, 14, 15], # Row 4
         ]                 # We don't do anything with this matrix.
                           # It's just here for clarification.

def fwt97_2d(m, nlevels=1):
    ''' Perform the CDF 9/7 transform on a 2D matrix signal m.
    nlevel is the desired number of times to recursively transform the
    signal. '''

    w = len(m[0])
    h = len(m)
    for i in range(nlevels):
        m = fwt97(m, w, h) # cols
        m = fwt97(m, w, h) # rows
        w /= 2
        h /= 2

    return m

def iwt97_2d(m, nlevels=1):
    ''' Inverse CDF 9/7 transform on a 2D matrix signal m.
        nlevels must be the same as the nlevels used to perform the fwt.
    '''

    w = len(m[0])
    h = len(m)

    # Find starting size of m:
    for i in range(nlevels-1):
        h /= 2
        w /= 2

    for i in range(nlevels):
        m = iwt97(m, w, h) # rows
        m = iwt97(m, w, h) # cols
        h *= 2
        w *= 2

    return m

def fwt97(s, width, height):
    ''' Forward Cohen-Daubechies-Feauveau 9 tap / 7 tap wavelet transform
    performed on all columns of the 2D n*n matrix signal s via lifting.
    The returned result is s, the modified input matrix.
    The highpass and lowpass results are stored on the left half and right
    half of s respectively, after the matrix is transposed. '''

    # 9/7 Coefficients:
    a1 = -1.586134342
    a2 = -0.05298011854
    a3 = 0.8829110762
    a4 = 0.4435068522

    # Scale coeff:
    k1 = 0.81289306611596146 # 1/1.230174104914
    k2 = 0.61508705245700002 # 1.230174104914/2
    # Another k used by P. Getreuer is 1.1496043988602418

    for col in range(width): # Do the 1D transform on all cols:
        ''' Core 1D lifting process in this loop. '''
        ''' Lifting is done on the cols. '''

        # Predict 1. y1
        for row in range(1, height-1, 2):
            s[row][col] += a1 * (s[row-1][col] + s[row+1][col])
        s[height-1][col] += 2 * a1 * s[height-2][col] # Symmetric extension

        # Update 1. y0
        for row in range(2, height, 2):
            s[row][col] += a2 * (s[row-1][col] + s[row+1][col])
        s[0][col] +=  2 * a2 * s[1][col] # Symmetric extension

        # Predict 2.
        for row in range(1, height-1, 2):
            s[row][col] += a3 * (s[row-1][col] + s[row+1][col])
        s[height-1][col] += 2 * a3 * s[height-2][col]

        # Update 2.
        for row in range(2, height, 2):
            s[row][col] += a4 * (s[row-1][col] + s[row+1][col])
        s[0][col] += 2 * a4 * s[1][col]

    # de-interleave
    temp_bank = [[0]*width for i in range(height)]
    for row in range(height):
        for col in range(width):
            # k1 and k2 scale the vals
            # simultaneously transpose the matrix when deinterleaving
            if row % 2 == 0: # even
                temp_bank[col][row/2] = k1 * s[row][col]
            else:            # odd
                temp_bank[col][row/2 + height/2] = k2 * s[row][col]

    # write temp_bank to s:
    for row in range(width):
        for col in range(height):
            s[row][col] = temp_bank[row][col]

    return s

def iwt97(s, width, height):
    ''' Inverse CDF 9/7. '''

    # 9/7 inverse coefficients:
    a1 = 1.586134342
    a2 = 0.05298011854
    a3 = -0.8829110762
    a4 = -0.4435068522

    # Inverse scale coeffs:
    k1 = 1.230174104914
    k2 = 1.6257861322319229

    # Interleave:
    temp_bank = [[0]*width for i in range(height)]
    for col in range(width/2):
        for row in range(height):
            # k1 and k2 scale the vals
            # simultaneously transpose the matrix when interleaving
            temp_bank[col * 2][row] = k1 * s[row][col]
            temp_bank[col * 2 + 1][row] = k2 * s[row][col + width/2]

    # write temp_bank to s:
    for row in range(width):
        for col in range(height):
            s[row][col] = temp_bank[row][col]

    for col in range(width): # Do the 1D transform on all cols:
        ''' Perform the inverse 1D transform. '''

        # Inverse update 2.
        for row in range(2, height, 2):
            s[row][col] += a4 * (s[row-1][col] + s[row+1][col])
        s[0][col] += 2 * a4 * s[1][col]

        # Inverse predict 2.
        for row in range(1, height-1, 2):
            s[row][col] += a3 * (s[row-1][col] + s[row+1][col])
        s[height-1][col] += 2 * a3 * s[height-2][col]

        # Inverse update 1.
        for row in range(2, height, 2):
            s[row][col] += a2 * (s[row-1][col] + s[row+1][col])
        s[0][col] +=  2 * a2 * s[1][col] # Symmetric extension

        # Inverse predict 1.
        for row in range(1, height-1, 2):
            s[row][col] += a1 * (s[row-1][col] + s[row+1][col])
        s[height-1][col] += 2 * a1 * s[height-2][col] # Symmetric extension

    return s

def seq_to_img(m, pix):
    ''' Copy matrix m to pixel buffer pix.
    Assumes m has the same number of rows and cols as pix. '''
    for row in range(len(m)):
        for col in range(len(m[row])):
            pix[col,row] = m[row][col]

And here is an example usage:

if __name__ == "__main__":
    # Load image.
    im = Image.open("test1_512.png") # Must be a single band image! (grey)

    # Create an image buffer object for fast access.
    pix = im.load()

    # Convert the 2d image to a 1d sequence:
    m = list(im.getdata())

    # Convert the 1d sequence to a 2d matrix.
    # Each sublist represents a row. Access is done via m[row][col].
    m = [m[i:i+im.size[0]] for i in range(0, len(m), im.size[0])]

    # Cast every item in the list to a float:
    for row in range(0, len(m)):
        for col in range(0, len(m[0])):
            m[row][col] = float(m[row][col])

    # Perform a forward CDF 9/7 transform on the image:
    m = fwt97_2d(m, 3)

    seq_to_img(m, pix) # Convert the list of lists matrix to an image.
    im.save("test1_512_fwt.png") # Save the transformed image.

    # Perform an inverse transform:
    m = iwt97_2d(m, 3)

    seq_to_img(m, pix) # Convert the inverse list of lists matrix to an image.
    im.save("test1_512_iwt.png") # Save the inverse transformation.

This is a first step in a larger task of compressing a certain type of vertex data, which can be decompressed entirely by a GPU via an implementation written in shader code.

The reason for a Python implementation is to create a working, easy to understand version of the implementation. Not only for myself, but for the interwebs.

The Python code does use “height” and “width” variables, although in order to use rectangular images, some adjustments need to be made. This code also assumes that your images have width and height of the form 2^n. E.g. 1024 x 1024, 512 x 512, etc.

Sample input image for transforming.

Image after a single level CDF 9/7 transform.

2 level CDF 9/7 wavelet transformed image