Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
1.1k views
in Technique[技术] by (71.8m points)

algorithm - Maximum subarray of size HxW within a 2D matrix

Given a 2-dimensional array of positive integers, find the subrectangle of size HxW with the largest sum. The sum of a rectangle is the sum of all the elements in that rectangle.

Input: A 2D array NxN with positive elements The HxW size of the subrectangle

Output: The submatrix of HxW size with the largest sum of its elements.

I've solved this using a brute-force method, however, I'm now looking for a better solution with better complexity (my brute-force method's complexity is O(n6)).

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

First create the cumulative sum of your matrix: O(n2)

Matrix
2 4 5 6
2 3 1 4
2 0 2 1

Cumulative sum
2 6  11 17
4 11 17 27
6 13 21 32

cumulative_sum(i,j) is the sum of all the elements in the submatrix (0:i,0:j). You can calculate the cumulative sum matrix using below logic:

cumulative_sum(i,j) = cumulative_sum(i-1,j) + cumulative_sum(i,j-1) - cumulative_sum(i-1,j-1) + matrix(i,j)

Using the cumulative sum matrix you can calculate sum of every sub-matrix in O(1):

calculating sum of submatrix (r1 ... r2 , c1 ... c2)
sum_sub = cumulative_sum(r2,c2) - cumulative_sum(r1-1,c2) - cumulative_sum(r2,c1-1) + cumulative_sum(r1-1,c1-1)

Inclusion-Exclusion

Then using two loops you can put the top-left of your HW rectangle on every point of the matrix and calculate the sum of that rectangle.

for r1=0->n_rows
   for c1=0->n_cols
       r2 = r1 + height - 1
       c2 = c1 + width - 1
       if valid(r1,c1,r2,c2) // doesn't exceed the original matrix
            sum_sub = ... // formula mentioned above
            best = max(sum_sub, best)
return best

This approach is in O(N2).

Here is the python implementation:

from copy import deepcopy

def findMaxSubmatrix(matrix, height, width):
    nrows = len(matrix)
    ncols = len(matrix[0])           

    cumulative_sum = deepcopy(matrix)

    for r in range(nrows):
        for c in range(ncols):
            if r == 0 and c == 0:
                cumulative_sum[r][c] = matrix[r][c]
            elif r == 0:
                cumulative_sum[r][c] = cumulative_sum[r][c-1] + matrix[r][c]
            elif c == 0:
                cumulative_sum[r][c] = cumulative_sum[r-1][c] + matrix[r][c]
            else:
                cumulative_sum[r][c] = cumulative_sum[r-1][c] + cumulative_sum[r][c-1] - cumulative_sum[r-1][c-1] + matrix[r][c]

    best = 0
    best_pos = None

    for r1 in range(nrows):
        for c1 in range(ncols):
            r2 = r1 + height - 1
            c2 = c1 + width - 1
            if r2 >= nrows or c2 >= ncols:
                continue
            if r1 == 0 and c1 == 0:
                sub_sum = cumulative_sum[r2][c2]
            elif r1 == 0:
                sub_sum = cumulative_sum[r2][c2] - cumulative_sum[r2][c1-1]
            elif c1 == 0:
                sub_sum = cumulative_sum[r2][c2] - cumulative_sum[r1-1][c2]
            else:
                sub_sum = cumulative_sum[r2][c2] - cumulative_sum[r1-1][c2] - cumulative_sum[r2][c1-1] + cumulative_sum[r1-1][c1-1]
            if best < sub_sum:
                best_pos = r1,c1
                best = sub_sum

    print "maximum sum is:", best
    print "top left corner on:", best_pos


matrix = [ [2,4,5,6],
           [2,3,1,4],
           [2,0,2,1] ]
findMaxSubmatrix(matrix,2,2)

output

maximum sum is: 16
top left corner on: (0, 2)

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...