Program to Find Out the Minimal Submatrices in Python

Python Server Side Programming Programming

Suppose we have a 2D matrix and another value k. Our goal is to return a matrix that contains the lowest values of all k x k sub-matrices.

So, if the input is like

3	5	6
8	6	5
4	3	12

and k =2,

then the output will be [[3, 5], [3, 3]] .

From the input, we can see that the top left submatrix has the lowest value of 3

3 5
8 6

The top right submatrix has the lowest value of 5

5 6
6 5

The bottom left submatrix has the lowest value of 3

8 6
4 3

The bottom right submatrix has the lowest value of 3

6 5
3 12

To solve this, we will follow these steps −

for each r, row in index r and item row in matrix, do
- q := a new double ended queue
- nrow := a new list
- for i in range 0 to size of row, do
  - if q and q[0] is same as i - k, then
    - pop leftmost item of q
  - while q and row[q[-1]] > row[i] is non-zero, do
    - pop rightmost item of q
  - insert i at the right end of q
  - insert row[q[0]] at the end of nrow
- matrix[r] := nrow
for j in range 0 to size of matrix[0], do
- q := a new double ended queue
- ncol := a new list
- for i in range 0 to size of matrix, do
  - if q and q[0] is same as i - k, then
    - pop leftmost item of q
  - while q and matrix[q[-1]][j] > matrix[i][j] is non-zero, do
    - pop rightmost item of q
  - insert i at the right end of q
  - insert matrix[q[0],j] at the right end of ncol
  - for i in range 0 to size of matrix, do
    - matrix[i, j] := ncol[i]
ret := a new list of the size of matrix - k + 1 initialized with 0
for i in range 0 to size of ret, do
- for j in range 0 to size of ret[0], do
  - ret[i, j] := matrix[i + k - 1, j + k - 1]
return ret

Example

Let us see the following implementation to get a better understanding −

Live Demo

import collections
class Solution:
   def solve(self, matrix, k):
      for r, row in enumerate(matrix):
         q = collections.deque()
         nrow = []
         for i in range(len(row)):
            if q and q[0] == i - k:
               q.popleft()
            while q and row[q[-1]] > row[i]:
               q.pop()
            q.append(i)
            nrow.append(row[q[0]])
         matrix[r] = nrow
      for j in range(len(matrix[0])):
         q = collections.deque()
         ncol = []
         for i in range(len(matrix)):
            if q and q[0] == i - k:
               q.popleft()
            while q and matrix[q[-1]][j] > matrix[i][j]:
               q.pop()
            q.append(i)
            ncol.append(matrix[q[0]][j])
         for i in range(len(matrix)):
            matrix[i][j] = ncol[i]
      ret = [[0] * (len(matrix[0]) - k + 1) for _ in range(len(matrix) - k + 1)]
      for i in range(len(ret)):
         for j in range(len(ret[0])):
            ret[i][j] = matrix[i + k - 1][j + k - 1]
         return ret
ob = Solution()
print(ob.solve(matrix = [
   [3, 5, 6],
   [8, 6, 5],
   [4, 3, 12]
], k = 2))