Program to Find Out the Minimal Submatrices in Python

PythonServer Side ProgrammingProgramming

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

356
865
4312

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))

Input

[[3, 5, 6],[8, 6, 5],[4, 3, 12]], 2

Output

[[3, 5], [3, 3]]
raja
Published on 23-Dec-2020 11:01:11
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