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 −
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]]