def largest in_matrix(matrix: list[list[int -> int: Returns the area of the largest rectangle in . The area of a rectangle is defined by number of 1's that it contains. Again, you MUST make use of the helper here. If you managed to code correctly, this function should be simple to implement. Similarly, do not modify (i.e., mutate) the input matrix. Precondition: will only contain the integers 1 and 0. >>> case1 = [[1, 0, 1, 0, 0], [1, 0, 1, 1, 1], [1, 1, 1, 1, 1], [1, в, в, 1, 0] >>> largest_in_matrix(case1) pass

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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given the fully functioned largest_at_position and longest_chain.

Do not use any imports, do not use dictionaries

- def longest_chatn(lst):
result - []
while(i < len(lst) and Ist[t] = 1):
result.append(Lst[1])
i + 1
return sur (result)
- def largest_at_postion(m, x, y):
x_nax = longest_chain(n[x][y:]);
result - x_max
while(i < len(m) and m[t][y] == 1):
x_nax - i (longest chaln(m[tjty:]), x_nax)
result = na (result, x_max (t-x+1))
return result
if
name
casei
main
[[1,0,1,0,01, [1,0,1,1,1], [1,1,1,1,1], [1,0,0,1,0]]
print(largest_at_position(case1, 0, 0))
print(largest_at_postion(case1, 2, 0))
print(largest_ at_position(case1, 1, 2))
Transcribed Image Text:- def longest_chatn(lst): result - [] while(i < len(lst) and Ist[t] = 1): result.append(Lst[1]) i + 1 return sur (result) - def largest_at_postion(m, x, y): x_nax = longest_chain(n[x][y:]); result - x_max while(i < len(m) and m[t][y] == 1): x_nax - i (longest chaln(m[tjty:]), x_nax) result = na (result, x_max (t-x+1)) return result if name casei main [[1,0,1,0,01, [1,0,1,1,1], [1,1,1,1,1], [1,0,0,1,0]] print(largest_at_position(case1, 0, 0)) print(largest_at_postion(case1, 2, 0)) print(largest_ at_position(case1, 1, 2))
def largest in_matrix(matrix: list[list[int]]) -> int:
Returns the area of the largest rectangle in <matrix>.
The area of a rectangle is defined by number of 1's that it contains.
Again, you MUST make use of the helper <largest_at_position> here. If you
managed to code <largest_at_position> correctly, this function should be
simple to implement.
Similarly, do not modify (i.e., mutate) the input matrix.
Precondition:
<matrix> will only contain the integers 1 and 0.
>>> case1 = [[1, 0, 1, 0, 0],
[1, 0, 1, 1, 1],
[1, 1, 1, 1, 1],
[1, 0, 0, 1, 0]]
>>> largest_in_matrix(case1)
pass
Transcribed Image Text:def largest in_matrix(matrix: list[list[int]]) -> int: Returns the area of the largest rectangle in <matrix>. The area of a rectangle is defined by number of 1's that it contains. Again, you MUST make use of the helper <largest_at_position> here. If you managed to code <largest_at_position> correctly, this function should be simple to implement. Similarly, do not modify (i.e., mutate) the input matrix. Precondition: <matrix> will only contain the integers 1 and 0. >>> case1 = [[1, 0, 1, 0, 0], [1, 0, 1, 1, 1], [1, 1, 1, 1, 1], [1, 0, 0, 1, 0]] >>> largest_in_matrix(case1) pass
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