"""
Given an matrix of numbers in which all rows and all columns are sorted in decreasing
order, return the number of negative numbers in grid.
Reference: https://leetcode.com/problems/count-negative-numbers-in-a-sorted-matrix
"""
def generate_large_matrix() -> list[list[int]]:
"""
>>> generate_large_matrix() # doctest: +ELLIPSIS
[[1000, ..., -999], [999, ..., -1001], ..., [2, ..., -1998]]
"""
return [list(range(1000 - i, -1000 - i, -1)) for i in range(1000)]
grid = generate_large_matrix()
test_grids = (
[[4, 3, 2, -1], [3, 2, 1, -1], [1, 1, -1, -2], [-1, -1, -2, -3]],
[[3, 2], [1, 0]],
[[7, 7, 6]],
[[7, 7, 6], [-1, -2, -3]],
grid,
)
def validate_grid(grid: list[list[int]]) -> None:
"""
Validate that the rows and columns of the grid is sorted in decreasing order.
>>> for grid in test_grids:
... validate_grid(grid)
"""
assert all(row == sorted(row, reverse=True) for row in grid)
assert all(list(col) == sorted(col, reverse=True) for col in zip(*grid))
def find_negative_index(array: list[int]) -> int:
"""
Find the smallest negative index
>>> find_negative_index([0,0,0,0])
4
>>> find_negative_index([4,3,2,-1])
3
>>> find_negative_index([1,0,-1,-10])
2
>>> find_negative_index([0,0,0,-1])
3
>>> find_negative_index([11,8,7,-3,-5,-9])
3
>>> find_negative_index([-1,-1,-2,-3])
0
>>> find_negative_index([5,1,0])
3
>>> find_negative_index([-5,-5,-5])
0
>>> find_negative_index([0])
1
>>> find_negative_index([])
0
"""
left = 0
right = len(array) - 1
if not array or array[0] < 0:
return 0
while right + 1 > left:
mid = (left + right) // 2
num = array[mid]
if num < 0 and array[mid - 1] >= 0:
return mid
if num >= 0:
left = mid + 1
else:
right = mid - 1
return len(array)
def count_negatives_binary_search(grid: list[list[int]]) -> int:
"""
An O(m logn) solution that uses binary search in order to find the boundary between
positive and negative numbers
>>> [count_negatives_binary_search(grid) for grid in test_grids]
[8, 0, 0, 3, 1498500]
"""
total = 0
bound = len(grid[0])
for i in range(len(grid)):
bound = find_negative_index(grid[i][:bound])
total += bound
return (len(grid) * len(grid[0])) - total
def count_negatives_brute_force(grid: list[list[int]]) -> int:
"""
This solution is O(n^2) because it iterates through every column and row.
>>> [count_negatives_brute_force(grid) for grid in test_grids]
[8, 0, 0, 3, 1498500]
"""
return len([number for row in grid for number in row if number < 0])
def count_negatives_brute_force_with_break(grid: list[list[int]]) -> int:
"""
Similar to the brute force solution above but uses break in order to reduce the
number of iterations.
>>> [count_negatives_brute_force_with_break(grid) for grid in test_grids]
[8, 0, 0, 3, 1498500]
"""
total = 0
for row in grid:
for i, number in enumerate(row):
if number < 0:
total += len(row) - i
break
return total
def benchmark() -> None:
"""Benchmark our functions next to each other"""
from timeit import timeit
print("Running benchmarks")
setup = (
"from __main__ import count_negatives_binary_search, "
"count_negatives_brute_force, count_negatives_brute_force_with_break, grid"
)
for func in (
"count_negatives_binary_search",
"count_negatives_brute_force_with_break",
"count_negatives_brute_force",
):
time = timeit(f"{func}(grid=grid)", setup=setup, number=500)
print(f"{func}() took {time:0.4f} seconds")
if __name__ == "__main__":
import doctest
doctest.testmod()
benchmark()