This post is part of the Algorithms Problem Solving series.
Problem description
This is the Odd in Matrix problem. The description looks like this:
Given n and m which are the dimensions of a matrix initialized by zeros and given an array indices where indices[i] = [ri, ci]. For each pair of [ri, ci] you have to increment all cells in row ri and column ci by 1.
Return the number of cells with odd values in the matrix after applying the increment to all indices.
Examples
Input: n = 2, m = 3, indices = [[0,1],[1,1]]
Output: 6
Input: n = 2, m = 2, indices = [[1,1],[0,0]]
Output: 0
Solution
- Initialize the matrix with all elements as zero
- For each pair of indices, increment for the row, and increment for the column
- Traverse the matrix counting all the odd numbers
- Return the counter
def init_matrix(rows, columns):
return [[0 for _ in range(columns)] for _ in range(rows)]
def odd_cells(n, m, indices):
matrix = init_matrix(n, m)
for [ri, ci] in indices:
for column in range(m):
matrix[ri][column] += 1
for row in range(n):
matrix[row][ci] += 1
odds = 0
for row in range(n):
for column in range(m):
if matrix[row][column] % 2 != 0:
odds += 1
return odds
Top comments (3)
Don't you think there can be a better approach to this problem?
"Better" is a subjective concept. Some could call better this approach:
You can say it's better because it's shorter and faster. But it's much harder to read. If I am asked to maintain it, I would prefer much more TK's approach.
Or something like this:
A lot more code, but way easier to read and understand, in my opinion.
Therefore, I ask you the same: don't you think there can be a better approach to this problem? How would you solve it?
I believe that there is a solution use only 4 numbers: number of columns (rows) which have even (odd) cells on each. Need time to test it...