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Program to find number of elements in matrix follows row column criteria in Python
Suppose we have a binary matrix; we have to find the number of elements in matrix that follows the following rules −
matrix[r, c] = 1
matrix[r, j] = 0 for every j when j is not equal to c and matrix[i, c] = 0 for every i when i is not equal to r.
So, if the input is like
| 0 | 0 | 1 |
| 1 | 0 | 0 |
| 0 | 1 | 0 |
then the output will be 3, because we have cells (0,2), (1,0) and (2,1) those meet the criteria.
To solve this, we will follow these steps −
if matrix is empty, then
return 0
row := a list of sum of all row entries in matrix
col := a list of sum of all column entries in matrix
m := row count of matrix
n := column count of matrix
res := 0
for r in range 0 to m - 1, do
for c in range 0 to n - 1, do
if matrix[r, c] is 1 and row[r] is 1 and col[c] is also 1, then
res := res + 1
return res
Example
Let us see the following implementation to get better understanding
def solve(matrix):
if not matrix:
return 0
row = [sum(r) for r in matrix]
col = [sum(c) for c in zip(*matrix)]
m, n = len(matrix), len(matrix[0])
res = 0
for r in range(m):
for c in range(n):
if matrix[r][c] == 1 and row[r] == 1 and col[c] == 1:
res += 1
return res
matrix = [
[0, 0, 1],
[1, 0, 0],
[0, 1, 0]
]
print(solve(matrix))Input
[[0, 0, 1],[1, 0, 0],[0, 1, 0]]
Output
3
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