# Program to check whether we can fill square where each row and column will hold distinct elements in Python

PythonServer Side ProgrammingProgramming

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Suppose we have one n × n matrix containing values from 0 to n. Here 0 represents an unfilled square, we have to check whether we can fill empty squares such that in each row and each column every number from 1 to n appears exactly once.

So, if the input is like

 0 0 2 2 0 1 1 2 3

then the output will be True, as we can set the matrix to

 3 1 2 2 3 1 1 2 3

To solve this, we will follow these steps −

• Define a function find_empty_cell() . This will take matrix, n

• for i in range 0 to n, do

• for j in range 0 to n, do

• if matrix[i, j] is same as 0, then

• return(i, j)

• return(-1, -1)

• Define a function is_feasible() . This will take matrix, i, j, x

• if x in ith row of matrix, then

• return False

• if x in jth column in any row of matrix, then

• return False

• return True

• Define a function is_complete() . This will take matrix, n

• for each row in matrix, do

• if row has some duplicate elements, then

• return False

• for col in range 0 to n, do

• if col has some duplicate elements, then

• return False

• return True

• From the main method do the following −

• n := row count of matrix

• (i, j) = find_empty_cell(matrix, n)

• if (i, j) is same as (-1, -1), then

• if is_complete(matrix, n) is true, then

• return True

• otherwise,

• return False

• for x in range 1 to n + 1, do

• if is_feasible(matrix, i, j, x) is true, then

• matrix[i, j] := x

• if solve(matrix) is true, then

• return True

• matrix[i, j] := 0

• return False

Let us see the following implementation to get better understanding −

## Example

Live Demo

class Solution:
def solve(self, matrix):
n = len(matrix)
def find_empty_cell(matrix, n):
for i in range(n):
for j in range(n):
if matrix[i][j] == 0:
return (i, j)
return (-1, -1)
def is_feasible(matrix, i, j, x):
if x in matrix[i]:
return False
if x in [row[j] for row in matrix]:
return False
return True
def is_complete(matrix, n):
for row in matrix:
if set(row) != set(range(1, n + 1)):
return False
for col in range(n):
if set(row[col] for row in matrix) != set(range(1, n + 1)):
return False
return True
(i, j) = find_empty_cell(matrix, n)

if (i, j) == (-1, -1):
if is_complete(matrix, n):
return True
else:
return False
for x in range(1, n + 1):
if is_feasible(matrix, i, j, x):
matrix[i][j] = x
if self.solve(matrix):
return True
matrix[i][j] = 0
return False
ob = Solution()
matrix = [
[0, 0, 2],
[2, 0, 1],
[1, 2, 3]
]
print(ob.solve(matrix))

## Input

matrix = [
[0, 0, 2],
[2, 0, 1],
[1, 2, 3] ]

## Output

True
Updated on 09-Oct-2020 15:39:48