Given the following details of a matrix with n
columns and 2
rows :
- The matrix is a binary matrix, which means each element in the matrix can be
0
or1
. - The sum of elements of the 0-th(upper) row is given as
upper
. - The sum of elements of the 1-st(lower) row is given as
lower
. - The sum of elements in the i-th column(0-indexed) is
colsum[i]
, wherecolsum
is given as an integer array with lengthn
.
Your task is to reconstruct the matrix with upper
, lower
and colsum
.
Return it as a 2-D integer array.
If there are more than one valid solution, any of them will be accepted.
If no valid solution exists, return an empty 2-D array.
Example 1:
Input: upper = 2, lower = 1, colsum = [1,1,1]
Output: [[1,1,0],[0,0,1]]
Explanation: [[1,0,1],[0,1,0]], and [[0,1,1],[1,0,0]] are also correct answers.
Example 2:
Input: upper = 2, lower = 3, colsum = [2,2,1,1]
Output: []
Example 3:
Input: upper = 5, lower = 5, colsum = [2,1,2,0,1,0,1,2,0,1]
Output: [[1,1,1,0,1,0,0,1,0,0],[1,0,1,0,0,0,1,1,0,1]]
Constraints:
-
1 <= colsum.length <= 10^5
-
0 <= upper, lower <= colsum.length
-
0 <= colsum[i] <= 2
SOLUTION:
class Solution:
def reconstructMatrix(self, upper: int, lower: int, colsum: List[int]) -> List[List[int]]:
n = len(colsum)
op = [[0 for i in range(n)] for j in range(2)]
ctr = []
for i in range(n):
if colsum[i] == 2:
op[0][i] = 1
upper -= 1
op[1][i] = 1
lower -= 1
elif colsum[i] == 1:
ctr.append(i)
if upper >= 0 and lower >= 0 and upper + lower == len(ctr):
if upper > 0:
for i in ctr[:upper]:
op[0][i] = 1
if lower > 0:
for i in ctr[-lower:]:
op[1][i] = 1
return op
return []
Top comments (0)