Reconstruct a 2-Row Binary Matrix

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 or 1.
• 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], where colsum is given as an integer array with length n.

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 []