Reconstruct a 2-Row Binary Matrix

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ArraysGreedy Algorithms

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

Try coding it in LeetCode

Reconstruct a 2-Row Binary Matrix

Medium

Grab

5 views

Topics:

ArraysGreedy Algorithms

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for `upper`, `lower`, and the elements within the `colsum` array?
Could `upper`, `lower`, or the `colsum` array be empty or null?
If multiple valid binary matrices exist, is any one acceptable, or is there a specific criterion for choosing between them?
What should be returned if no valid binary matrix can be constructed?
Is the length of `colsum` guaranteed to be the number of columns in the desired matrix, or might it be inconsistent with `upper` and `lower`?

Brute Force Solution

Approach

We are given clues about a grid with two rows. The brute force approach is to try every possible arrangement of 0s and 1s in the grid, making sure that the clues we are given hold true for each arrangement. This is done by systematically testing all possibilities.

Here's how the algorithm would work step-by-step:

Start by considering the first cell in the top row and trying both 0 and 1.
For each choice in the top row, consider the first cell in the bottom row, again trying both 0 and 1.
Continue filling the grid cell by cell, always trying both 0 and 1 for each position.
After each complete grid is filled, check if it satisfies all the conditions described in the problem (row sums, column sums).
If the grid satisfies all conditions, save it as a possible solution.
Repeat the process until all possible grid configurations have been tried.
If any valid solutions were found, return one. If not, the problem has no solution.

Code Implementation

def reconstruct_matrix_brute_force(upper_value, lower_value, column_sums):
    number_of_columns = len(column_sums)
    solution = []

    def solve(current_matrix, column_index):
        nonlocal solution

        # If a solution is already found, stop
        if solution:
            return

        # If we've filled all columns, check if the solution is valid
        if column_index == number_of_columns:
            upper_sum = sum(current_matrix[0])
            lower_sum = sum(current_matrix[1])

            if upper_sum == upper_value and lower_sum == lower_value:
                solution.append([row[:] for row in current_matrix])
            return

        # Try placing a 0 in the top row
        current_matrix[0][column_index] = 0
        solve(current_matrix, column_index + 1)

        # Try placing a 1 in the top row
        current_matrix[0][column_index] = 1

        # Prune possibilities if the column sum is greater than 1.
        if column_sums[column_index] > 0:
            current_matrix[1][column_index] = column_sums[column_index] - current_matrix[0][column_index]
            solve(current_matrix, column_index + 1)

    # Initialize an empty 2xN matrix
    initial_matrix = [[0] * number_of_columns for _ in range(2)]

    solve(initial_matrix, 0)

    # Return a solution if found, otherwise return an empty list
    if solution:
        return solution[0]
    else:
        return []

Big(O) Analysis

Time Complexity

O(2^(2n)) – The brute force approach explores every possible combination of 0s and 1s in the 2 x n grid. Since each cell can have two possible values (0 or 1), and there are 2 * n cells in total, the algorithm essentially generates 2^(2n) possible grids. For each of these 2^(2n) grids, we need to check if it satisfies the given row and column sum conditions. The check itself takes O(n) time, iterating through each column to verify the sums. Therefore, the overall time complexity is O(n * 2^(2n)), which simplifies to O(2^(2n)) as the exponential term dominates. The 'n' factor from checking the sum is less significant than the exponential growth from exploring all possible combinations.

Space Complexity

O(N) – The brute force approach, as described, constructs and checks complete 2-row grids of size 2xN, where N is the number of columns (and the length of the input arrays). Each such grid requires storing 2*N integers. While only one solution is stored at a time if found, the temporary grid used for checking each potential solution accounts for auxiliary space usage. Therefore, the space complexity is O(N) due to the temporary grid.

Optimal Solution

Approach

The goal is to build two rows that add up to a given total while respecting constraints on the sum of each row. We build the matrix in a greedy fashion, prioritizing placing values where they are forced, simplifying the remaining choices.

Here's how the algorithm would work step-by-step:

First, create the two rows of the matrix, filling them initially with zeros.
Iterate through the given totals for each column.
If a column total is two, both rows in that column must be one.
If a column total is zero, both rows in that column must be zero.
After filling the forced values, try to greedily fill in values of one in the first row until that row meets its sum constraint.
Next, fill any remaining values of one in the second row until it meets its sum constraint.
If at any point you run out of ones to place, or the row sums become negative, it's impossible to create the matrix: return an empty result to indicate that.
If the process succeeds in filling all required ones without violating constraints, the matrix is successfully constructed. Return this result.

Code Implementation

def reconstruct_matrix(upper_sum, lower_sum, column_sums):
    number_of_columns = len(column_sums)
    upper_row = [0] * number_of_columns
    lower_row = [0] * number_of_columns

    for column_index in range(number_of_columns):
        if column_sums[column_index] == 2:
            # Both rows must be 1 in this column
            upper_row[column_index] = 1
            lower_row[column_index] = 1
            upper_sum -= 1
            lower_sum -= 1
        elif column_sums[column_index] == 0:
            # Both rows must be 0 in this column
            upper_row[column_index] = 0
            lower_row[column_index] = 0

    for column_index in range(number_of_columns):
        if column_sums[column_index] == 1:
            # Greedily place 1 in the upper row if possible
            if upper_sum > 0:
                upper_row[column_index] = 1
                lower_row[column_index] = 0
                upper_sum -= 1
            elif lower_sum > 0:
                lower_row[column_index] = 1
                upper_row[column_index] = 0
                lower_sum -= 1
            else:
                return []

    # Check if the row sums match the constraints
    if upper_sum == 0 and lower_sum == 0:
        return [upper_row, lower_row]
    else:
        return []

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input column array of size n exactly once to determine the values for each column of the binary matrix. Within this single loop, the operations performed for each column (setting matrix values and decrementing row sums) take constant time. Therefore, the time complexity is directly proportional to the size of the input array, resulting in a linear time complexity of O(n).

Space Complexity

O(N) – The algorithm creates two lists, representing the rows of the binary matrix. Each list has a length equal to the number of columns in the input, denoted as N. Thus, the auxiliary space required to store these lists grows linearly with the input size N. No other significant data structures are used, resulting in a space complexity of O(N).

Edge Cases

Case	How to Handle
Null or empty colsum array	Return an empty list immediately, as no matrix can be constructed.
upper or lower is negative	Return an empty list because ones cannot be negative.
Sum of colsum is not equal to upper + lower	Return an empty list because the total number of ones doesn't match the constraints.
colsum contains values other than 0, 1, or 2	Return an empty list as the column sums can only be 0, 1, or 2.
Impossible case: too many 2's for the available upper/lower values	Return an empty list if the count of 2's in colsum exceeds min(upper, lower).
Impossible case: too many 1's and insufficient remaining upper/lower values after accommodating 2's.	Return an empty list if allocating 1's after handling 2's and 0's results in upper or lower becoming negative.
upper or lower greater than the length of colsum	Return an empty list as it's impossible to have more ones than columns.
Very large colsum array (scalability check)	The solution should iterate through colsum only once, making it O(n) and scalable.