1605. Find Valid Matrix Given Row and Column Sums
Description
You are given two arrays rowSum
and colSum
of non-negative integers where rowSum[i]
is the sum of the elements in the ith
row and colSum[j]
is the sum of the elements of the jth
column of a 2D matrix. In other words, you do not know the elements of the matrix, but you do know the sums of each row and column.
Find any matrix of non-negative integers of size rowSum.length x colSum.length
that satisfies the rowSum
and colSum
requirements.
Return a 2D array representing any matrix that fulfills the requirements. It's guaranteed that at least one matrix that fulfills the requirements exists.
Example 1:
Input: rowSum = [3,8], colSum = [4,7] Output: [[3,0], [1,7]] Explanation: 0th row: 3 + 0 = 3 == rowSum[0] 1st row: 1 + 7 = 8 == rowSum[1] 0th column: 3 + 1 = 4 == colSum[0] 1st column: 0 + 7 = 7 == colSum[1] The row and column sums match, and all matrix elements are non-negative. Another possible matrix is: [[1,2], [3,5]]
Example 2:
Input: rowSum = [5,7,10], colSum = [8,6,8] Output: [[0,5,0], [6,1,0], [2,0,8]]
Constraints:
1 <= rowSum.length, colSum.length <= 500
0 <= rowSum[i], colSum[i] <= 108
sum(rowSum) == sum(colSum)
Solutions
Solution 1: Greedy + Construction
We can first initialize an $m$ by $n$ answer matrix $ans$.
Next, we traverse each position $(i, j)$ in the matrix, set the element at this position to $x = \min(rowSum[i], colSum[j])$, and subtract $x$ from $rowSum[i]$ and $colSum[j]$ respectively. After traversing all positions, we can get a matrix $ans$ that meets the requirements of the problem.
The correctness of the above strategy is explained as follows:
According to the requirements of the problem, we know that the sum of $rowSum$ and $colSum$ is equal, so $rowSum[0]$ must be less than or equal to $\sum_{j = 0}^{n - 1} colSum[j]$. Therefore, after $n$ operations, $rowSum[0]$ can definitely be made $0$, and for any $j \in [0, n - 1]$, $colSum[j] \geq 0$ is guaranteed.
Therefore, we reduce the original problem to a subproblem with $m-1$ rows and $n$ columns, continue the above operations, until all elements in $rowSum$ and $colSum$ are $0$, we can get a matrix $ans$ that meets the requirements of the problem.
The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Where $m$ and $n$ are the lengths of $rowSum$ and $colSum$ respectively.
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