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 <= 5000 <= rowSum[i], colSum[i] <= 108sum(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 ∑_{j = 0}n - 1 colSum[j]. Therefore, after n operations, rowSum[0] can definitely be made 0, and for any j ∈ [0, n - 1], colSum[j] ≥ 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 × n), and the space complexity is O(m × n). Where m and n are the lengths of rowSum and colSum respectively.
class Solution: def restoreMatrix(self, rowSum: List[int], colSum: List[int]) -> List[List[int]]: m, n = len(rowSum), len(colSum) ans = [[0] * n for _ in range(m)] for i in range(m): for j in range(n): x = min(rowSum[i], colSum[j]) ans[i][j] = x rowSum[i] -= x colSum[j] -= x return ans(code-box)
