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 = 0 == 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]]
Example 3:
Input: rowSum = [14,9], colSum = [6,9,8] Output: [[0,9,5], [6,0,3]]
Example 4:
Input: rowSum = [1,0], colSum = [1] Output: [[1], [0]]
Example 5:
Input: rowSum = [0], colSum = [0] Output: [[0]]
Constraints:
1 <= rowSum.length, colSum.length <= 500
0 <= rowSum[i], colSum[i] <= 108
sum(rows) == sum(columns)
Solution: Greedy
Let a = min(row[i], col[j]), m[i][j] = a, row[i] -= a, col[j] -=a
Time complexity: O(m*n)
Space complexity: O(m*n)
C++
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// Author: Huahua class Solution { public: vector<vector<int>> restoreMatrix(vector<int>& rowSum, vector<int>& colSum) { const int m = rowSum.size(); const int n = colSum.size(); vector<vector<int>> ans(m, vector<int>(n)); for (int i = 0; i < m; ++i) for (int j = 0; j < n; ++j) { ans[i][j] = min(rowSum[i], colSum[j]); rowSum[i] -= ans[i][j]; colSum[j] -= ans[i][j]; } return ans; } }; |
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