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Formatted question description: https://leetcode.ca/all/1605.html
1605. Find Valid Matrix Given Row and Column Sums (Medium)
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 <= 5000 <= rowSum[i], colSum[i] <= 108sum(rows) == sum(columns)
Related Topics:
Greedy
Similar Questions:
Solution 1.
// OJ: https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/
// Time: O(MN)
// Space: O(1)
class Solution {
public:
vector<vector<int>> restoreMatrix(vector<int>& R, vector<int>& C) {
int M = R.size(), N = C.size(), d;
vector<vector<int>> ans(M, vector<int>(N));
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
d = ans[i][j] = min(R[i], C[j]);
R[i] -= d;
C[j] -= d;
}
}
return ans;
}
};
Solution 2.
We just need to go from the top-left to the bottom-right. Once R[i] or C[j] becomes 0, the rest of the elements in row i or column j must be 0s, so we can increment i or j.
// OJ: https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/
// Time: O(MN) for initialization, O(M + N) for processing
// Space: O(MN)
// Ref: https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/discuss/876845/JavaC%2B%2BPython-Easy-and-Concise-with-Prove
class Solution {
public:
vector<vector<int>> restoreMatrix(vector<int>& R, vector<int>& C) {
int M = R.size(), N = C.size(), i = 0, j = 0, d;
vector<vector<int>> ans(M, vector<int>(N));
while (i < M && j < N) {
d = ans[i][j] = min(R[i], C[j]);
if ((R[i] -= d) == 0) ++i;
if ((C[j] -= d) == 0) ++j;
}
return ans;
}
};
-
class Solution { public int[][] restoreMatrix(int[] rowSum, int[] colSum) { int rows = rowSum.length, columns = colSum.length; int[][] matrix = new int[rows][columns]; for (int i = 0; i < rows; i++) { for (int j = 0; j < columns; j++) { int num = Math.min(rowSum[i], colSum[j]); matrix[i][j] = num; rowSum[i] -= num; colSum[j] -= num; } } return matrix; } } ############ class Solution { public int[][] restoreMatrix(int[] rowSum, int[] colSum) { int m = rowSum.length; int n = colSum.length; int[][] ans = new int[m][n]; for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { int x = Math.min(rowSum[i], colSum[j]); ans[i][j] = x; rowSum[i] -= x; colSum[j] -= x; } } return ans; } } -
// OJ: https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/ // Time: O(MN) // Space: O(1) extra space class Solution { public: vector<vector<int>> restoreMatrix(vector<int>& R, vector<int>& C) { int M = R.size(), N = C.size(), d; vector<vector<int>> ans(M, vector<int>(N)); for (int i = 0; i < M; ++i) { for (int j = 0; j < N; ++j) { d = ans[i][j] = min(R[i], C[j]); R[i] -= d; C[j] -= d; } } return ans; } }; -
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 ############ # 1605. Find Valid Matrix Given Row and Column Sums # https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/ class Solution: def restoreMatrix(self, rs: List[int], cs: List[int]) -> List[List[int]]: r = len(rs) c = len(cs) lst = [[0]*c for _ in range(r)] for i in range(r): for j in range(c): cur = min(rs[i], cs[j]) lst[i][j] = cur rs[i] -= cur cs[j] -= cur return lst -
func restoreMatrix(rowSum []int, colSum []int) [][]int { m, n := len(rowSum), len(colSum) ans := make([][]int, m) for i := range ans { ans[i] = make([]int, n) } for i := range rowSum { for j := range colSum { x := min(rowSum[i], colSum[j]) ans[i][j] = x rowSum[i] -= x colSum[j] -= x } } return ans } func min(a, b int) int { if a < b { return a } return b } -
function restoreMatrix(rowSum: number[], colSum: number[]): number[][] { const m = rowSum.length; const n = colSum.length; const ans = Array.from(new Array(m), () => new Array(n).fill(0)); for (let i = 0; i < m; i++) { for (let j = 0; j < n; j++) { const x = Math.min(rowSum[i], colSum[j]); ans[i][j] = x; rowSum[i] -= x; colSum[j] -= x; } } return ans; } -
/** * @param {number[]} rowSum * @param {number[]} colSum * @return {number[][]} */ var restoreMatrix = function (rowSum, colSum) { const m = rowSum.length; const n = colSum.length; const ans = Array.from(new Array(m), () => new Array(n).fill(0)); for (let i = 0; i < m; i++) { for (let j = 0; j < n; j++) { const x = Math.min(rowSum[i], colSum[j]); ans[i][j] = x; rowSum[i] -= x; colSum[j] -= x; } } return ans; };