Question

1605. Find Valid Matrix Given Row and Column Sums

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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

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

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;
  }
}

class Solution {
public:
  vector<vector<int>> restoreMatrix(vector<int>& rowSum, vector<int>& colSum) {
    int m = rowSum.size(), 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) {
        int x = min(rowSum[i], colSum[j]);
        ans[i][j] = x;
        rowSum[i] -= x;
        colSum[j] -= x;
      }
    }
    return ans;
  }
};

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
}

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;
};