FindValidMatrixGivenRowAndColumnSums.cpp

// Source : https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/
// Author : Sudesh Chaudhary
// Date   : 2020-10-03
/*******************************************************************************
* 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]]
*********************************************************************************/
class Solution {
public:
    vector<vector<int>> restoreMatrix(vector<int>& row, vector<int>& col) {
        int n = row.size();
        int m = col.size();
        if(n==0 ||m==0)
            return {};
        vector<vector<int>> res(n,vector<int>(m,0));
        priority_queue<pair<int,int>> p,q;
        for(int i=0;i<n;i++){
            p.push({row[i],i});
        }
        for(int j=0;j<m;j++){
            q.push({col[j],j});
        }

        while(!q.empty() && !p.empty()){
            auto a = p.top();
            auto b = q.top();
            p.pop();
            q.pop();
            int t = min(a.first,b.first);
            res[a.second][b.second]=t;
            a.first-=t;
            b.first-=t;
            if(a.first>0)
                p.push(a);
            if(b.first>0)
                q.push(b);
        }
        return res;
    }
};