81 lines
2.5 KiB
C++
81 lines
2.5 KiB
C++
// Source : https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/
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// Author : Sudesh Chaudhary
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// Date : 2020-10-03
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/*******************************************************************************
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* You are given two arrays rowSum and colSum of non-negative integers where rowSum[i] is the sum
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* of the elements in the ith row and colSum[j] is the sum of the elements of the jth column of a 2D matrix.
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* In other words, you do not know the elements of the matrix, but you do know the sums of each row and column.
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*
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* Find any matrix of non-negative integers of size rowSum.length x colSum.length that satisfies the rowSum and
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* colSum requirements.
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*
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* Return a 2D array representing any matrix that fulfills the requirements. It's guaranteed that at
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* least one matrix that fulfills the requirements exists.
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* Example 1:
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*
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* Input: rowSum = [3,8], colSum = [4,7]
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* Output: [[3,0],
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* [1,7]]
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* Explanation:
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* 0th row: 3 + 0 = 0 == rowSum[0]
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* 1st row: 1 + 7 = 8 == rowSum[1]
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* 0th column: 3 + 1 = 4 == colSum[0]
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* 1st column: 0 + 7 = 7 == colSum[1]
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* The row and column sums match, and all matrix elements are non-negative.
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* Another possible matrix is: [[1,2],
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* [3,5]]
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* Example 2:
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*
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* Input: rowSum = [5,7,10], colSum = [8,6,8]
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* Output: [[0,5,0],
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* [6,1,0],
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* [2,0,8]]
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* Example 3:
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*
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* Input: rowSum = [14,9], colSum = [6,9,8]
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* Output: [[0,9,5],
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* [6,0,3]]
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* Example 4:
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*
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* Input: rowSum = [1,0], colSum = [1]
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* Output: [[1],
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* [0]]
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* Example 5:
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*
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* Input: rowSum = [0], colSum = [0]
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* Output: [[0]]
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*********************************************************************************/
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class Solution {
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public:
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vector<vector<int>> restoreMatrix(vector<int>& row, vector<int>& col) {
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int n = row.size();
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int m = col.size();
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if(n==0 ||m==0)
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return {};
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vector<vector<int>> res(n,vector<int>(m,0));
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priority_queue<pair<int,int>> p,q;
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for(int i=0;i<n;i++){
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p.push({row[i],i});
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}
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for(int j=0;j<m;j++){
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q.push({col[j],j});
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}
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while(!q.empty() && !p.empty()){
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auto a = p.top();
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auto b = q.top();
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p.pop();
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q.pop();
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int t = min(a.first,b.first);
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res[a.second][b.second]=t;
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a.first-=t;
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b.first-=t;
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if(a.first>0)
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p.push(a);
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if(b.first>0)
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q.push(b);
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}
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return res;
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}
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};
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