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Given the following details of a matrix with n columns and 2 rows :

  • The matrix is a binary matrix, which means each element in the matrix can be 0 or 1.
  • The sum of elements of the 0-th(upper) row is given as upper.
  • The sum of elements of the 1-st(lower) row is given as lower.
  • The sum of elements in the i-th column(0-indexed) is colsum[i], where colsum is given as an integer array with length n.

Your task is to reconstruct the matrix with upper, lower and colsum.

Return it as a 2-D integer array.

If there are more than one valid solution, any of them will be accepted.

If no valid solution exists, return an empty 2-D array.

 

Example 1:

Input: upper = 2, lower = 1, colsum = [1,1,1]
Output: [[1,1,0],[0,0,1]]
Explanation: [[1,0,1],[0,1,0]], and [[0,1,1],[1,0,0]] are also correct answers.

Example 2:

Input: upper = 2, lower = 3, colsum = [2,2,1,1]
Output: []

Example 3:

Input: upper = 5, lower = 5, colsum = [2,1,2,0,1,0,1,2,0,1]
Output: [[1,1,1,0,1,0,0,1,0,0],[1,0,1,0,0,0,1,1,0,1]]

 

Constraints:

  • 1 <= colsum.length <= 10^5
  • 0 <= upper, lower <= colsum.length
  • 0 <= colsum[i] <= 2

Related Topics:
Math, Greedy

Similar Questions:

Solution 1.

// OJ: https://leetcode.com/problems/reconstruct-a-2-row-binary-matrix/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    vector<vector<int>> reconstructMatrix(int upper, int lower, vector<int>& colsum) {
        if (accumulate(begin(colsum), end(colsum), 0) != upper + lower) return {};
        vector<vector<int>> ans(2, vector<int>(colsum.size()));
        for (int i = 0; i < colsum.size(); ++i) {
            if (colsum[i] == 2) --upper, --lower, ans[0][i] = ans[1][i] = 1;
            if (upper < 0 || lower < 0) return {};
        }
        for (int i = 0; i < colsum.size(); ++i) {
            if (colsum[i] != 1) continue;
            if (upper) --upper, ans[0][i] = 1;
            else --lower, ans[1][i] = 1;
            if (upper < 0 || lower < 0) return {};
        }
        return ans;
    }
};