Formatted question description: https://leetcode.ca/all/1253.html

1253. Reconstruct a 2-Row Binary Matrix (Medium)

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:

Find Valid Matrix Given Row and Column Sums (Medium)

Solution 1.

class Solution {
    public List<List<Integer>> reconstructMatrix(int upper, int lower, int[] colsum) {
        List<List<Integer>> matrix = new ArrayList<List<Integer>>();
        int sum = 0;
        int upperRemain = upper, lowerRemain = lower;
        for (int num : colsum) {
            sum += num;
            if (num == 2) {
                upperRemain--;
                lowerRemain--;
            }
        }
        if (sum != upper + lower || upperRemain < 0 || lowerRemain < 0)
            return matrix;
        for (int i = 0; i < 2; i++)
            matrix.add(new ArrayList<Integer>());
        int columns = colsum.length;
        for (int i = 0; i < columns; i++) {
            int curSum = colsum[i];
            if (curSum == 2) {
                matrix.get(0).add(1);
                matrix.get(1).add(1);
            } else if (curSum == 1) {
                if (upperRemain > 0) {
                    matrix.get(0).add(1);
                    matrix.get(1).add(0);
                    upperRemain--;
                } else {
                    matrix.get(0).add(0);
                    matrix.get(1).add(1);
                    lowerRemain--;
                }
            } else {
                matrix.get(0).add(0);
                matrix.get(1).add(0);
            }
        }
        return matrix;
    }
}

############

class Solution {
    public List<List<Integer>> reconstructMatrix(int upper, int lower, int[] colsum) {
        int n = colsum.length;
        List<List<Integer>> ans = new ArrayList<>();
        List<Integer> first = new ArrayList<>();
        List<Integer> second = new ArrayList<>();
        for (int j = 0; j < n; ++j) {
            if (colsum[j] == 2) {
                first.add(1);
                second.add(1);
                upper--;
                lower--;
            } else if (colsum[j] == 1) {
                if (upper > lower) {
                    upper--;
                    first.add(1);
                    second.add(0);
                } else {
                    lower--;
                    first.add(0);
                    second.add(1);
                }
            } else {
                first.add(0);
                second.add(0);
            }
            if (upper < 0 || lower < 0) {
                return ans;
            }
        }
        if (upper != 0 || lower != 0) {
            return ans;
        }
        ans.add(first);
        ans.add(second);
        return ans;
    }
}

// OJ: https://leetcode.com/problems/reconstruct-a-2-row-binary-matrix/
// 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;
    }
};

# 1253. Reconstruct a 2-Row Binary Matrix
# https://leetcode.com/problems/reconstruct-a-2-row-binary-matrix/

class Solution:
    def reconstructMatrix(self, upper: int, lower: int, colsum: List[int]) -> List[List[int]]:
        cols = len(colsum)
        
        res = [[0]*cols for _ in range(2)]
        
        for j in range(cols):
            if colsum[j] == 2:
                res[0][j] = 1
                res[1][j] = 1
                colsum[j] -= 2
                upper -= 1
                lower -= 1
            
            if upper < 0 or lower < 0: return []
        
        
        for i in range(2):
            for j in range(cols):
                rowsum = upper if i == 0 else lower

                if rowsum > 0 and colsum[j] > 0:
                    res[i][j] = 1
                
                    if i == 0: upper -= 1
                    else: lower -= 1

                    colsum[j] -= 1
                

        return res if upper + lower + sum(colsum) == 0 else []

func reconstructMatrix(upper int, lower int, colsum []int) [][]int {
	n := len(colsum)
	ans := make([][]int, 2)
	for i := range ans {
		ans[i] = make([]int, n)
	}
	for j, v := range colsum {
		if v == 2 {
			ans[0][j], ans[1][j] = 1, 1
			upper--
			lower--
		}
		if v == 1 {
			if upper > lower {
				upper--
				ans[0][j] = 1
			} else {
				lower--
				ans[1][j] = 1
			}
		}
		if upper < 0 || lower < 0 {
			return [][]int{}
		}
	}
	if upper != 0 || lower != 0 {
		return [][]int{}
	}
	return ans
}

1253 - Reconstruct a 2-Row Binary Matrix

1253. Reconstruct a 2-Row Binary Matrix (Medium)

Solution 1.

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1253. Reconstruct a 2-Row Binary Matrix (Medium)

Solution 1.

All Problems

All Solutions