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:

Solution 1.

// 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;
    }
};

Java

  • 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;
        }
    }
    
  • // 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 []
    

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