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.

// 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

• Java
• C++
• Python

• 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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