code: def setZeroesSecondMoreEfficient(self, matrix):
      """
      :type matrix: List[List[int]]
      :rtype: void Do not return anything, modify matrix in-place instead.
      """
      is_col = False
      R = len(matrix)
      C = len(matrix[0])
      for i in range(R):
          # Since first cell for both first row and first column is the same i.e. matrix[0][0]
          # We can use an additional variable for either the first row/column.
          # For this solution we are using an additional variable for the first column
          # and using matrix[0][0] for the first row.
          if matrix[i][0] == 0:
              is_col = True
          for j in range(1, C):
              # If an element is zero, we set the first element of the corresponding row and column to 0
              if matrix[i][j] == 0:
                  matrix[0][j] = 0
                  matrix[i][0] = 0
  
      # Iterate over the array once again and using the first row and first column, update the elements.
      for i in range(1, R):
          for j in range(1, C):
              if not matrix[i][0] or not matrix[0][j]:
                  matrix[i][j] = 0
  
      # See if the first row needs to be set to zero as well
      if matrix[0][0] == 0:
          for j in range(C):
              matrix[0][j] = 0
  
      # See if the first column needs to be set to zero as well
      if is_col:
          for i in range(R):
              matrix[i][0] = 0
  
label_desc: Use image node_id 4 for calling the Solution obj's underlying member method code with example usage: obj.setZeroesSecondMoreEfficient(matrix) without return types