Solution with explanation
Approach\n Describe your approach to solving the problem. \n\n- Check the basic conditions: The sum of all elements in the colsum array must be equal to upper +
# Approach\n<!-- Describe your approach to solving the problem. -->\n\n- Check the basic conditions: The sum of all elements in the colsum array must be equal to upper + lower. If this condition is not met, there is no valid solution.\n- Initialize a 2-D matrix with dimensions 2xn.\n- Iterate through each column, and decide the values of the matrix based on the value of colsum[i]:\n- If colsum[i] == 2, both elements in the upper and lower row of this column must be 1.\n- If colsum[i] == 1, one element should be 1 and the other should be 0. Prioritize setting the upper row to 1 until upper reaches 0, then set the lower row.\n- If colsum[i] == 0, both elements should be 0.\nAfter processing all columns, verify if the sums of the upper and lower rows are as expected. If not, return an empty array.\n\n\n# Code\n```\nfunc reconstructMatrix(upper int, lower int, colsum []int) [][]int {\n\tn := len(colsum)\n\tresult := make([][]int, 2)\n\tresult[0] = make([]int, n)\n\tresult[1] = make([]int, n)\n\n\tsumOfCols := 0\n\tfor _, val := range colsum {\n\t\tsumOfCols += val\n\t}\n\n\tif sumOfCols != upper+lower {\n\t\treturn [][]int{}\n\t}\n\n\tfor i, col := range colsum {\n\t\tswitch col {\n\t\tcase 2:\n\t\t\tif upper > 0 && lower > 0 {\n\t\t\t\tresult[0][i] = 1\n\t\t\t\tresult[1][i] = 1\n\t\t\t\tupper--\n\t\t\t\tlower--\n\t\t\t} else {\n\t\t\t\treturn [][]int{}\n\t\t\t}\n\t\tcase 1:\n\t\t\tif upper > lower {\n\t\t\t\tresult[0][i] = 1\n\t\t\t\tupper--\n\t\t\t} else if lower > 0 {\n\t\t\t\tresult[1][i] = 1\n\t\t\t\tlower--\n\t\t\t} else {\n\t\t\t\treturn [][]int{}\n\t\t\t}\n\t\tcase 0:\n\t\t\tresult[0][i] = 0\n\t\t\tresult[1][i] = 0\n\t\tdefault:\n\t\t\treturn [][]int{}\n\t\t}\n\t}\n\n\tif upper != 0 || lower != 0 {\n\t\treturn [][]int{}\n\t}\n\n\treturn result\n}\n```
['Go']