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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.

• 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++)
int columns = colsum.length;
for (int i = 0; i < columns; i++) {
int curSum = colsum[i];
if (curSum == 2) {
} else if (curSum == 1) {
if (upperRemain > 0) {
upperRemain--;
} else {
lowerRemain--;
}
} else {
}
}
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) {
upper--;
lower--;
} else if (colsum[j] == 1) {
if (upper > lower) {
upper--;
} else {
lower--;
}
} else {
}
if (upper < 0 || lower < 0) {
return ans;
}
}
if (upper != 0 || lower != 0) {
return ans;
}
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
}