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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
or1
. - 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]
, wherecolsum
is given as an integer array with lengthn
.
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
Similar Questions:
Solution 1.
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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; } } ############ 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) { first.add(1); second.add(1); upper--; lower--; } else if (colsum[j] == 1) { if (upper > lower) { upper--; first.add(1); second.add(0); } else { lower--; first.add(0); second.add(1); } } else { first.add(0); second.add(0); } if (upper < 0 || lower < 0) { return ans; } } if (upper != 0 || lower != 0) { return ans; } ans.add(first); ans.add(second); return ans; } }
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// 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; } };
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# 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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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 }