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1253. Reconstruct a 2-Row Binary Matrix

Description

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

Solutions

Solution 1: Greedy

First, we create an answer array $ans$, where $ans[0]$ and $ans[1]$ represent the first and second rows of the matrix, respectively.

Next, we traverse the array $colsum$ from left to right. For the current element $colsum[j]$, we have the following cases:

  • If $colsum[j] = 2$, then we set both $ans[0][j]$ and $ans[1][j]$ to $1$. In this case, both $upper$ and $lower$ are reduced by $1$.
  • If $colsum[j] = 1$, then we set either $ans[0][j]$ or $ans[1][j]$ to $1$. If $upper \gt lower$, then we prefer to set $ans[0][j]$ to $1$; otherwise, we prefer to set $ans[1][j]$ to $1$. In this case, either $upper$ or $lower$ is reduced by $1$.
  • If $colsum[j] = 0$, then we set both $ans[0][j]$ and $ans[1][j]$ to $0$.
  • If $upper \lt 0$ or $lower \lt 0$, then it is impossible to construct a matrix that meets the requirements, and we return an empty array.

At the end of the traversal, if both $upper$ and $lower$ are $0$, then we return $ans$; otherwise, we return an empty array.

The time complexity is $O(n)$, where $n$ is the length of the array $colsum$. Ignoring the space consumption of the answer array, the space complexity is $O(1)$.

  • class Solution { public List<List<Integer>> reconstructMatrix(int upper, int lower, int[] colsum) { int n = colsum.length; List<Integer> first = new ArrayList<>(); List<Integer> second = new ArrayList<>(); for (int j = 0; j < n; ++j) { int a = 0, b = 0; if (colsum[j] == 2) { a = b = 1; upper--; lower--; } else if (colsum[j] == 1) { if (upper > lower) { upper--; a = 1; } else { lower--; b = 1; } } if (upper < 0 || lower < 0) { break; } first.add(a); second.add(b); } return upper == 0 && lower == 0 ? List.of(first, second) : List.of(); } } 
  • class Solution { public: vector<vector<int>> reconstructMatrix(int upper, int lower, vector<int>& colsum) { int n = colsum.size(); vector<vector<int>> ans(2, vector<int>(n)); for (int j = 0; j < n; ++j) { if (colsum[j] == 2) { ans[0][j] = ans[1][j] = 1; upper--; lower--; } if (colsum[j] == 1) { if (upper > lower) { upper--; ans[0][j] = 1; } else { lower--; ans[1][j] = 1; } } if (upper < 0 || lower < 0) { break; } } return upper || lower ? vector<vector<int>>() : ans; } }; 
  • class Solution: def reconstructMatrix( self, upper: int, lower: int, colsum: List[int] ) -> List[List[int]]: n = len(colsum) ans = [[0] * n for _ in range(2)] for j, v in enumerate(colsum): if v == 2: ans[0][j] = ans[1][j] = 1 upper, lower = upper - 1, lower - 1 if v == 1: if upper > lower: upper -= 1 ans[0][j] = 1 else: lower -= 1 ans[1][j] = 1 if upper < 0 or lower < 0: return [] return ans if lower == upper == 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 { break } } if upper != 0 || lower != 0 { return [][]int{} } return ans } 
  • function reconstructMatrix(upper: number, lower: number, colsum: number[]): number[][] { const n = colsum.length; const ans: number[][] = Array(2) .fill(0) .map(() => Array(n).fill(0)); for (let j = 0; j < n; ++j) { if (colsum[j] === 2) { ans[0][j] = ans[1][j] = 1; upper--; lower--; } else if (colsum[j] === 1) { if (upper > lower) { ans[0][j] = 1; upper--; } else { ans[1][j] = 1; lower--; } } if (upper < 0 || lower < 0) { break; } } return upper || lower ? [] : ans; }

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