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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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